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  • How do I reference a pointer from a different class?

    - by Justagruvn
    First off, I despise singletons with a passion. Though I should probably be trying to use one, I just don't want to. I want to create a data class (that is instantiated only once by a view controller on loading), and then using a different class, message the crap out of that data instance until it is brimming with so much data, it smiles. So, how do I do that? I made a pointer to the instance of the data class when I instantiated it. I'm now over in a separate view controller, action occurs, and I want to update the initial data object. I think I need to reference that object by way of pointer, but I have no idea how to do that. Yes, I've set properties and getters and setters, which seem to work, but only in the initial view controller class. Peace Love applesauce.

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  • Rendering HTML in rails without actually displaying it

    - by Kevin Whitaker
    Hello all, My current project requires me to assemble a .zip file containing HTML and text-only templates for a user to download, for importing into an email marketing program. I've inherited this project, and currently the code uses a "fake" model (that is a model that does not directly correlate to a database table), in which it stores the entire template in a string, using dynamic variables to populate certain areas. The "fake" model then has a method for creating a zip file. It seems to me that there has to be a better way to do this. I was wondering if there was a way to move the template into a .erb/haml file, and then write a method that would populate the file in preparation for being zipped up? Basically, is there a way to render an HTML and text file, without actually having to display them? Thanks for any help.

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  • Problem in building a tab bar inside navigation controller

    - by Heba
    Hello Everyone! I am really frustrated and I rally hope that you will help me to solve this problem! I'm trying to build a tab bar inside a navigation controller. I used this template provided by WiredBob. My problem is that I want to add more bar items to the tab bar, but I keep getting crash! From the log: 2010-05-24 00:15:43.469 NavTab[9315:207] * Terminating app due to uncaught exception 'NSInternalInconsistencyException', reason: '-[UITableViewController loadView] loaded the "AnnView" nib but didn't get a UITableView.' Also, I tried to fix the size of the a view in IB to fit in with the tab bar, but I couldn't! It was unchangeable. Thanks in advance :-)

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  • rootview controller in Interface builder?

    - by senthilmuthu
    hi, i have done in didapplicationfinishing function tabBarController = [[UITabBarController alloc] init] ; tabBarController.navigationItem.title = @"News"; SimpleTableViewController *rtbfViewController = [[SimpleTableViewController alloc] init]; //initWithStyle:UITableViewStyleGrouped]; rtbfViewController.tabBarItem.title = @"News1"; rtbfViewController.tabBarItem.image = [UIImage imageNamed:@"home.png"];; UINavigationController *table2NavController = [[[UINavigationController alloc] initWithRootViewController:rtbfViewController] autorelease]; [rtbfViewController release]; it works fine .suppose if i have Navigation controller in Interface Builder,how can i set initWithRootViewController in Interface builder?

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  • Codeigniter: Using data in a controller

    - by Kevin Brown
    I'm new to php and CI, and I'm having some trouble in my controller. I feel that I'm doing this the wrong way, and it could be easier, I just don't know the syntax: $data['members'] = $this->home_model->getUser($id); $credit = $this->home_model->getCredit($id); if ($credit == '0'){stuff...} So I'm getting the user's data that has their the same information as "getCredit" does, but I don't know how to get the single variable that I need for my if statement... How can I just use the "getUser" function so that I'm not pulling redundant information?

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  • Connecting SceneBuilder edited FXML to Java code

    - by daniel
    Recently I had to answer several questions regarding how to connect an UI built with the JavaFX SceneBuilder 1.0 Developer Preview to Java Code. So I figured out that a short overview might be helpful. But first, let me state the obvious. What is FXML? To make it short, FXML is an XML based declaration format for JavaFX. JavaFX provides an FXML loader which will parse FXML files and from that construct a graph of Java object. It may sound complex when stated like that but it is actually quite simple. Here is an example of FXML file, which instantiate a StackPane and puts a Button inside it: -- <?xml version="1.0" encoding="UTF-8"?> <?import java.lang.*?> <?import java.util.*?> <?import javafx.scene.control.*?> <?import javafx.scene.layout.*?> <?import javafx.scene.paint.*?> <StackPane prefHeight="150.0" prefWidth="200.0" xmlns:fx="http://javafx.com/fxml"> <children> <Button mnemonicParsing="false" text="Button" /> </children> </StackPane> ... and here is the code I would have had to write if I had chosen to do the same thing programatically: import javafx.scene.control.*; import javafx.scene.layout.*; ... final Button button = new Button("Button"); button.setMnemonicParsing(false); final StackPane stackPane = new StackPane(); stackPane.setPrefWidth(200.0); stackPane.setPrefHeight(150.0); stacPane.getChildren().add(button); As you can see - FXML is rather simple to understand - as it is quite close to the JavaFX API. So OK FXML is simple, but why would I use it?Well, there are several answers to that - but my own favorite is: because you can make it with SceneBuilder. What is SceneBuilder? In short SceneBuilder is a layout tool that will let you graphically build JavaFX user interfaces by dragging and dropping JavaFX components from a library, and save it as an FXML file. SceneBuilder can also be used to load and modify JavaFX scenegraphs declared in FXML. Here is how I made the small FXML file above: Start the JavaFX SceneBuilder 1.0 Developer Preview In the Library on the left hand side, click on 'StackPane' and drag it on the content view (the white rectangle) In the Library, select a Button and drag it onto the StackPane on the content view. In the Hierarchy Panel on the left hand side - select the StackPane component, then invoke 'Edit > Trim To Selected' from the menubar That's it - you can now save, and you will obtain the small FXML file shown above. Of course this is only a trivial sample, made for the sake of the example - and SceneBuilder will let you create much more complex UIs. So, I have now an FXML file. But what do I do with it? How do I include it in my program? How do I write my main class? Loading an FXML file with JavaFX Well, that's the easy part - because the piece of code you need to write never changes. You can download and look at the SceneBuilder samples if you need to get convinced, but here is the short version: Create a Java class (let's call it 'Main.java') which extends javafx.application.Application In the same directory copy/save the FXML file you just created using SceneBuilder. Let's name it "simple.fxml" Now here is the Java code for the Main class, which simply loads the FXML file and puts it as root in a stage's scene. /* * Copyright (c) 2012, Oracle and/or its affiliates. All rights reserved. */ package simple; import java.util.logging.Level; import java.util.logging.Logger; import javafx.application.Application; import javafx.fxml.FXMLLoader; import javafx.scene.Scene; import javafx.scene.layout.StackPane; import javafx.stage.Stage; public class Main extends Application { /** * @param args the command line arguments */ public static void main(String[] args) { Application.launch(Main.class, (java.lang.String[])null); } @Override public void start(Stage primaryStage) { try { StackPane page = (StackPane) FXMLLoader.load(Main.class.getResource("simple.fxml")); Scene scene = new Scene(page); primaryStage.setScene(scene); primaryStage.setTitle("FXML is Simple"); primaryStage.show(); } catch (Exception ex) { Logger.getLogger(Main.class.getName()).log(Level.SEVERE, null, ex); } } } Great! Now I only have to use my favorite IDE to compile the class and run it. But... wait... what does it do? Well nothing. It just displays a button in the middle of a window. There's no logic attached to it. So how do we do that? How can I connect this button to my application logic? Here is how: Connection to code First let's define our application logic. Since this post is only intended to give a very brief overview - let's keep things simple. Let's say that the only thing I want to do is print a message on System.out when the user clicks on my button. To do that, I'll need to register an action handler with my button. And to do that, I'll need to somehow get a handle on my button. I'll need some kind of controller logic that will get my button and add my action handler to it. So how do I get a handle to my button and pass it to my controller? Once again - this is easy: I just need to write a controller class for my FXML. With each FXML file, it is possible to associate a controller class defined for that FXML. That controller class will make the link between the UI (the objects defined in the FXML) and the application logic. To each object defined in FXML we can associate an fx:id. The value of the id must be unique within the scope of the FXML, and is the name of an instance variable inside the controller class, in which the object will be injected. Since I want to have access to my button, I will need to add an fx:id to my button in FXML, and declare an @FXML variable in my controller class with the same name. In other words - I will need to add fx:id="myButton" to my button in FXML: -- <Button fx:id="myButton" mnemonicParsing="false" text="Button" /> and declare @FXML private Button myButton in my controller class @FXML private Button myButton; // value will be injected by the FXMLLoader Let's see how to do this. Add an fx:id to the Button object Load "simple.fxml" in SceneBuilder - if not already done In the hierarchy panel (bottom left), or directly on the content view, select the Button object. Open the Properties sections of the inspector (right panel) for the button object At the top of the section, you will see a text field labelled fx:id. Enter myButton in that field and validate. Associate a controller class with the FXML file Still in SceneBuilder, select the top root object (in our case, that's the StackPane), and open the Code section of the inspector (right hand side) At the top of the section you should see a text field labelled Controller Class. In the field, type simple.SimpleController. This is the name of the class we're going to create manually. If you save at this point, the FXML will look like this: -- <?xml version="1.0" encoding="UTF-8"?> <?import java.lang.*?> <?import java.util.*?> <?import javafx.scene.control.*?> <?import javafx.scene.layout.*?> <?import javafx.scene.paint.*?> <StackPane prefHeight="150.0" prefWidth="200.0" xmlns:fx="http://javafx.com/fxml" fx:controller="simple.SimpleController"> <children> <Button fx:id="myButton" mnemonicParsing="false" text="Button" /> </children> </StackPane> As you can see, the name of the controller class has been added to the root object: fx:controller="simple.SimpleController" Coding the controller class In your favorite IDE, create an empty SimpleController.java class. Now what does a controller class looks like? What should we put inside? Well - SceneBuilder will help you there: it will show you an example of controller skeleton tailored for your FXML. In the menu bar, invoke View > Show Sample Controller Skeleton. A popup appears, displaying a suggestion for the controller skeleton: copy the code displayed there, and paste it into your SimpleController.java: /** * Sample Skeleton for "simple.fxml" Controller Class * Use copy/paste to copy paste this code into your favorite IDE **/ package simple; import java.net.URL; import java.util.ResourceBundle; import javafx.fxml.FXML; import javafx.fxml.Initializable; import javafx.scene.control.Button; public class SimpleController implements Initializable { @FXML // fx:id="myButton" private Button myButton; // Value injected by FXMLLoader @Override // This method is called by the FXMLLoader when initialization is complete public void initialize(URL fxmlFileLocation, ResourceBundle resources) { assert myButton != null : "fx:id=\"myButton\" was not injected: check your FXML file 'simple.fxml'."; // initialize your logic here: all @FXML variables will have been injected } } Note that the code displayed by SceneBuilder is there only for educational purpose: SceneBuilder does not create and does not modify Java files. This is simply a hint of what you can use, given the fx:id present in your FXML file. You are free to copy all or part of the displayed code and paste it into your own Java class. Now at this point, there only remains to add our logic to the controller class. Quite easy: in the initialize method, I will register an action handler with my button: () { @Override public void handle(ActionEvent event) { System.out.println("That was easy, wasn't it?"); } }); ... -- ... // initialize your logic here: all @FXML variables will have been injected myButton.setOnAction(new EventHandler<ActionEvent>() { @Override public void handle(ActionEvent event) { System.out.println("That was easy, wasn't it?"); } }); ... That's it - if you now compile everything in your IDE, and run your application, clicking on the button should print a message on the console! Summary What happens is that in Main.java, the FXMLLoader will load simple.fxml from the jar/classpath, as specified by 'FXMLLoader.load(Main.class.getResource("simple.fxml"))'. When loading simple.fxml, the loader will find the name of the controller class, as specified by 'fx:controller="simple.SimpleController"' in the FXML. Upon finding the name of the controller class, the loader will create an instance of that class, in which it will try to inject all the objects that have an fx:id in the FXML. Thus, after having created '<Button fx:id="myButton" ... />', the FXMLLoader will inject the button instance into the '@FXML private Button myButton;' instance variable found on the controller instance. This is because The instance variable has an @FXML annotation, The name of the variable exactly matches the value of the fx:id Finally, when the whole FXML has been loaded, the FXMLLoader will call the controller's initialize method, and our code that registers an action handler with the button will be executed. For a complete example, take a look at the HelloWorld SceneBuilder sample. Also make sure to follow the SceneBuilder Get Started guide, which will guide you through a much more complete example. Of course, there are more elegant ways to set up an Event Handler using FXML and SceneBuilder. There are also many different ways to work with the FXMLLoader. But since it's starting to be very late here, I think it will have to wait for another post. I hope you have enjoyed the tour! --daniel

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  • Can't view other computers on Network

    - by Darkart
    Systems: My Machine: Windows 7 Ultimate connected through ethernet into router Her Machine--Other Machine: Windows 7 Ultimate connected through wireless Router:F5D8236-4 N Wireless Router Version 1 Firmware: 2.01.03 (Apr 28 2009) ISP: Comcast Problem: I can not view the "Other Machine" on the network at all. I opened command prompt and ran net view and saw the pc name. I tried pinging the pc and it times out. Went inside the router and tried viewing the computer on the DHCP list and it can not be seen. I restored the router back to default settings and firmware and completely reset the modem and router, and created home group. I went to the other machine to configure home group settings and made sure that both PC's had identical settings. She was able to see my machine but I could not see hers. I restarted both machines and now we cant see each other at all. Also her PC ("Other Machine") had exclamation mark in the wireless icon but was connected just fine. There is no firewalls on currently or anti-virus enabled, and still can not see each other. Right now I am checking for updated drivers for the wireless card, but my question is could it be the router or something hardware related? I have went through all the settings in the Home group and visited most FAQ's and still no luck. Also as it stands I can not view her machine inside the router DHCP Client List :(

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  • If a raid controller changes, are the drives still usable without re-formatting?

    - by Jeremy
    I've been wanting to do a raid 1 setup in my home with a pair of sata drives. Someone told me that if the controller fails, you can't just get a new controller because you'll have to reformat the drives. Or is that true only in some implementations? I was originally just looking at an onboard raid controller, or an entry level nas drvice like the intel SS4200-E, but If the hardware (controller) ever fails, will I be out of luck accessing the data if I can't get the exact same hardware to replace it?

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  • Freebsd 7.2: View firmware version for disks?

    - by Stefan Lasiewski
    I'm running FreeBSD 7.2, with Seagate Cheetah (Model ST####) drives. We are having some problems with the SCSI drives on these machines. Our vendor says that updating the firmware on the drives may fix the problems, and a firmware update did seem to fix some SCSI problems on another FreeBSD host. How can I view the firmware version of these drives? I tried some tips from nixCraft, but nothing has worked so far. In dmesg, I see the Make and Model, but In Linux, this information is often in /var/log/dmesg (Although /var/log/dmesg is sometimes out of date), or I often find this information with something like sudo lshw -class disk, lshal or dmidecode.

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  • Administrator view all mapped drives

    - by kskid19
    In my understanding of security, an administrator should be able to view all connections to and from a computer - just as they can view all processes/owner, network connections/owning process. However, Windows 8 seems to have disabled this. As administrator running an elevated in Win Vista+ when you run net use you get back all drives mapped, listed as unavailable. In Windows 8, the same command run from an elevated prompt returns "There are no entries in the list". The behavior is identical for powershell Get-WmiObject Win32_LogonSessionMappedDisk. A workaround for persistent mappings is to run Get-ChildItem Registry::HKU*\Network*. This does not include temporary mappings (in my particular example it was created through explorer on an administrator account and I did not select "Reconnect at sign-in") Is there a direct/simple way for Administrator to view connections of any user (short of a script that runs under each user context)? I have read Some Programs Cannot Access Network Locations When UAC Is Enabled but I do not think it particularly applies. ServerFault has an answer, but it still does not address non-persistent drives How can I tell what network drives users have mapped?

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  • Administrator view ALL mapped drives

    - by kskid19
    In my understanding of security, an administrator should be able to view all connections to and from a computer - just as they can view all processes/owner, network connections/owning process. However, Windows 8 seems to have disabled this. As administrator running an elevated in Win Vista+ when you run net use you get back all drives mapped, listed as unavailable. In Windows 8, the same command run from an elevated prompt returns "There are no entries in the list". The behavior is identical for powershell Get-WmiObject Win32_LogonSessionMappedDisk. A workaround for persistent mappings is to run Get-ChildItem Registry::HKU*\Network*. This does not include temporary mappings (in my particular example it was created through explorer on an administrator account and I did not select "Reconnect at sign-in") Is there a direct/simple way for Administrator to view connections of any user (short of a script that runs under each user context)? I have read Some Programs Cannot Access Network Locations When UAC Is Enabled but I do not think it particularly applies. I have seen this answer, but it still does not address non-persistent drives How can I tell what network drives users have mapped?

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  • How do you enable view source in ie8 when it gets magically diabled

    - by Tim Meers
    I have multiple computers that all seem to have View Source disabled from the content menu when you right click on a web page. Now I know it's not that the web page is some how disabling it, I'm pretty sure thats not even possible. But alas I have at least 3 machines in my office (not on AD) that have this problem. I have also worked on clients computers that have this same issue. It's down right maddening! I tried to Google for it, but it just shows results from the dawn of IE6 in all of it's "glory" with a bug where if the cache was full it would be disabled. But this is not the case in IE8. Any body have a clue why this is happening, or a fix for it? Maybe a reg setting? Update: So I got a little closer to solving it, but there was still an issue on one computer where it allowed it not is HTTP, but not in HTTPS. One other computer works correctly in both. I Found these two keys missing in the registry: [-HKEY_CURRENT_USER\SOFTWARE\Microsoft\Internet Explorer\View Source Editor] [-HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Internet Explorer\View Source Editor]

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  • Accessing clearcase view drive from virtual machine is slow

    - by PermanentGuest
    I have a windows XP virtual machine running under a Windows XP host. On the host : On the host clearcase 7.1.1.2 is installed. I have a dynamic view mapped onto some drive. The view has certain VOB/directory structure where my application DLLs from the nightly build and config files are stored. I run my application on the host machine which uses the DLLs and config files from the VOB and everything runs smooth. Now I want to move this set-up to a virtual machine. On the guest : I'm running the guest with a vm-player. I don't want to install clear-case on this as I don't want to expose this machine onto the network. The network setting in the guest is 'host-only'. I have mapped the host's clearcase view drive as a shared folder and I'm able to access this drive from the virtual machine. Also, the application is running. However, the problem is that the access of the clearcase drive from the virtual machine is very slow. I can experience this from the windows explorer. Due to this, the starting of my application takes several seconds in the virtual machine while on the guest it comes up pretty fast. My question is : Is there any way to speed up the performance? I have managed to copy some of the DLLs which don't change frequently to the virtual machine to improve the performance. However, there are still lot of DLLs which have to be taken from the clearcase drive as they change frequently. VMplayer version is : VM Player 3.0.1 build-227600 Both guest and host is : Windows XP service pack 3 Host clearcase is : clearcase 7.1.1.2

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  • Model M Keyboard inputs incorrect characters after logging in to Fedora

    - by mickburkejnr
    I recently bought a 24 year old IBM Model M keyboard. From what I gather, it'd been left on a shelf for the last 5 years, so you can imagine the amount of dust dirt and crap that was on it. Before cleaning it, I plugged it in to my laptop (running Fedora 17) using a PS/2 to USB adapter. What I found was, while it still works, the keys I press don't correspond to what is displayed on the screen. So for example, when I type S on the keyboard, I get ß display on the screen instead. At the time, I put this down to the adapter not working properly. Since then, I stripped the keys off the keyboard and cleaned the whole thing. It looks like it's just come out of a box! I then plugged it in to my computer (also running Fedora 17) via a standard PS/2 plug. The computer loaded up to the login screen, and I typed in my password. Pressed enter, and I logged straight in to my machine. At this point, I opened up a text editor and started typing some stuff. To my horror, the keystrokes I was entering weren't coming up as intended. What came up instead were characters that would map to the pressed key but only under a different keyboard language setting. I opened up a program to see what keyboard language had been selected, and the correct one for the keyboard was selected (which is UK in my case). I opened up a window that would show what characters mapped to what keys, and I pressed every single key on the keyboard, and every corresponding block representing each key lit up. I went back to the text editor to try again, but I was still getting these random characters. Whats more is that the backspace key would not work, although in the other utility it would flash when pressed. What I know is that at the login screen the keyboard must have entered the correct characters, otherwise I wouldn't have been able to log in. Further more, keys that don't respond while using a text editor as sending signals to the computer, as illustrated in that keyboard utility. The question is why random characters are displayed when they really shouldn't be? Would this be a hardware fault or a software issue?

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  • Copy Database Wizard fails on creation of view into another not-yet-copied database

    - by user22037
    Update - I found that doing a manual detach/reattach using MSDN article "How to: Move a Database Using Detach and Attach (Transact-SQL)" got around this issue. I'll just be creating a script to dettach and reattach but do the file copies manually. Any info on how to overcome the problems with the wizard would be helpful in the future. I am in the process of moving around 20 databases from our current server to a new one. When performing the copies however I have found that some databases can not copy if they have views into other databases that have not yet been copied to the target system. The log file generated says "failed with the following error: "Invalid object name" in reference to the database in the view. If I first copy just the database referenced in the view and then in a separate step copy the database over containing the view it is successful. However some other database have views into each other so can't just adjust the order in which the copy occurs. Is there any way to ignore this error and just allow everything to copy?

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  • WPF DataContext does not refresh the DataGrid using MVVM model

    - by vikram bhatia
    Project Overview I have a view which binds to a viewmodel containing 2 ObserverableCollection. The viewmodel constructor populates the first ObserverableCollection and the view datacontext is collected to bind to it through a public property called Sites. Later the 2ed ObserverableCollection is populated in the LoadOrders method and the public property LoadFraudResults is updated for binding it with datacontext. I am using WCF to pull the data from the database and its getting pulled very nicely. VIEWMODEL SOURCE class ManageFraudOrderViewModel:ViewModelBase { #region Fields private readonly ICollectionView collectionViewSites; private readonly ICollectionView collectionView; private ObservableCollection<GeneralAdminService.Website> _sites; private ObservableCollection<FraudService.OrderQueue> _LoadFraudResults; #endregion #region Properties public ObservableCollection<GeneralAdminService.Website> Sites { get { return this._sites; } } public ObservableCollection<FraudService.OrderQueue> LoadFraudResults { get { return this._LoadFraudResults;} } #endregion public ManageFraudOrderViewModel() { //Get values from wfc service model GeneralAdminService.GeneralAdminServiceClient generalAdminServiceClient = new GeneralAdminServiceClient(); GeneralAdminService.Website[] websites = generalAdminServiceClient.GetWebsites(); //Get values from wfc service model if (websites.Length > 0) { _sites = new ObservableCollection<Wqn.Administration.UI.GeneralAdminService.Website>(); foreach (GeneralAdminService.Website website in websites) { _sites.Add((Wqn.Administration.UI.GeneralAdminService.Website)website); } this.collectionViewSites= CollectionViewSource.GetDefaultView(this._sites); } generalAdminServiceClient.Close(); } public void LoadOrders(Wqn.Administration.UI.FraudService.Website website) { //Get values from wfc service model FraudServiceClient fraudServiceClient = new FraudServiceClient(); FraudService.OrderQueue[] OrderQueue = fraudServiceClient.GetFraudOrders(website); //Get values from wfc service model if (OrderQueue.Length > 0) { _LoadFraudResults = new ObservableCollection<Wqn.Administration.UI.FraudService.OrderQueue>(); foreach (FraudService.OrderQueue orderQueue in OrderQueue) { _LoadFraudResults.Add(orderQueue); } } this.collectionViewSites= CollectionViewSource.GetDefaultView(this._LoadFraudResults); fraudServiceClient.Close(); } } VIEW SOURCE public partial class OrderQueueControl : UserControl { private ManageFraudOrderViewModel manageFraudOrderViewModel ; private OrderQueue orderQueue; private ButtonAction ButtonAction; private DispatcherTimer dispatcherTimer; public OrderQueueControl() { LoadOrderQueueForm(); } #region LoadOrderQueueForm private void LoadOrderQueueForm() { //for binding the first observablecollection manageFraudOrderViewModel = new ManageFraudOrderViewModel(); this.DataContext = manageFraudOrderViewModel; } #endregion private void cmbWebsite_SelectionChanged(object sender, SelectionChangedEventArgs e) { BindItemsSource(); } #region BindItemsSource private void BindItemsSource() { using (OverrideCursor cursor = new OverrideCursor(Cursors.Wait)) { if (!string.IsNullOrEmpty(Convert.ToString(cmbWebsite.SelectedItem))) { Wqn.Administration.UI.FraudService.Website website = (Wqn.Administration.UI.FraudService.Website)Enum.Parse(typeof(Wqn.Administration.UI.FraudService.Website),cmbWebsite.SelectedItem.ToString()); //for binding the second observablecollection******* manageFraudOrderViewModel.LoadOrders(website); this.DataContext = manageFraudOrderViewModel; //for binding the second observablecollection******* } } } #endregion } XAML ComboBox x:Name="cmbWebsite" ItemsSource="{Binding Sites}" Margin="5" Width="100" Height="25" SelectionChanged="cmbWebsite_SelectionChanged" DataGrid ItemsSource ={Binding Path = LoadFraudResults} PROBLEM AREA: When I call the LoadOrderQueueForm to bind the first observablecollection and later BindItemsSource to bind 2ed observable collection, everything works fine and no problem for the first time binding. But, when I call BindItemsSource again to repopulate the obseravablecollection based on changed selected combo value via cmbWebsite_SelectionChanged, the observalblecollection gets populated with new value and LoadFraudResults property in viewmodule is populated with new values; but when i call the datacontext to rebind the datagrid,the datagrid does not reflect the changed values. In other words the datagrid doesnot get changed when the datacontext is called the 2ed time in BindItemsSource method of the view. manageFraudOrderViewModel.LoadOrders(website); this.DataContext = manageFraudOrderViewModel; manageFraudOrderViewModel values are correct but the datagrid is not relected with changed values. Please help as I am stuck with this thing for past 2 days and the deadline is approaching near. Thanks in advance

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  • ViewPager cycle between views?

    - by Erdem Azakli
    I want my ViewPager implementation to cycle between views instead of stopping at the last view. For example, if I have 3 views to display via a ViewPager, it should return back to the first View after the third View on fling instead of stopping at that third view. I want it to return to the first page/view when the user flings forward on the last page Thanks, Mypageradapter; package com.example.pictures; import android.content.Context; import android.media.AudioManager; import android.os.Parcelable; import android.support.v4.view.PagerAdapter; import android.support.v4.view.ViewPager; import android.view.LayoutInflater; import android.view.View; public class MyPagerAdapter extends PagerAdapter{ SoundManager snd; int sound1,sound2,sound3; boolean loaded = false; public int getCount() { return 6; } public Object instantiateItem(View collection, int position) { View view=null; LayoutInflater inflater = (LayoutInflater) collection.getContext() .getSystemService(Context.LAYOUT_INFLATER_SERVICE); this.setVolumeControlStream(AudioManager.STREAM_MUSIC); int resId = 0; switch (position) { case 0: resId = R.layout.picture1; view = inflater.inflate(resId, null); break; case 1: resId = R.layout.picture2; view = inflater.inflate(resId, null); break; case 2: resId = R.layout.picture3; view = inflater.inflate(resId, null); break; case 3: resId = R.layout.picture4; view = inflater.inflate(resId, null); break; case 4: resId = R.layout.picture5; view = inflater.inflate(resId, null); break; case 5: resId = R.layout.picture6; view = inflater.inflate(resId, null); break; } ((ViewPager) collection).addView(view, 0); return view; } @SuppressWarnings("unused") private Context getApplicationContext() { // TODO Auto-generated method stub return null; } private void setVolumeControlStream(int streamMusic) { // TODO Auto-generated method stub } @SuppressWarnings("unused") private Context getBaseContext() { // TODO Auto-generated method stub return null; } @SuppressWarnings("unused") private PagerAdapter findViewById(int myfivepanelpager) { // TODO Auto-generated method stub return null; } @Override public void destroyItem(View arg0, int arg1, Object arg2) { ((ViewPager) arg0).removeView((View) arg2); } @Override public boolean isViewFromObject(View arg0, Object arg1) { return arg0 == ((View) arg1); } @Override public Parcelable saveState() { return null; } public static Integer getItem(int position) { // TODO Auto-generated method stub return null; } } OnPageChangeListener; package com.example.pictures; import android.app.Activity; import android.content.Intent; import android.os.Bundle; import android.support.v4.view.ViewPager; import android.support.v4.view.ViewPager.OnPageChangeListener; import android.view.View; import android.widget.Button; import android.widget.Toast; public class Pictures extends Activity implements OnPageChangeListener{ SoundManager snd; int sound1,sound2,sound3; View view=null; @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.picturespage); MyPagerAdapter adapter = new MyPagerAdapter(); ViewPager myPager = (ViewPager) findViewById(R.id.myfivepanelpager); myPager.setAdapter(adapter); myPager.setCurrentItem(0); myPager.setOnPageChangeListener(this); snd = new SoundManager(this); sound1 = snd.load(R.raw.sound1); sound2 = snd.load(R.raw.sound2); sound3 = snd.load(R.raw.sound3); } public void onPageScrollStateChanged(int arg0) { // TODO Auto-generated method stub } public void onPageScrolled(int arg0, float arg1, int arg2) { // TODO Auto-generated method stub } public void onPageSelected(int position) { // TODO Auto-generated method stub switch (position) { case 0: snd.play(sound1); break; case 1: snd.play(sound2); break; case 2: snd.play(sound3); break; case 3: Toast.makeText(this, "1", Toast.LENGTH_SHORT).show(); break; case 4: Toast.makeText(this, "2", Toast.LENGTH_SHORT).show(); break; case 5: Toast.makeText(this, "3", Toast.LENGTH_SHORT).show(); break; } } };

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  • XForms and multiple inputs for same model tag

    - by iHeartGreek
    Hi! I apologize ahead of time if I am not asking this properly.. it is hard to put into words what I am asking.. I have XForms model such as: <file> <criteria> <criterion></criterion> </criteria> </file> I want to have multiple input text boxes that create a new criterion tag. user interface such as: <xf:input ref="/file/criteria/criterion" model="select_data"> <xf:label>Select</xf:label> </xf:input> <xf:input ref="/file/criteria/criterion" model="select_data"> <xf:label>Select</xf:label> </xf:input> <xf:input ref="/file/criteria/criterion" model="select_data"> <xf:label>Select</xf:label> </xf:input> And I would like the XML output to look like this (once user has entered in info): <file> <criteria> <criterion>AAA</criterion> <criterion>BBB</criterion> <criterion>CCC</criterion> </criteria> </file> The way I have it doesn't work, as it sees the 3 input fields to be referring all to the same criterion tag. How do I differentiate? Thanks! I hope that made some sense! BEGIN FIRST EDIT Thanks for the responses for the basic text box! However, I now need to do this with a listbox. But for the life of me, I can't figure out how. I read somewhere to use with the xforms:select and deselect events.. but I didn't know where to place them, and the places I tried gave me very weird behaviour. I am currently implementing the following: <xf:select ref="instance('criteria_data')/criteria/criterion" selection="" appearance="compact" > <xf:label>Choose criteria</xf:label> <xf:itemset nodeset="instance('criteria_choices')/choice"> <xf:label ref="@label"></xf:label> <xf:value ref="."></xf:value> </xf:itemset> </xf:select> However when multiple choices are submitted, all selection values are inserted into the same node, separated by spaces. For example: If AAA and BBB and FFF were selected from listbox, it would result in the following XML: <criterion>AAA BBB FFF</criterion> How do I change my code to have each selection be in a separate node? i.e. I want it to look like this: <criterion>AAA</criterion> <criterion>BBB</criterion> <criterion>FFF</criterion> Thanks! END FIRST EDIT BEGIN SECOND EDIT: For the listboxes (ie xf:select appearance="compact") I ended up allowing the spaces to occur in the same node and then just transformed that xml using xsl to generate a properly formatted new xml doc (with separate individual nodes). Unfortunately, I did not find a less cumbersome solution by inserting them originally into separate nodes. The selected answer works very well for text boxes however, hence why I selected it as the answer. END SECOND EDIT

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  • Spring security - Reach users ID without passing it through every controller

    - by nilsi
    I have a design issue that I don't know how to solve. I'm using Spring 3.2.4 and Spring security 3.1.4. I have a Account table in my database that looks like this: create table Account (id identity, username varchar unique, password varchar not null, firstName varchar not null, lastName varchar not null, university varchar not null, primary key (id)); Until recently my username was just only a username but I changed it to be the email address instead since many users want to login with that instead. I have a header that I include on all my pages which got a link to the users profile like this: <a href="/project/users/<%= request.getUserPrincipal().getName()%>" class="navbar-link"><strong><%= request.getUserPrincipal().getName()%></strong></a> The problem is that <%= request.getUserPrincipal().getName()%> returns the email now, I don't want to link the user's with thier emails. Instead I want to use the id every user have to link to the profile. How do I reach the users id's from every page? I have been thinking of two solutions but I'm not sure: Change the principal to contain the id as well, don't know how to do this and having problem finding good information on the topic. Add a model attribute to all my controllers that contain the whole user but this would be really ugly, like this. Account account = entityManager.find(Account.class, email); model.addAttribute("account", account); There are more way's as well and I have no clue which one is to prefer. I hope it's clear enough and thank you for any help on this. ====== Edit according to answer ======= I edited Account to implement UserDetails, it now looks like this (will fix the auto generated stuff later): @Entity @Table(name="Account") public class Account implements UserDetails { @Id private int id; private String username; private String password; private String firstName; private String lastName; @ManyToOne private University university; public Account() { } public Account(String username, String password, String firstName, String lastName, University university) { this.username = username; this.password = password; this.firstName = firstName; this.lastName = lastName; this.university = university; } public String getUsername() { return username; } public String getPassword() { return password; } public String getFirstName() { return firstName; } public String getLastName() { return lastName; } public void setUsername(String username) { this.username = username; } public void setPassword(String password) { this.password = password; } public void setFirstName(String firstName) { this.firstName = firstName; } public void setLastName(String lastName) { this.lastName = lastName; } public University getUniversity() { return university; } public void setUniversity(University university) { this.university = university; } public int getId() { return id; } public void setId(int id) { this.id = id; } @Override public Collection<? extends GrantedAuthority> getAuthorities() { // TODO Auto-generated method stub return null; } @Override public boolean isAccountNonExpired() { // TODO Auto-generated method stub return false; } @Override public boolean isAccountNonLocked() { // TODO Auto-generated method stub return false; } @Override public boolean isCredentialsNonExpired() { // TODO Auto-generated method stub return false; } @Override public boolean isEnabled() { // TODO Auto-generated method stub return true; } } I also added <%@ taglib prefix="sec" uri="http://www.springframework.org/security/tags" %> To my jsp files and trying to reach the id by <sec:authentication property="principal.id" /> This gives me the following org.springframework.beans.NotReadablePropertyException: Invalid property 'principal.id' of bean class [org.springframework.security.authentication.UsernamePasswordAuthenticationToken]: Bean property 'principal.id' is not readable or has an invalid getter method: Does the return type of the getter match the parameter type of the setter? ====== Edit 2 according to answer ======= I based my application on spring social samples and I never had to change anything until now. This are the files I think are relevant, please tell me if theres something you need to see besides this. AccountRepository.java public interface AccountRepository { void createAccount(Account account) throws UsernameAlreadyInUseException; Account findAccountByUsername(String username); } JdbcAccountRepository.java @Repository public class JdbcAccountRepository implements AccountRepository { private final JdbcTemplate jdbcTemplate; private final PasswordEncoder passwordEncoder; @Inject public JdbcAccountRepository(JdbcTemplate jdbcTemplate, PasswordEncoder passwordEncoder) { this.jdbcTemplate = jdbcTemplate; this.passwordEncoder = passwordEncoder; } @Transactional public void createAccount(Account user) throws UsernameAlreadyInUseException { try { jdbcTemplate.update( "insert into Account (firstName, lastName, username, university, password) values (?, ?, ?, ?, ?)", user.getFirstName(), user.getLastName(), user.getUsername(), user.getUniversity(), passwordEncoder.encode(user.getPassword())); } catch (DuplicateKeyException e) { throw new UsernameAlreadyInUseException(user.getUsername()); } } public Account findAccountByUsername(String username) { return jdbcTemplate.queryForObject("select username, firstName, lastName, university from Account where username = ?", new RowMapper<Account>() { public Account mapRow(ResultSet rs, int rowNum) throws SQLException { return new Account(rs.getString("username"), null, rs.getString("firstName"), rs.getString("lastName"), new University("test")); } }, username); } } security.xml <?xml version="1.0" encoding="UTF-8"?> <beans:beans xmlns="http://www.springframework.org/schema/security" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:beans="http://www.springframework.org/schema/beans" xsi:schemaLocation="http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd"> <http pattern="/resources/**" security="none" /> <http pattern="/project/" security="none" /> <http use-expressions="true"> <!-- Authentication policy --> <form-login login-page="/signin" login-processing-url="/signin/authenticate" authentication-failure-url="/signin?error=bad_credentials" /> <logout logout-url="/signout" delete-cookies="JSESSIONID" /> <intercept-url pattern="/addcourse" access="isAuthenticated()" /> <intercept-url pattern="/courses/**/**/edit" access="isAuthenticated()" /> <intercept-url pattern="/users/**/edit" access="isAuthenticated()" /> </http> <authentication-manager alias="authenticationManager"> <authentication-provider> <password-encoder ref="passwordEncoder" /> <jdbc-user-service data-source-ref="dataSource" users-by-username-query="select username, password, true from Account where username = ?" authorities-by-username-query="select username, 'ROLE_USER' from Account where username = ?"/> </authentication-provider> <authentication-provider> <user-service> <user name="admin" password="admin" authorities="ROLE_USER, ROLE_ADMIN" /> </user-service> </authentication-provider> </authentication-manager> </beans:beans> And this is my try of implementing a UserDetailsService public class RepositoryUserDetailsService implements UserDetailsService { private final AccountRepository accountRepository; @Autowired public RepositoryUserDetailsService(AccountRepository repository) { this.accountRepository = repository; } @Override public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException { Account user = accountRepository.findAccountByUsername(username); if (user == null) { throw new UsernameNotFoundException("No user found with username: " + username); } return user; } } Still gives me the same error, do I need to add the UserDetailsService somewhere? This is starting to be something else compared to my initial question, I should maybe start another question. Sorry for my lack of experience in this. I have to read up.

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  • Entity Data Model Wizzard not creating tables in EDMX file

    - by Shawn
    I'm trying the database first approach by creating an ADO.NET Entity Data Model using the Wizard with the Adventureworks2012 DB. Testing DB connection works, and the connection string is added to the App.Config. I'm selecting all the tables except the ones marked as (dbo) AWBuildVersion, DatabaseLog, and ErrorLog. When the Wizard finishes the .edmx file is blank, and if I view the file in XML view the EntityContainer is empty. I'm using VS 2010 & .NET Framework 4.0

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  • MVC : Does Code to save data in cache or session belongs in controller?

    - by newbie
    I'm a bit confused if saving the information to session code below, belongs in the controller action as shown below or should it be part of my Model? I would add that I have other controller methods that will read this session value later. public ActionResult AddFriend(FriendsContext viewModel) { if (!ModelState.IsValid) { return View(viewModel); } // Start - Confused if the code block below belongs in Controller? Friend friend = new Friend(); friend.FirstName = viewModel.FirstName; friend.LastName = viewModel.LastName; friend.Email = viewModel.UserEmail; httpContext.Session["latest-friend"] = friend; // End Confusion return RedirectToAction("Home"); } I thought about adding a static utility class in my Model which does something like below, but it just seems stupid to add 2 lines of code in another file. public static void SaveLatestFriend(Friend friend, HttpContextBase httpContext) { httpContext.Session["latest-friend"] = friend; } public static Friend GetLatestFriend(HttpContextBase httpContext) { return httpContext.Session["latest-friend"] as Friend; }

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  • Why is String Templating Better Than String Concatenation from an Engineering Perspective?

    - by stephen
    I once read (I think it was in "Programming Pearls") that one should use templates instead of building the string through the use of concatenation. For example, consider the template below (using C# razor library) <in a properties file> Browser Capabilities Type = @Model.Type Name = @Model.Browser Version = @Model.Version Supports Frames = @Model.Frames Supports Tables = @Model.Tables Supports Cookies = @Model.Cookies Supports VBScript = @Model.VBScript Supports Java Applets = @Model.JavaApplets Supports ActiveX Controls = @Model.ActiveXControls and later, in a separate code file private void Button1_Click(object sender, System.EventArgs e) { BrowserInfoTemplate = Properties.Resources.browserInfoTemplate; // see above string browserInfo = RazorEngine.Razor.Parse(BrowserInfoTemplate, browser); ... } From a software engineering perspective, how is this better than an equivalent string concatentation, like below: private void Button1_Click(object sender, System.EventArgs e) { System.Web.HttpBrowserCapabilities browser = Request.Browser; string s = "Browser Capabilities\n" + "Type = " + browser.Type + "\n" + "Name = " + browser.Browser + "\n" + "Version = " + browser.Version + "\n" + "Supports Frames = " + browser.Frames + "\n" + "Supports Tables = " + browser.Tables + "\n" + "Supports Cookies = " + browser.Cookies + "\n" + "Supports VBScript = " + browser.VBScript + "\n" + "Supports JavaScript = " + browser.EcmaScriptVersion.ToString() + "\n" + "Supports Java Applets = " + browser.JavaApplets + "\n" + "Supports ActiveX Controls = " + browser.ActiveXControls + "\n" ... }

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  • How to get distinct values from the List&lt;T&gt; with LINQ

    - by Vincent Maverick Durano
    Recently I was working with data from a generic List<T> and one of my objectives is to get the distinct values that is found in the List. Consider that we have this simple class that holds the following properties: public class Product { public string Make { get; set; } public string Model { get; set; } }   Now in the page code behind we will create a list of product by doing the following: private List<Product> GetProducts() { List<Product> products = new List<Product>(); Product p = new Product(); p.Make = "Samsung"; p.Model = "Galaxy S 1"; products.Add(p); p = new Product(); p.Make = "Samsung"; p.Model = "Galaxy S 2"; products.Add(p); p = new Product(); p.Make = "Samsung"; p.Model = "Galaxy Note"; products.Add(p); p = new Product(); p.Make = "Apple"; p.Model = "iPhone 4"; products.Add(p); p = new Product(); p.Make = "Apple"; p.Model = "iPhone 4s"; products.Add(p); p = new Product(); p.Make = "HTC"; p.Model = "Sensation"; products.Add(p); p = new Product(); p.Make = "HTC"; p.Model = "Desire"; products.Add(p); p = new Product(); p.Make = "Nokia"; p.Model = "Some Model"; products.Add(p); p = new Product(); p.Make = "Nokia"; p.Model = "Some Model"; products.Add(p); p = new Product(); p.Make = "Sony Ericsson"; p.Model = "800i"; products.Add(p); p = new Product(); p.Make = "Sony Ericsson"; p.Model = "800i"; products.Add(p); return products; }   And then let’s bind the products to the GridView. protected void Page_Load(object sender, EventArgs e) { if (!IsPostBack) { Gridview1.DataSource = GetProducts(); Gridview1.DataBind(); } }   Running the code will display something like this in the page: Now what I want is to get the distinct row values from the list. So what I did is to use the LINQ Distinct operator and unfortunately it doesn't work. In order for it work is you must use the overload method of the Distinct operator for you to get the desired results. So I’ve added this IEqualityComparer<T> class to compare values: class ProductComparer : IEqualityComparer<Product> { public bool Equals(Product x, Product y) { if (Object.ReferenceEquals(x, y)) return true; if (Object.ReferenceEquals(x, null) || Object.ReferenceEquals(y, null)) return false; return x.Make == y.Make && x.Model == y.Model; } public int GetHashCode(Product product) { if (Object.ReferenceEquals(product, null)) return 0; int hashProductName = product.Make == null ? 0 : product.Make.GetHashCode(); int hashProductCode = product.Model.GetHashCode(); return hashProductName ^ hashProductCode; } }   After that you can then bind the GridView like this: protected void Page_Load(object sender, EventArgs e) { if (!IsPostBack) { Gridview1.DataSource = GetProducts().Distinct(new ProductComparer()); Gridview1.DataBind(); } }   Running the page will give you the desired output below: As you notice, it now eliminates the duplicate rows in the GridView. Now what if we only want to get the distinct values for a certain field. For example I want to get the distinct “Make” values such as Samsung, Apple, HTC, Nokia and Sony Ericsson and populate them to a DropDownList control for filtering purposes. I was hoping the the Distinct operator has an overload that can compare values based on the property value like (GetProducts().Distinct(o => o.PropertyToCompare). But unfortunately it doesn’t provide that overload so what I did as a workaround is to use the GroupBy,Select and First LINQ query operators to achieve what I want. Here’s the code to get the distinct values of a certain field. protected void Page_Load(object sender, EventArgs e) { if (!IsPostBack) { DropDownList1.DataSource = GetProducts().GroupBy(o => o.Make).Select(o => o.First()); DropDownList1.DataTextField = "Make"; DropDownList1.DataValueField = "Model"; DropDownList1.DataBind(); } } Running the code will display the following output below:   That’s it! I hope someone find this post useful!

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