Search Results

Search found 14875 results on 595 pages for 'resource controller'.

Page 541/595 | < Previous Page | 537 538 539 540 541 542 543 544 545 546 547 548  | Next Page >

  • Passing user_id, site_id, and question_id to same table on create...

    - by bgadoci
    I can't seem to figure out how to do this. I am trying to pass four different variables to a single table. To be more specific I am trying to create a "like" in a likes table that also captures the site_id (like an answer), user_id, and the question_id. Here is the set up. class Like < ActiveRecord::Base belongs_to :site belongs_to :user belongs_to :question end I will spare you the reverse, has_many associations but they are there. Here is the likes controller where I think the problem is. class LikesController < ApplicationController def create @user = current_user @site = Site.find(params[:site_id]) @like = @site.likes.create!(params[:like]) @like.user = current_user @like.save respond_to do |format| format.html { redirect_to @site} format.js end end end This code successfully passes the like and site_id but after many different variations of trying I can't get it to pass the question id. Here is my form: /views/sites/_site.html.erb (though the partial is being displayed in the /views/questions/show.html.erb file). <% remote_form_for [site, Like.new] do |f| %> <%= f.hidden_field :site_name, :value => "#{site.name}" %> <%= f.hidden_field :ip_address, :value => "#{request.remote_ip}" %> <%= f.hidden_field :like, :value => "1" %> <%= submit_tag "^" , :class => 'voteup' %> <% end %>

    Read the article

  • Representing versions of objects with CakePHP

    - by user636901
    Hi people, Have just started to get into CakePHP since a couple of weeks back. I have some experience with MVC-frameworks, but this problem holds me back a bit. I am currently working on a model foo, containing a primary id and some attributes. Since a complete history of the changes of foo is necessary, the content of foo is saved in the table foo_content. The two tables are connected through foo_content.foo_id = foo.id, in Cake with a foo hasMany foo_content-relationship. To track the versions of foo, foo_content also contains the column version, and foo itself the field currentVersion. The version is an number incremented by one everytime the user updates foo. This is an older native PHP-app btw, to be rewritten on top of Cake. 9 times out of 10 in the app, the most recent version (foo.currentVersion) is the db-entry that need to be represented in the frontend. My question is simply: is there someway of representing this directly in the model? Or does this kind of logic simply need to be defined in the controller? Most grateful for your help!

    Read the article

  • Custom bean instantiation logic in Spring MVC

    - by Michal Bachman
    I have a Spring MVC application trying to use a rich domain model, with the following mapping in the Controller class: @RequestMapping(value = "/entity", method = RequestMethod.POST) public String create(@Valid Entity entity, BindingResult result, ModelMap modelMap) { if (entity== null) throw new IllegalArgumentException("An entity is required"); if (result.hasErrors()) { modelMap.addAttribute("entity", entity); return "entity/create"; } entity.persist(); return "redirect:/entity/" + entity.getId(); } Before this method gets executed, Spring uses BeanUtils to instantiate a new Entity and populate its fields. It uses this: ... ReflectionUtils.makeAccessible(ctor); return ctor.newInstance(args); Here's the problem: My entities are Spring managed beans. The reason for this is to inject DAOs on them. Instead of calling new, I use EntityFactory.createEntity(). When they're retrieved from the database, I have an interceptor that overrides the public Object instantiate(String entityName, EntityMode entityMode, Serializable id) method and hooks the factories into that. So the last piece of the puzzle missing here is how to force Spring to use the factory rather than its own BeanUtils reflective approach? Any suggestions for a clean solution? Thanks very much in advance.

    Read the article

  • query not displaying proper result

    - by ravindra
    In my Rails 3 project I have the following code for My controller: class TasksController < ApplicationController def today @tasks = Task.today @task = Task.new respond_to do |format| format.html { render :text=> "Sorry , you don't have any task pending today." } format.html # new.html.erb format.xml { render :xml => @tasks } end end def this_week @tasks = Task.this_week @task = Task.new respond_to do |format| format.html { render :text => "Sorry , No content for selected period." } format.html # new.html.erb format.xml { render :xml => @tasks } end end end My model: class Task < ActiveRecord::Base def self.today Task.where(:due_date => "Date.today" , :task_status => "open").order("due_date ASC") end def self.this_week Task.where(:due_date =>"Time.now.this_week" , :task_status => "open" ).order("due_date ASC") end end Why it does not displaying anything in the relative view. Please help me. Thanks

    Read the article

  • codeigniter cron job with http access

    - by user1313850
    Sorry if this is a duplicate question...I've searched around and found similar advice but nothing that helps my exact problem. And please excuse the noob questions, CRON is a new thing for me. I have a codeigniter script that scrapes the html DOM of another site and stores some of that in a database. I'd like to run this script at a regular interval. This has lead me to looking into cron jobs. The page I have is at myserver.com/index.php/update I realize I can run a cron job with curl and run this page. If I want to be a bit more secure I can put a string at the end like: myserver.com/index.php/update/asdfh2784fufds And check for that in my CI controller. This seems like it would be mostly secure, but doesn't seem like the "right" way to do things. I've looked into running CI from the command line, and can execute basic pages like: php index.php mycontroller But when I try to do: php index.php update It doesn't work. I suspect this is because it needs to use HTTP to scrape the DOM of the outside page. So, my question: How do I securely run a codeigniter script with a cron job that needs HTTP access?

    Read the article

  • Sanitizing User Input with Ruby on Rails

    - by phreakre
    I'm writing a very simple CRUD app that takes user stories and stores them into a database so another fellow coder can organize them for a project we're both working on. However, I have come across a problem with sanitizing user input before it is saved into the database. I cannot call the sanitize() function from within the Story model to strip out all of the html/scripting. It requires me to do the following: def sanitize_inputs self.name = ActionController::Base.helpers.sanitize(self.name) unless self.name.nil? self.story = ActionController::Base.helpers.sanitize(self.story) unless self.story.nil? end I want to validate that the user input has been sanitized and I am unsure of two things: 1) When should the user input validation take place? Before the data is saved is pretty obvious, I think, however, should I be processing this stuff in the Controller, before validation, or some other non-obvious area before I validate that the user input has no scripting/html tags? 2) Writing a unit test for this model, how would I verify that the scripting/html is removed besides comparing "This is a malicious code example" to the sanitize(example) output? Thanks in advance.

    Read the article

  • Stop lazy-loading images?

    - by Greg
    Here's the issue – I followed along with the Apple lazy-load image sample code to handle my graphical tables. It works great. However, my lazy-load image tables are being stacked within a navigation controller so that you can move in and out of menus. While all this works, I'm getting a persistent crash when I move into a table then move immediately back out of it using the "back" button. This appears to be a result of the network connections loading content not being closed properly, or calling back to their released delegates. Now, I've tried working through this and carefully setting all delegates to nil and calling close on all open network connections before releasing a menu. However, I'm still getting the error. Also – short posting my entire application code into this post, I can't really post a specific code snippet that illustrates this situation. I wonder if anyone has ideas for tasks that I may be missing? Do I need to do anything to close a network connection other than closing it and setting it to nil? Also, I'm new to debugging tools – can anyone suggest a good tool to use to watch network connections and see what's causing them to fail? Thanks!

    Read the article

  • asp.net mvc getting id of button clicked

    - by mazhar kaunain baig
    <div id="4591" > <input type="text" id="Title1" name="Title1" value="a" /> <input type="submit" name="button" value="Save" /> </div> <div id="4592" > <input type="text" id="Title2" name="Title2" value="a" /> <input type="submit" name="button" value="Save" /> </div> <div id="4593" > <input type="text" id="Title3" name="Title3" value="a" /> <input type="submit" name="button" value="Save" /> </div> This is the copy paste version of the html source generated by the browser which is making it clear that i am generating the dynamic fields on the page. name in the textbox is the field in the database. After pressing the one of the save buttons how would i send the particular textbox name and value to the controller action to be updated.

    Read the article

  • How to model and handle presentation DTO's to abstract from complicated domain model?

    - by arrages
    Hi I am developing an application that needs to work with a complex domain model using Hibernate. This application uses Spring MVC and using the domain objects in the presentation layer is very messy so I think I should use DTO's that go to and from my service layer so that these match what I need in my views. Now lets assume I have a CarLease entity whose properties are not simple java primitives but it's composed with other entities like Make, Model, etc public class CarLease { private Make make; Private Model model; . . . } most properties are in this fashion and they are selectable using drop down selects on the jsp view, each will post back an ID to the controller. Now considering some standard use cases: create, edit, display How would you go about modeling the presentation DTO's to be used as form backing objects and communication between presentation and service layers?? Would you create a different DTO for each case (create, edit, display), would you make DTO's for the complex attributes? if so where would you translate the ID to entity? how and where would you handle validation, DTO/Domain assembly, what would you return from service layer methods? (create, edit, get) As you can see, I now I will benefit by separating my view from the domain objects (very complex with lots of stuff I don't need.) but I am having a hard time finding any real world examples and best practices for this. I need some architecture guidance from top to bottom, please keep in mind I will use Spring MVC in case that may leverage on your anwser. thanks in advance.

    Read the article

  • Get Form Input via Ajax

    - by user3651491
    I have a jqgrid plugin which I call via Ajax. I have index.php and a getGridData.php. How will I pass form input in getGridData.php via ajax and use it in getGridData.php? I tried serialize but I can't pass or access it on getGridData.php. I need it as parameters for mysql. Here's my code. <script language="javascript" type="text/javascript"> function jgGrid() { $(document).ready(function () { $("#grid").jqGrid({ url: "inc/Controller/getGridData.php"+$("#thisForm").serialize(), data : formData, datatype: "json", mtype: "POST", colNames: ["SiteID", "TerminalID", "TransactionType", "Amount", "ServiceStatus"], colModel: [ { name: "SiteID"}, { name: "TerminalID"}, { name: "TransactionType"}, { name: "Amount"}, { name: "ServiceStatus"}, ], pager: "#pager", rowNum: 10, rowList: [10,20], sortname: "SiteID", sortorder: "asc", height: 'auto', viewrecords: true, gridview: true, caption: "" }); }); } </script> getGridData.php include('../Model/Queries.php'); $cardnumber = $_POST['cardnumber']; $transact_type = $_POST['transact_type']; $fromdate = $_POST['fromdate']; $todate = $_POST['todate']; $loyalty = new Queries(); $get_mid = $loyalty->loyaltyConn($cardnumber); $somedata = json_encode($loyalty->nposConn($get_mid, $transact_type, $fromdate, $todate)); echo $somedata;

    Read the article

  • CI + Joomla 1.5

    - by DMin
    Hi, This is something that I just cooked up with Joomla and CodeIgniter(CI). I Wrote my database intensive application in CodeIgniter and frontend is Joomla. I'm using Jumi(Joomla Extention) so I can include the CI files inside joomla to basically insert the content generated by CI into Joomla articles. Problem is, you can't include CI files directly using JUMI from joomla because CI tends to route the pages so instead of seeing your joomla page with the CI content, you be redirected to the CI page itself. I did a little work around for this : Made an additional page that just basically does cURL to the CI page - gets the data and echos it out. From jumi, I include this cURL page instead. Couple of questions: I've seen at least a few posts that CI + Joomla is difficult to do(link). 1) Do you see any glaring security issues or possible performance issues? 2) Do you know of a better way to implement this? 3) What do you think of this? Do you think this is a good way to do this? There is one component out there that plugs CI with Joomla but it requires you to have a fresh CI install. It allows only one controller & the download link is down as well.

    Read the article

  • where to store temporary data in MVC 2.0 project

    - by StuffHappens
    Hello! I'm starting to learn MVC 2.0 and I'm trying to create a site with a quiz: user is asked a question and given several options of answer. If he chooses the right answer he gets some points, if he doesn't, he looses them. I tried to do this the following way public class HomeController : Controller { private ITaskGenerator taskGenerator = new TaskGenerator(); private string correctAnswer; public ActionResult Index() { var task = taskGenerator .GenerateTask(); ViewData["Task"] = task.Task; ViewData["Options"] = task.Options; correctAnswer= task.CorrectAnswer; return View(); } public ActionResult Answer(string id) { if (id == correctAnswer) return View("Correct") return View("Incorrect"); } } But I have a problem: when user answers the cotroller class is recreated and I loose correct answer. So what is the best place to store correct answer? Should I create a static class for this purpose? Thanks for your help!

    Read the article

  • Error after redirection using CakePHP

    - by Praveen kalal
    I have created some code called LoginController. Whenever Admin gets successfully logged in I redirect the page to index. However, I got an error like "problem on loading page". This is my code: <?php class LoginController extends AdminAppController { var $name = 'Login'; var $uses = array('Admin.Login'); var $sessionkey= ''; /*function beforeFilter() { if($this->Session->read('user')=='Admin' || $this->params['action']=='login') { echo "in"; exit; } else { echo "else"; exit; $this->Session->setFlash('Login first','flash_failure'); $this->redirect(array('action'=>'login')); } }*/ function index() { } function login() { //pr($this->data); exit; if(!empty($this->data)) { $results = $this->Login->findByEmail($this->data['Login']['email']); if(!empty($results) && $results['Login']['password']== md5($this->data['Login']['password'])) { $this->Session->write('user', 'Admin'); $results['Login']['last_login']=date("Y-m-d H:i:s"); $this->Login->save($results); $this->Session->setFlash('Login successfully.', 'flash_success'); $this->redirect(array('controller'=>'login','action' => 'index')); } } } } ?> Can anyone help me? Thanks.

    Read the article

  • cakephp - form for belongsTo Model

    - by user1511579
    I createt the following model to link 2 relational tables: class Ficha extends AppModel { //public $useTable = 'ficha_seg'; var $primaryKey = 'id_ficha'; var $name = 'Ficha'; var $belongsTo = array( 'Perigo' => array( 'className' => 'Perigo', 'foreignKey' => false, 'conditions' => 'Perigo.id_fichas = Ficha.id_ficha' ) ); } Now, i have a form that requires data from the class Ficha, and then is redirected to another ctp page where i will input the data for the table "Perigos". However, since i'm still a newbie in cakephp i'm having difficult building that second form to insert the data on the table "Perigos". Here goes the code i built at the moment related to the second form: FichasController.php (the method where is it supposed to save the data on the table "Perigos": public function preencher_ficha(){ if ($this->request->is('ficha')) { $this->Ficha->create(); if ($this->Ficha->Perigo->save($this->request->data)) { $last_id=$this->Ficha->getLastInsertID(); $this->Session->setFlash('Your post has been updated '.$last_id.'.'); //$this->redirect(array('action' => 'preencher_ficha')); } else { $this->Session->setFlash('Unable to qualquer coisa your post.'); } } } The preencher_ficha.ctp file with the form: echo $this->Form->create('Ficha->Perigo', array('action' => 'index')); echo $this->Form->input('class_subst', array('label' => 'Classificação:')); echo $this->Form->input('simbolos_perigo', array('label' => 'Símbolos:')); echo $this->Form->input('frases_r', array('label' => 'Frases:')); echo $this->Form->end('Finalizar Ficha'); Here i guess the create part is wrong, but i think i have errors too in the controller part.

    Read the article

  • Adding UIView on top of all views dynamically during runtime?

    - by Easwaramoorthy Kanagaraj
    Team, I am trying to bring a menu in top of all visible views during runtime. This menu should be easily addable and removable dynamically in certain conditions. To do this, I have tried adding a button view to the UIWindow as a subview during runtime. UIButton *button = [UIButton buttonWithType:UIButtonTypeRoundedRect]; [button addTarget:self action:nil forControlEvents:UIControlEventTouchDown]; [button setTitle:@"Show View" forState:UIControlStateNormal]; button.frame = CGRectMake(80.0, 210.0, 160.0, 40.0); [window addSubview:button]; [window makeKeyAndVisible]; [window bringSubviewToFront:button]; But it doesnt worked. Also I have tried to place this button in the root view controller, but no luck again. Edit - Note: This code is not from a UIViewController. I am trying to build a library this will be in that library code. Use case be like you could post NSNotification to enable and disable this menu dynamically during runtime. Please suggest. Thanks !

    Read the article

  • Dropdownlist value not being set when there is a default blank option

    - by uriDium
    I am using ASP.Net MVC. I have a partial view which has a form with dropdownlists. The dropdownlists are set via ViewData. The partial view is used in a Create and Edit page. The create works fine. I get the dropdownlists and the blank option is a "Please select", like so <%= Html.DropDownList("ContactNrType", ViewData["ContactNrType"] as SelectList, "Please Select") %> But this doesn't seem to work for my edit. If I have that extra "Please select" parameter then it does not select the value for the drop down. I am setting the value of the drop down in the controller like so ViewData["ContactNrType"] = new SelectList(new List<string> { "Mobile", "Home", "Work", "Friend" }, candidate.ContactNrType); Any idea as to what I am doing wrong? I want to share the partial view which contains the form between the two pages. So I need the "Please Select" option for the Create. And I need the value set for the Edit (I don't mind that it has an option that still says "Please Select").

    Read the article

  • Frame sizing of tableview within nested child controllers/subviews

    - by jwoww
    I'm a bit confused by the proper frame sizing of a table view to fit within my screen. Here's my setup of view controllers within view controllers: UITabBarController UINavigationController as one of the tab bar viewcontrollers; title bar hidden ViewController - a container view controller because I need the option to place some controls beneath the UITableView, sometimes (but not in the current scenario) UITableViewController Now, my question is what the proper frame dimensions of the UITableview should be. Here's what I've got in the ViewController viewDidLoad method. I used subtracted 49.0 (the size of the tab bar) from 480.0. However, this leaves a black bar at the bottom. 20.0 appears to do it (coincidentally?) the size of the status bar, but I don't understand why that would be. Wouldn't the true pixel dimensions of the tableview be 480-49? // MessageTableViewController is my subclass of UITableViewController MessagesTableViewController *vcMessagesTable = [[MessagesTableViewController alloc] init]; CGRect tableViewFrame = CGRectMake(0, 0, 320.0, 480.0 - 49.0); [[vcMessagesTable view] setFrame:tableViewFrame]; self.tableViewController = vcMessagesTable; [self addChildViewController:vcMessagesTable]; [[self view] addSubview:vcMessagesTable.view]; Here's how it looks:

    Read the article

  • Cakephp 1.3, Weird behavior on firefox when using $this->Html->link ...

    - by ion
    Greetings, I am getting a very weird and unpredictable result in firefox when using the following syntax: $this->Html->link($this->Html->div('p-cpt',$project['Project']['name']) . $this->Html->div('p-img',$this->Html->image('/img/projects/'.$project['Project']['slug'].'/project.thumb.jpg', array('alt'=>$project['Project']['name'],'width'=>100,'height'=>380))),array('controller' => 'projects', 'action' => 'view', $project['Project']['slug']),array('title' => $project['Project']['name'], 'escape' => false),false); OK I know it is big but bear with me. The point is to get the following output: <a href="x" title="x"> <div class="p-ctp">Name</div> <div class="p-img"><img src="z width="y" height="a" alt="d" /></div> </a> I'm not sure if this validates correctly both on cakephp and html but it works everywhere else apart from firefox. You can actually see the result here: http://www.gnomonconstructions.com/projects/browser To reproduce the result use the form with different categories and press search. At some point it will happen!! Although most of the time it renders the way it should, sometimes it produces an invalid output like that: <a href="x" title="x"></a> <div class="p-cpt"> <a href="x" title="x">name</a> </div> <div class="p-img"> <a href="x" title="x"><img src="x" width="x" height="x" alt="x" /></a> </div> Looks like it repeats the link inside each element. To be honest the only reason I used this syntax was because cakephp encourages it. Any help will be much appreciated :)

    Read the article

  • C#: Determine Type for (De-)Serialization

    - by dbemerlin
    Hi, i have a little problem implementing some serialization/deserialization logic. I have several classes that each take a different type of Request object, all implementing a common interface and inheriting from a default implementation: This is how i think it should be: Requests interface IRequest { public String Action {get;set;} } class DefaultRequest : IRequest { public String Action {get;set;} } class LoginRequest : DefaultRequest { public String User {get;set;} public String Pass {get;set;} } Handlers interface IHandler<T> { public Type GetRequestType(); public IResponse HandleRequest(IModel model, T request); } class DefaultHandler<T> : IHandler<T> // Used as fallback if the handler cannot be determined { public Type GetRequestType() { return /* ....... how to get the Type of T? ((new T()).GetType()) ? .......... */ } public IResponse HandleRequest(IModel model, T request) { /* ... */ } } class LoginHandler : DefaultHandler<LoginRequest> { public IResponse HandleRequest(IModel mode, LoginRequest request) { } } Calling class Controller { public ProcessRequest(String action, String serializedRequest) { IHandler handler = GetHandlerForAction(action); IRequest request = serializer.Deserialize<handler.GetRequestType()>(serializedRequest); handler(this.Model, request); } } Is what i think of even possible? My current Solution is that each handler gets the serialized String and is itself responsible for deserialization. This is not a good solution as it contains duplicate code, the beginning of each HandleRequest method looks the same (FooRequest request = Deserialize(serializedRequest); + try/catch and other Error Handling on failed deserialization). Embedding type information into the serialized Data is not possible and not intended. Thanks for any Hints.

    Read the article

  • ViewController Navigating

    - by Kobe.o4
    I have 4 ViewControllers, startViewController as the initial View Controller. This contains my intro. After its finish, it will [self presentViewController:vc animated:YES completion:NULL]; into my menuViewController. startViewController ------> menuViewController ------> C1ViewController \ \ ------> ImportantViewController In my menuViewController are buttons for the two ViewController like the above illustration. Also I presented them in the View with this: [self presentViewController:vc animated:YES completion:NULL]; I return tomenuViewController with this [self dismissModalViewControllerAnimated:YES]; What I wanted is to make the ImportantViewController to be the like the parent view or the mainVIew even if I go to other Views. What I need is when ImportantViewController is presented when I go to either C1ViewController or menuViewController it wont be deallocated, or its content there will still be retained. Is it possible? And How? I don't know much about what parent and child view controllers for so I dont know what to implement in my problem. Thank you. BTW, Im using Storyboard.

    Read the article

  • Remove in between folder structure from the url in a php website

    - by pabz
    I have a php website having following folder structure (basic structure). project_name app controller model view css js img index.php So when I view index.php in WAMP the url is http://localhost/project_name/ But when I go inside the site (eg. login.php which resides under view folder) url is like this. http://localhost/project_name/app/view/login.php I found that using .htaccess we can change the urls. So I tried this (in .htaccess). RewriteEngine on RewriteBase / Redirect 301 /project_name/app/view/login.php /project_name/login.php RewriteRule ^/project_name/login.php$ /project_name/app/view/login.php [L] Now url is http://localhost/project_name/login.php It is correct. But it seems php does not use the original link to grab the file (ie. from /project_name/app/view/login.php) but from here /project_name/login.php So it throws 404 error. What should I change? Please help me, i am just trying to hide /app/view/ part from the url so that user won't see my folder structure. I have read about various ways of doing that for about 9hrs today but still couldn't get anything working correctly. Hope my question is clear enough. Any help is greatly appreciated!

    Read the article

  • understanding the ORM models in MVC

    - by fayer
    i cant fully understand the ORM models in MVC. so i am using symfony with doctrine. the doctrine models are created. does this mean that i don't have to create any models? are the doctrine models the only models i need? where should i put the code that uses the doctrine models: eg. $phoneIds = array(); $phone1 = new Phonenumber(); $phone1['phonenumber'] = '555 202 7890'; $phone1->save(); $phoneIds[] = $phone1['id']; $phone2 = new Phonenumber(); $phone2['phonenumber'] = '555 100 7890'; $phone2->save(); $phoneIds[] = $phone2['id']; $user = new User(); $user['username'] = 'jwage'; $user['password'] = 'changeme'; $user->save(); $user->link('Phonenumbers', $phoneIds); should this code be in the controller or in another model? and where should i validate these fields (check if it exists in database, that email is email etc)? could someone please shed a light on this. thanks.

    Read the article

  • paginate the class

    - by small
    please help me with pagination of this method my method is this one controller # def show @topic = Topic.find(params[:id]) @posts = @topic.posts.find(:all ,:order=> 'id') end views # %div{:style=>"margin: 0 auto;"} %table.sortable.style2{:cellpadding=>5} %thead %tr %td{:width => "25%",:align => "center",:style => "font-weight: bold;"}Posted By %td{:width => "75%",:style => "font-weight: bold;"}Comments %tbody - for post in @posts %tr %td{:align => "center"} &nbsp %div{:width=>"5px" , :style=>"border: 1px solid rgb(232, 232, 232);background-color: rgb(248, 248, 248);width: 60px; height:60px;" } - if post.posted_by.image = image_tag(post.posted_by.image.public_filename(),:width => "60px", :height => "60px",:align=>"center") %div{:style => "font-weight: bold; font-style: italic;"} = post.posted_by ? post.posted_by.firstname : "<em>Unknown User</em>" %br %div{:style => "font-style: italic;"} Posted on = post.created_at.strftime('%d of %B %Y ') = post.created_at.strftime('at %H:%M') %td{:valign => "top"} =post.body

    Read the article

  • NoMethodError / undefined method `foobar_path' when using form_for

    - by user1850886
    I'm using form_for to create a chatroom and when I view the page I get the following error: NoMethodError in Chatrooms#new undefined method `chatrooms_path' for #<#<Class:0xa862b94>:0xa5307f0> Here's the code for the view, located in app/views/chatrooms/new.html.erb: <div class="center"> <%= form_for(@chatroom) do |f| %> <%=f.text_field :topic%> <br> <%=f.submit "Start a discussion", class: "btn btn-large btn-primary"%> <% end %> </div> Here's the relevant controller: class ChatroomsController < ApplicationController def new @chatroom = Chatroom.new end def show @chatroom = Chatroom.find(params[:id]) end end If I change the line <%= form_for(@chatroom) do |f| %> to <%= form_for(:chatroom) do |f| %> it works fine. I've searched around for similar questions but none of the solutions have worked for me. Help?

    Read the article

  • Jquery conditionals, window locations, and viewdata. Oh my!

    - by John Stuart
    I have one last thing left on a project and its a doozy. Not only is this my first web application, but its the first app i used Jquery, CSS and MVC. I have no idea on how to proceed with this. What i am trying to do is: In my controller, a waste item is validated, and based on the results one of these things can happen. The validation is completed, nothing bad happens, which sets ViewData["FailedWasteId"] to -9999. Its a new waste item and the validation did not pass, which sets ViewData["FailedWasteId"] to 0. Its an existing waste item and the validation did not pass, which sets ViewData["FailedWasteId"] to the id of the waste item. This ViewData["FailedWasteId"] is set on page load using <%=Html.Hidden("wFailId", int.Parse(ViewData["WasteFailID"].ToString()))%> When the validations do not pass, then the page zooms (by window.location) to an invisible div, opens the invisible div etc. Hopefully my intentions are clear with this poor attempt at jquery. The new waste div is and the existing item divs are dynamically generated (this i know works) " So my question here is... Help? I cant even get the data to parse correctly, nor can i even get the conditionals to work. And since this happens after post, i cant get firebug to help my step through the debugger, as the script isnt loaded yet. $(document).ready(function () { var wasteId = parseInt($('#wFailId').text()); if (wasteId == -9999) { //No Issue } else if (wasteId < 0) { //Waste not saved to database } else if (wasteId == 0) { //New Waste window.location = '#0'; $('.editPanel').hide(); $('#GeneratedWasteGrid:first').before(newRow); $('.editPanel').appendTo('#edit-panel-row').slideDown('slow'); } else if (wasteId > 0) { //Waste saved to database } });

    Read the article

< Previous Page | 537 538 539 540 541 542 543 544 545 546 547 548  | Next Page >