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  • How and where to store user uploaded files in high traffic web farm scenario website?

    - by Inam Jameel
    i am working on a website which deploy on web farms to serve high traffic. where should i store user uploaded files? is it wise to store uploaded files in the file system of the same website and synchronize these files in all web servers(web farm)? or should i use another server to store all uploaded files in this server to store files in a central location? if separate file server will be a better choice, than how can i pass files from web server to that file server efficiently? or should i upload files directly to that file server?

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  • How much user data should be required to grant a password reset?

    - by Andrew Heath
    I'm looking to add password-reset functionality to my site and have been browsing the numerous threads discussing various aspects of that issue here on SO. One thing I haven't really seen clarified is how much information to require from the user for confirmation before sending out the reset email. is email alone enough? email + account username? email + account username + some other identifying value all accounts must input? I don't want my site to seem like an old wrinkly nun with a ruler, but I don't want people to be able to abuse the password reset system willy-nilly. Suggestions?

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  • PHP / MYSQL: Sanitizing user input - is this a bad idea?

    - by Greg
    I have one "go" script that fetches any other script requested and this is what I wrote to sanitize user input: foreach ($_REQUEST as $key => $value){ if (get_magic_quotes_gpc()) $_REQUEST[$key] = mysql_real_escape_string(stripslashes($value)); else $_REQUEST[$key] = mysql_real_escape_string($value); } I haven't seen anyone else use this approach. Is there any reason not to? EDIT - amended for to work for arrays: function mysql_escape($thing) { if (is_array($thing)) { $escaped = array(); foreach ($thing as $key => $value) { $escaped[$key] = mysql_escape($value); } return $escaped; } // else if (get_magic_quotes_gpc()) $thing = stripslashes($thing); return mysql_real_escape_string($thing); } foreach ($_REQUEST as $key => $value){ $_REQUEST[$key] = mysql_escape($value); }

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  • How can I tag a user in a photo using the Facebook Graph API?

    - by fr6
    I tried: $args = array( 'access_token' => $access_token, 'id' => $uid ); $url = "https://graph.facebook.com/{$idPhoto}/tags"; $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_HEADER, false); curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); curl_setopt($ch, CURLOPT_POST, true); curl_setopt($ch, CURLOPT_POSTFIELDS, $args); curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false); curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 2); $data = curl_exec($ch); It has returned me: {"error":{"type":"QueryParseException","message":"Unknown path components: \/tags"}} It does not seem possible because its not in the Facebook documentation: http://developers.facebook.com/docs/api#publishing Can someone confirm me that it's not possible to tag a user in a recently uploaded photo?

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  • How to close the Facebook pop-up login window after the user Connects?

    - by alex
    Below is my code. For some reason, after the user logs into the little pop-up window, the little window will redirect back to '/" with a lot of session JSON junk at the end of the URL. How do I make it so that the little window closes, and my parent window refreshes? <script src="http://static.ak.connect.facebook.com/js/api_lib/v0.4/FeatureLoader.js.php/en_US" type="text/javascript"></script> <fb:login-button v="2" onlogin='window.location("/test");' size="medium">Connect</fb:login-button> <script type="text/javascript">FB.init("XXXXX",'/xd_receiver.htm');</script>

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  • Killing Mysql prcoesses staying in sleep command.

    - by Shino88
    Hey I am connecting a MYSQL database through hibernate and i seem to have processes that are not being killed after they are finished in the session. I have called flush and close on each session but when i check the server the last processes are still there with a sleep command. This is a new problem which i am having and was not the case yesterday. Is there any way i can ensure the killng of theses processes when i am done with a session. Below is an example of one of my classes. public JSONObject check() { //creates a new session needed to add elements to a database Session session = null; //holds the result of the check in the database JSONObject check = new JSONObject(); try{ //creates a new session needed to add elements to a database SessionFactory sessionFactory = new Configuration().configure().buildSessionFactory(); session = sessionFactory.openSession(); if (justusername){ //query created to select a username from user table String hquery = "Select username from User user Where username = ? "; //query created Query query = session.createQuery(hquery); //sets the username of the query the values JSONObject contents query.setString(0, username); // executes query and adds username string variable String user = (String) query.uniqueResult(); //checks to see if result is found (null if not found) if (user == null) { //adds false to Jobject if not found check.put("indatabase", "false"); } else { check.put("indatabase", "true"); } //adds check to Jobject to say just to check username check.put("justusername", true); } else { //query created to select a username and password from user table String hquery = "Select username from User user Where username = :user and password = :pass "; Query query = session.createQuery(hquery); query.setString("user", username); query.setString("pass", password); String user = (String) query.uniqueResult(); if(user ==null) { check.put("indatabase", false); } else { check.put("indatabase", true); } check.put("justusername", false); } }catch(Exception e){ System.out.println(e.getMessage()); //logg.log(Level.WARNING, " Exception", e.getMessage()); }finally{ // Actual contact insertion will happen at this step session.flush(); session.close(); } //returns Jobject return check; }

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  • What is the best euphemism for a non-developer?

    - by Edward Tanguay
    I'm writing a description for a piece of software that targets the user who is "not technically minded", i.e. a person who uses "browser/office/email" and has a low tolerance for anything technical, he just "wants it to work" without being involved in any of the technical details. What is the best non-disparaging term you have seen to describe this kind of user? non-technical user low-tech user office user normal user technically challenged user non-developer computer joe Surely there is some official, politically-correct retronym for this kind of user that the press and software marketing use.

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  • Why can't my main class see the array in my calender class

    - by Rocky Celltick Eadie
    This is a homework problem. I'm already 5 days late and can't figure out what I'm doing wrong.. this is my 1st semester in Java and my first post on this site Here is the assignment.. Create a class called Calendar. The class should contain a variable called events that is a String array. The array should be created to hold 5 elements. Use a constant value to specify the array size. Do not hard code the array size. Initialize the array in the class constructor so that each element contains the string “ – No event planned – “. The class should contain a method called CreateEvent. This method should accept a String argument that contains a one-word user event and an integer argument that represents the day of the week. Monday should be represented by the number 1 and Friday should be represented by the number 5. Populate the events array with the event info passed into the method. Although the user will input one-word events, each event string should prepend the following string to each event: event_dayAppoinment: (where event_day is the day of the week) For example, if the user enters 1 and “doctor” , the first array element should read: Monday Appointment: doctor If the user enters 2 and “PTA” , the second array element should read: Tuesday Appointment: PTA Write a driver program (in a separate class) that creates and calls your Calendar class. Then use a loop to gather user input. Ask for the day (as an integer) and then ask for the event (as a one word string). Pass the integer and string to the Calendar object’s CreateEvent method. The user should be able enter 0 – 5 events. If the user enters -1, the loop should exit and your application should print out all the events in a tabular format. Your program should not allow the user to enter invalid values for the day of the week. Any input other than 1 – 5 or -1 for the day of the week would be considered invalid. Notes: When obtaining an integer from the user, you will need to use the nextInt() method on your Scanner object. When obtaining a string from a user, you will need to use the next() method on your Scanner object. Here is my code so far.. //DRIVER CLASS /** * * @author Rocky */ //imports scanner import java.util.Scanner; //begin class driver public class driver { /** * @paramargs the command line arguments */ //begin main method public static void main(String[] args) { //initiates scanner Scanner userInput = new Scanner (System.in); //declare variables int dayOfWeek; String userEvent; //creates object for calender class calendercalenderObject = new calender(); //user prompt System.out.println("Enter day of week for your event in the following format:"); System.out.println("Enter 1 for Monday"); System.out.println("Enter 2 for Tuesday"); System.out.println("Enter 3 for Wednsday"); System.out.println("Enter 4 for Thursday"); System.out.println("Enter 5 for Friday"); System.out.println("Enter -1 to quit"); //collect user input dayOfWeek = userInput.nextInt(); //user prompt System.out.println("Please type in the name of your event"); //collect user input userEvent = userInput.next(); //begin while loop while (dayOfWeek != -1) { //test for valid day of week if ((dayOfWeek>=1) && (dayOfWeek<=5)){ //calls createEvent method in calender class and passes 2 variables calenderObject.createEvent(userEvent,dayOfWeek); } else { //error message System.out.println("You have entered an invalid number"); //user prompts System.out.println("Press -1 to quit or enter another day"); System.out.println("Enter 1 for Monday"); System.out.println("Enter 2 for Tuesday"); System.out.println("Enter 3 for Wednsday"); System.out.println("Enter 4 for Thursday"); System.out.println("Enter 5 for Friday"); System.out.println("Enter -1 to quit"); //collect user input dayOfWeek = userInput.nextInt(); //end data validity test } //end while loop } //prints array to screen int i=0; for (i=0;i<events.length;i++){ System.out.println(events[i]); } //end main method } } /** * * @author Rocky */ //imports scanner import java.util.Scanner; //begin calender class public class calender { //creates events array String[] events = new String[5]; //begin calender class constructor public calender() { //Initializes array String[] events = {"-No event planned-","-No event planned-","-No event planned-","-No event planned-","-No event planned-"}; //end calender class constructor } //begin createEvent method public String[] createEvent (String userEvent, int dayOfWeek){ //Start switch test switch (dayOfWeek){ case 1: events[0] = ("Monday Appoinment:") + userEvent; break; case 2: events[1] = ("Tuesday Appoinment:") + userEvent; break; case 3: events[2] = ("WednsdayAppoinment:") + userEvent; break; case 4: events[3] = ("Thursday Appoinment:") + userEvent; break; case 5: events[4] = ("Friday Appoinment:") + userEvent; break; default: break; //End switch test } //returns events array return events; //end create event method } //end calender class }

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  • How to make script paths inside user control independent of page hierarchy.

    - by Rohit
    My website is organized in the following folder structure Root Pages CustomerManagement DepartMentManagement Script UserControls Now i have a user control inside userControl Folder. I have to use a external js file. So i use it this way <script type="text/javascript" language="javascript" src="../Script/slider.js"></script> My understanding is that since both are under root so i have used a single .. Now what happens is that when this usercontrol is used inside DepartMentManagement it checks for script folder inside CustomerManagement as .. refers to one hierarchy above and script file is not found.Even using a ~ doesnot work. I want to make this script path independent of the path where this control is used. I don't want to move script reference code to the page as control requires script mot page.Please tell me how to achieve this.

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  • Can we ask user for credit card number or paypal details,store credit card number on our server, can

    - by Hiren Gujarati
    Can we get credit card number from user or paypal details & use them for premium service of our application ? is apple accept this application if we directly get this information & use it in our api on server for transaction. Using ssl will be accpted by apple ? I have check from 1) http://stackoverflow.com/questions/1763306/credit-card-purchase-of-physical-goods-via-an-iphone-application 2) http://stackoverflow.com/questions/1707701/receiving-payments-trough-paypal-and-credit-card 3) http://stackoverflow.com/questions/1366864/using-the-paypal-api-in-an-iphone-application but not clear about all...

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  • How to make an ISO copy of Linux-filesystem and user files of VPS Debian based?

    - by moogeek
    Hello! I have a Debian-Based VPS on some hosting. I want to migrate from it and i need to make a full copy of all Linux-filesystem (and installed packages) + all home directory with website files. And then pack/convert it to ISO image so that to use it on cloud hostings like Amazon. The problem is that i have only ssh root access. Hosting support can't do that for me. Another part of the question - is it possible to enlarge the Linux-filesystem by not re-installing it and using the free space of home directory? Is it possible to do? I guess it is possible with rsync or something like that. Will my Mysql databes copy together with all other data? Thanks in advance!

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  • How to make an ISO copy of Linux-filesystem and user files of VPS Debian based?

    - by moogeek
    Hello! I have a Debian-Based VPS on some hosting. I want to migrate from it and i need to make a full copy of all Linux-filesystem (and installed packages) + all home directory with website files. And then pack/convert it to ISO image so that to use it on cloud hostings like Amazon. The problem is that i have only ssh root access. Hosting support can't do that for me. Another part of the question - is it possible to enlarge the Linux-filesystem by not re-installing it and using the free space of home directory? Is it possible to do? I guess it is possible with rsync or something like that. Will my Mysql databes copy together with all other data? Thanks in advance!

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  • How to change page author in wordpress?

    - by GaVrA
    Ofc i know how to do that, but the thing is that i am using "User Role Editor" and i have one user group that can read and edit published pages. Now, i will be adding all the pages on that site, and we will have several users that will need to have only one page they can edit, so i need for that page to change "Page author" to that user. In case you didnt know, when user have "Edit published pages" enabled they can edit only pages where they are listed as author. Problem is i can only do that by going in phpmyadmin and changing the page_author field to the id of that user because that user group, like i said, can only read and edit published pages. That is why i can not change page author from "Edit page" page to user from that user group. So my question is: does anyone know any solution to this problem which does not involve me going to phpmyadmin and changing the id for page_author there?

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  • Rails nested models and data separation by scope

    - by jobrahms
    I have Teacher, Student, and Parent models that all belong to User. This is so that a Teacher can create Students and Parents that can or cannot log into the app depending on the teacher's preference. Student and Parent both accept nested attributes for User so a Student and User object can be created in the same form. All four models also belong to Studio so I can do data separation by scope. The current studio is set in application_controller.rb by looking up the current subdomain. In my students controller (all of my controllers, actually) I'm using @studio.students.new instead of Student.new, etc, to scope the new student to the correct studio, and therefore the correct subdomain. However, the nested User does not pick up the studio from its parent - it gets set to nil. I was thinking that I could do something like params[:student][:user_attributes][:studio_id] = @student.studio.id in the controller, but that would require doing attr_accessible :studio_id in User, which would be bad. How can I make sure that the nested User picks up the same scope that the Student model gets when it's created? student.rb class Student < ActiveRecord::Base belongs_to :studio belongs_to :user, :dependent => :destroy attr_accessible :user_attributes accepts_nested_attributes_for :user, :reject_if => :all_blank end students_controller.rb def create @student = @studio.students.new @student.attributes = params[:student] if @student.save redirect_to @student, :notice => "Successfully created student." else render :action => 'new' end end user.rb class User < ActiveRecord::Base belongs_to :studio accepts_nested_attributes_for :studio attr_accessible :email, :password, :password_confirmation, :remember_me, :studio_attributes devise :invitable, :database_authenticatable, :recoverable, :rememberable, :trackable end

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  • Horizontal and vertical center text in html

    - by Christophe Herreman
    I have a div with a background image that needs to be centered horizontally and vertically. On top of that image, I also want to display a 1-line text, also centered horizontally and vertically. I managed to get the image centered, but the text is not centered vertically. I thought vertical-align:middle would do the trick. Here's the code I have: <div style="background: url('background.png') no-repeat center; width:100%; height:100%; text-align:center;"> <div style="color:#ffffff; text-align: center; vertical-align:middle;" > Some text here. </div> </div> Any ideas? Workaround: I actually got this to work by using a table. (I'll probably be cursed to hell by the HTML community.) Is there any significant reason not to use this btw? I'm still interested in the solution using divs though. <table width="100%" height="100%"> <tr> <td align="center" style="background: url('background.png') no-repeat center; color:#ffffff;">Some text here.</td> </tr> </table>

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  • How can I dynamically clear all controls in a user control?

    - by Michael
    Is it possible to dynamically (and generically) clear the state of all of a user control's child controls? (e.g., all of its TextBoxes, DropDrownLists, RadioButtons, DataGrids, Repeaters, etc -- basically anything that has ViewState) I'm trying to avoid doing something like this: foreach (Control c in myUserControl.Controls) { if (c is TextBox) { TextBox tb = (TextBox)c; tb.Text = ""; } else if (c is DropDownList) { DropDownList ddl = (DropDownList)c; ddl.SelectedIndex = -1; } else if (c is DataGrid) { DataGrid dg = (DataGrid)c; dg.Controls.Clear(); } // etc. } I'm looking for something like this: foreach (Control c in myUserControl.Controls) c.Clear(); ...but obviously that doesn't exist. Is there any easy way to accomplish this dynamically/generically?

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  • Best way to handle Many-to-Many relationships in PHP MySQL

    - by Jayrox
    I am looking for the best way to handle a database of many-to-many relationships in PHP and MySQL. Right now I have 2 tables: Users (id, user_name, first_name, last_name) Connections (id_1, id_2) In the User table id is auto incremented on add and user_name is unique, but can be changed. Unfortunately, I don't have control over the user_name and its ability to be changed, but I must account for it. The Connections table is obviously, user1 and user2's id. The connection table needs to account for these possible relations: user1 --> user2 (user 1 friends with user 2 but not user2 friends with user1) user2 --> user1 (user 2 friends with user 1 but not user1 friends with user2) user1 <--> user2 (user 1 and user 2 mutually friends) user1 <-!-> user2 (user 1 and user 2 not friends) That part is not the problem, The problem I am having with is keeping these relations unique when and if they change in batches. Possible solution 1: delete all of user 1's relations and readd them with the updated list. I think this might be too slow for my needs. Solution 2? Anyone else encounter this problem? How should I best handle this? update: distinguishing relationships: i handle relationships like this: user1, user2 user1, user3 user2, user1 in that example the following is true: user1 follows user2 and user3 user2 only follows user1 but doesn't follow user3 user3 doesn't follow either user1 or user2

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  • How to subString a block of user generated HTML while preserving formatting?

    - by Chad
    I'd like to create the typical preview paragraph with a [read more] link. Problem is, the content that I'd like to SubString() contains text and html, written by a user with a WYSIWYG editor. Of course, I check to make sure the string is not null or empty, then SubString() it, problem is that I could end up breaking the html tags, throwing off the rendering of the entire site. The WYSIWYG editor doesn't seem to create perfectly formatted HTML, and many times seems to use <br /> tags instead of <p></p>, etc... basically, I can't rely on well-formed tags, etc. My workaround was to just strip out all HTML and substring the leftover text. This works, but loses any of the formatting that was in the HTML. What's the best method of SubString()'ing a block of non-well-formed HTML while maintaining HTML that won't break the rendering of the site?

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  • JSF 2.1 Spring 3.0 Integration

    - by danny.lesnik
    I'm trying to make very simple Spring 3 + JSF2.1 integration according to examples I googled in the web. So here is my code: My HTML submitted to actionController.actionSubmitted() method: <h:form> <h:message for="textPanel" style="color:red;" /> <h:panelGrid columns="3" rows="5" id="textPanel"> //all my bean prperties mapped to HTML code. </h:panelGrid> <h:commandButton value="Submit" action="#{actionController.actionSubmitted}" /> </h:form> now the Action Controller itself: @ManagedBean(name="actionController") @SessionScoped public class ActionController implements Serializable{ @ManagedProperty(value="#{user}") User user; @ManagedProperty(value="#{mailService}") MailService mailService; public void setMailService(MailService mailService) { this.mailService = mailService; } public void setUser(User user) { this.user = user; } private static final long serialVersionUID = 1L; public ActionController() {} public String actionSubmitted(){ System.out.println(user.getEmail()); mailService.sendUserMail(user); return "success"; } } Now my bean Spring: public interface MailService { void sendUserMail(User user); } public class MailServiceImpl implements MailService{ @Override public void sendUserMail(User user) { System.out.println("Mail to "+user.getEmail()+" sent." ); } } This is my web.xml <listener> <listener-class> org.springframework.web.context.ContextLoaderListener </listener-class> </listener> <listener> <listener-class> org.springframework.web.context.request.RequestContextListener </listener-class> </listener> <!-- Welcome page --> <welcome-file-list> <welcome-file>index.xhtml</welcome-file> </welcome-file-list> <!-- JSF mapping --> <servlet> <servlet-name>Faces Servlet</servlet-name> <servlet-class>javax.faces.webapp.FacesServlet</servlet-class> <load-on-startup>1</load-on-startup> </servlet> my applicationContext.xml <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd"> <bean id="mailService" class="com.vanilla.jsf.services.MailServiceImpl"> </bean> </beans> my faces-config.xml is the following: <application> <el-resolver> org.springframework.web.jsf.el.SpringBeanFacesELResolver </el-resolver> <message-bundle> com.vanilla.jsf.validators.MyMessages </message-bundle> </application> <managed-bean> <managed-bean-name>actionController</managed-bean-name> <managed-bean-class>com.vanilla.jsf.controllers.ActionController</managed-bean-class> <managed-bean-scope>session</managed-bean-scope> <managed-property> <property-name>mailService</property-name> <value>#{mailService}</value> </managed-property> </managed-bean> <navigation-rule> <from-view-id>index.xhtml</from-view-id> <navigation-case> <from-action>#{actionController.actionSubmitted}</from-action> <from-outcome>success</from-outcome> <to-view-id>submitted.xhtml</to-view-id> <redirect /> </navigation-case> </navigation-rule> My Problem is that I'm getting NullPointerExeption because my mailService Spring bean is null. public String actionSubmitted(){ System.out.println(user.getEmail()); //mailService is null Getting NullPointerException mailService.sendUserMail(user); return "success"; }

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  • How do I pass the currently logged in user's credentials to a web service using Integrated Windows A

    - by Chris Smith
    I am having a frustrating time trying to do something with Perl that would take a couple of lines of code in C#, namely to call a web service on a Windows server that requires Integrated Windows Authentication. The most likely candidate I've found for success is a module called LWP::Authen::Ntlm, but all the examples I've googled require you to explicitly supply username, password and domain. I don't want to do that - I just want the request to use the credentials of the currently logged in user, a la CredentialCache.DefaultCredentials in .NET. Have any of you Perl gurus out there ever had to do this? Thanks.

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  • Best way to implement refusing a value change by the user in Swing?

    - by Michael Borgwardt
    I have a JCheckBox that should not be checked by the user when a certain other field is empty. So now I want to have an error popup and then reset the checkbox (I've considered disabling the checkbox, but the connection to the other field is non-obvious, and a tooltip text IMO not visible enough). What's the correct way to do that in Swing? Through a PropertyVetoException? Where do I throw it and where do I catch it? My first (probably ugly) idea would be to add a ChangeListener that itself shows the popup and resets the value.

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  • How can I have MySQL write outfiles as a different user?

    - by David Locke
    I'm working with a MySQL query that writes into an outfile. I run this query once every day or two and so I want to be able to remove the outfile without having to resort to su or sudo. The only way I can think of making that happen is to have the outfile written as owned by someone other than the mysql user. Is this possible? Edit: I am not redirecting output to a file, I am using the INTO OUTFILE part of a select query to output to a file. If it helps: mysql --version mysql Ver 14.12 Distrib 5.0.32, for pc-linux-gnu (x86_64) using readline 5.2

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  • How to encrypt and save a binary stream after serialization and read it back?

    - by Anindya Chatterjee
    I am having some problems in using CryptoStream when I want to encrypt a binary stream after binary serialization and save it to a file. I am getting the following exception System.ArgumentException : Stream was not readable. Can anybody please show me how to encrypt a binary stream and save it to a file and deserialize it back correctly? The code is as follows: class Program { public static void Main(string[] args) { var b = new B {Name = "BB"}; WriteFile<B>(@"C:\test.bin", b, true); var bb = ReadFile<B>(@"C:\test.bin", true); Console.WriteLine(b.Name == bb.Name); Console.ReadLine(); } public static T ReadFile<T>(string file, bool decrypt) { T bObj = default(T); var _binaryFormatter = new BinaryFormatter(); Stream buffer = null; using (var stream = new FileStream(file, FileMode.OpenOrCreate)) { if(decrypt) { const string strEncrypt = "*#4$%^.++q~!cfr0(_!#$@$!&#&#*&@(7cy9rn8r265&$@&*E^184t44tq2cr9o3r6329"; byte[] dv = {0x12, 0x34, 0x56, 0x78, 0x90, 0xAB, 0xCD, 0xEF}; CryptoStream cs; DESCryptoServiceProvider des = null; var byKey = Encoding.UTF8.GetBytes(strEncrypt.Substring(0, 8)); using (des = new DESCryptoServiceProvider()) { cs = new CryptoStream(stream, des.CreateEncryptor(byKey, dv), CryptoStreamMode.Read); } buffer = cs; } else buffer = stream; try { bObj = (T) _binaryFormatter.Deserialize(buffer); } catch(SerializationException ex) { Console.WriteLine(ex.Message); } } return bObj; } public static void WriteFile<T>(string file, T bObj, bool encrypt) { var _binaryFormatter = new BinaryFormatter(); Stream buffer; using (var stream = new FileStream(file, FileMode.Create)) { try { if(encrypt) { const string strEncrypt = "*#4$%^.++q~!cfr0(_!#$@$!&#&#*&@(7cy9rn8r265&$@&*E^184t44tq2cr9o3r6329"; byte[] dv = {0x12, 0x34, 0x56, 0x78, 0x90, 0xAB, 0xCD, 0xEF}; CryptoStream cs; DESCryptoServiceProvider des = null; var byKey = Encoding.UTF8.GetBytes(strEncrypt.Substring(0, 8)); using (des = new DESCryptoServiceProvider()) { cs = new CryptoStream(stream, des.CreateEncryptor(byKey, dv), CryptoStreamMode.Write); buffer = cs; } } else buffer = stream; _binaryFormatter.Serialize(buffer, bObj); buffer.Flush(); } catch(SerializationException ex) { Console.WriteLine(ex.Message); } } } } [Serializable] public class B { public string Name {get; set;} } It throws the serialization exception as follows The input stream is not a valid binary format. The starting contents (in bytes) are: 3F-17-2E-20-80-56-A3-2A-46-63-22-C4-49-56-22-B4-DA ...

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  • Can a user-chosen image be inserted directly into a JEditorPane?

    - by JIM
    What I am trying to do is open up a JFilechooser that filters jpeg,gif and png images, then gets the user's selection and inserts it into the JEditorPane. Can this be done? or am i attempting something impossible? Here is a sample of my program.(insert is a JMenuItem and mainText is a JEditorPane) insert.addActionListener(new ActionListener(){ public void actionPerformed(ActionEvent e){ JFileChooser imageChooser = new JFileChooser(); imageChooser.setFileFilter(new FileNameExtensionFilter("Image Format","jpg","jpeg","gif","png")); int choice = imageChooser.showOpenDialog(mainText); if (choice == JFileChooser.APPROVE_OPTION) { mainText.add(imageChooser.getSelectedFile()); } } }); What i tried to do is use the add method, i know it's wrong but just to give you an idea of what i'm trying to do. Before you complain, i'm sorry about the code formatting, i don't really know all the conventions of what is considered good or bad style. Thank you very much.

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