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  • Nested attributes form for model which belongs_to few models

    - by ExiRe
    I have few models - User, Teacher and TeacherLeader. class User < ActiveRecord::Base attr_accessible ..., :teacher_attributes has_one :teacher has_one :teacher_leader accepts_nested_attributes_for :teacher_leader end class Teacher < ActiveRecord::Base belongs_to :user has_one :teacher_leader end class TeacherLeader < ActiveRecord::Base belongs_to :user belongs_to :teacher end I would like to fill TeacherLeader via nested attributes. So, i do such things in controller: class TeacherLeadersController < ApplicationController ... def new @user = User.new @teacher_leader = @user.build_teacher_leader @teachers_collection = Teacher.all.collect do |t| [ "#{t.teacher_last_name} #{t.teacher_first_name} #{t.teacher_middle_name}", t.id ] end @choosen_teacher = @teachers_collection.first.last unless @teachers_collection.empty? end end And also have such view (new.html.erb): <%= form_for @user, :url => teacher_leaders_url, :html => {:class => "form-horizontal"} do |f| %> <%= field_set_tag do %> <% f.fields_for :teacher_leader do |tl| %> <div class="control-group"> <%= tl.label :teacher_id, "Teacher names", :class => "control-label" %> <div class="controls"> <%= select_tag( :teacher_id, options_for_select( @teachers_collection, @choosen_teacher )) %> </div> </div> <% end %> <div class="control-group"> <%= f.label :user_login, "Login", :class => "control-label" %> <div class="controls"> <%= f.text_field :user_login, :placeholder => @everpresent_field_placeholder %> </div> </div> <div class="control-group"> <%= f.label :password, "Pass", :class => "control-label" %> <div class="controls"> <%= f.text_field :password, :placeholder => @everpresent_field_placeholder %> </div> </div> <% end %> <%= f.submit "Create", :class => "btn btn-large btn-success" %> <% end %> Problem is that select form here does NOT appear. Why? Do i do something wrong?

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  • How to use input type="tel" in a contact form

    - by user1678664
    PHP newbie here trying to use a simple contact form. The fields I have are: "Name", "Email", "Phone", and "Message". I am trying to figure out what to do with the "Phone" field – I want it to output to the same place that "Message" does, so that when the email is received it will output both the message and the phone number to the body of the email. How do I do this? if( isset( $_POST['submit'] ) ) : wp_mail( get_option( 'admin_email' ), 'Website Contact Form', ( isset($_POST['message'] ) ? $_POST['message'] : '(blank)' ), 'From: ' . ( isset( $_POST['from_name'] ) ? $_POST['from_name'] : 'Website Contact Form' ) . ' <' . ( isset( $_POST['from_email'] ) ? $_POST['from_email'] : get_option( 'admin_email' ) ). '>' . "\r\n" ); $output = '<p>Thank you! Your message has been sent.</p>'; else : $output = ' <form action="" method="post"><ul> <li> <label for="from_name">Name</label> <input type="text" name="from_name" id="from_name" /> </li> <li> <label for="from_email">Email</label> <input type="email" name="from_email" id="from_email" /> </li> <li> <label for="from_tel">Phone</label> <input type="tel" name="from_tel" id="from_tel" /> </li> <li> <textarea name="message" id="message"></textarea> </li> <li class="submit"> <input name="submit" type="submit" value="Say Hello!" /> </li> </ul></form>'; endif; return $output;

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  • User authentication in Django. Problems with is_authenticated

    - by tim
    I have one problem with users menu. So, I want, that authenticated user can see his/her profile page and logout (links) in menu. It works (when I logging in) on index page: index, page1, profile, logout ,but, if I go to the, for example, page1 I can see in menu: index, page1, login, not profile and logout. How to fix it? in urls: url(r'^accounts/login/$', 'django.contrib.auth.views.login' ), url(r'^accounts/logout/$', 'django.contrib.auth.views.logout_then_login' ), url(r'^accounts/profile/$', 'my_app.views.profile' ), in views: def profile(request): if not request.user.is_authenticated(): return HttpResponseRedirect("/accounts/login/") else: user = request.user.is_authenticated() return render_to_response('profile.html',locals()) Part of index.html: {% if user.is_authenticated or request.user.is_authenticated %} <li><a href="/accounts/profile/">Profile</a></li> <li><a href="/accounts/logout/">logout</a></li> {% else %} <li><a href="/accounts/login/">login</a></li> {% endif %} login.html: {% extends "index.html" %} {% load url from future %} {% block application %} {% if form.errors %} <p>Try one more time</p> {% endif %} <form method="post" action="{% url 'django.contrib.auth.views.login' %}"> {% csrf_token %} <table> <tr> <td>{{ form.username.label_tag }}</td> <td>{{ form.username }}</td> </tr> <tr> <td>{{ form.password.label_tag }}</td> <td>{{ form.password }}</td> </tr> </table> <input type="submit" value="Login" /> <input type="hidden" name="next" value="{{ next }}" /> </form> {% endblock %} profile.html: {% extends "index.html" %} {% block application %} {% if request.user.is_authenticated %} <p>Welcome, {{ request.user.username }}. Thanks for logging in.</p> {% else %} <p>Welcome, new user. Please log in.</p> {% endif %} {% endblock %}

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  • Sliding panel in the middle of the page. Z-index given not working

    - by Nehal Rupani
    Hi all, I am implementing sliding panel element but problem is when i slide out other div element is floating down. I guess and tried to give z-index to element which i am sliding but it doesn't seems to work. Let me put code for both div. <div class="vrcontrol"> <div class="slide-out-div"> <a class="handle" href="http://link-for-non-js-users.html">Content</a> <h3>Contact me</h3> <p>Thanks for checking out my jQuery plugin, I hope you find this useful. </p> <p>This can be a form to submit feedback, or contact info</p> </div> This is div which i am sliding in and out and beneath is code of effective div. <div class="askform"> <p class="titletext">Ask an Expert Trade Forum</p> <p class="detailtext">WD-40’s leading source for DIY tips and tricks.</p> <span> <form id="askform" name="askform" action="" method="post"> <span class="left"><input name="input" type="text" class="askinputbox"/></span><span class="marginleft"><input type="image" src="images/search_icon.gif" /></span> </form> </span> <div class="followus"> <span class="followtext">Follow us on</span><span class="right"><img src="images/bookmark.jpg" width="121" height="45" alt="Bookmark" /></span> </div> </div> Sliding div is in left portion of the page and effective div is in right portion of the page. I guess something with z-index, positioning element and overflow properties will do something.

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  • Displaying pic for user through a question's answer

    - by bgadoci
    Ok, I am trying to display the profile pic of a user. The application I have set up allows users to create questions and answers (I am calling answers 'sites' in the code) the view in which I am trying to do so is in the /views/questions/show.html.erb file. It might also be of note that I am using the Paperclip gem. Here is the set up: Associations Users class User < ActiveRecord::Base has_many :questions, :dependent => :destroy has_many :sites, :dependent => :destroy has_many :notes, :dependent => :destroy has_many :likes, :through => :sites , :dependent => :destroy has_many :pics, :dependent => :destroy has_many :likes, :dependent => :destroy end Questions class Question < ActiveRecord::Base has_many :sites, :dependent => :destroy has_many :notes, :dependent => :destroy has_many :likes, :dependent => :destroy belongs_to :user end Answers (sites) class Site < ActiveRecord::Base belongs_to :question belongs_to :user has_many :notes, :dependent => :destroy has_many :likes, :dependent => :destroy has_attached_file :photo, :styles => { :small => "250x250>" } end Pics class Pic < ActiveRecord::Base has_attached_file :profile_pic, :styles => { :small => "100x100" } belongs_to :user end The /views/questions/show.html.erb is rendering the partial /views/sites/_site.html.erb which is calling the Answer (site) with: <% div_for site do %> <%=h site.description %> <% end %> I have been trying to do things like: <%=image_tag site.user.pic.profile_pic.url(:small) %> <%=image_tag site.user.profile_pic.url(:small) %> etc. But that is obviously wrong. My error directs me to the Questions#show action so I am imagining that I need to define something in there but not sure what. Is is possible to call the pic given the current associations, placement of the call, and if so what Controller additions do I need to make, and what line of code will call the pic? UPDATE: Here is the QuestionsController#show code: def show @question = Question.find(params[:id]) @sites = @question.sites.all(:select => "sites.*, SUM(likes.like) as like_total", :joins => "LEFT JOIN likes AS likes ON likes.site_id = sites.id", :group => "sites.id", :order => "like_total DESC") respond_to do |format| format.html # show.html.erb format.xml { render :xml => @question } end end

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  • On Click alert if $.get returns a value, if not, submit the form

    - by bradenkeith
    If the submit button is clicked, prevent the default action and see if the field 'account_name' is already in use. If the $.get() returns a result, alert the user of the results. If it doesn't, submit form with id="add_account_form". My problem is that my else{} statement is not submitting the form. I get no response when submit is clicked & there is no value returned. Also I would like to change my code where it goes $("#add_account_form").submit(..) instead of .click() however, would that cause a problem when trying to submit the form later in the script? <script type="text/javascript"> $(document).ready( function () { $("#submit").click( function () { var account_name = $("input[name=account_name]").val(); $.get( "'.url::site("ajax/check_account_name").'", {account_name: account_name}, function(data){ if(data.length > 0){ confirm( "The account name you entered looks like the following:\n" +data+ "Press cancel if this account already exists or ok to create it." ); }else{ $("#add_account_form").submit(); } }); return false; }); }); </script> <p> <input type="submit" id="submit" class="submit small" name="submit" value="Submit" /> </p> </form> Thanks for your help. EDIT So anyone who runs into my problems, it's that $.get() is asynchronous, so it will always return false, or true depending on what submitForm is defined as. $.ajax() however, allows async to be set as false, which allows the function to finish before moving on. See what I mean: $(document).ready( function () { $("#add_account_form").submit( function () { var submitForm = true; var account_name = $("input[name=account_name]").val(); $.ajax({ type: "GET", async: false, url: "'.url::site("ajax/check_account_name").'", data: ({account_name: account_name}), success: function(data){ if(data.length > 0){ if(!confirm( "The account name you entered looks like the following:\n" +data+ "Press cancel if this account already exists or ok to create it." )){ submitForm = false; } } } }); if (submitForm == false ) { return false; } }); }); Thanks for your help @Dan

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  • Connecting form to database errors

    - by Russell Ehrnsberger
    Hello I am trying to connect a page to a MySQL database for newsletter signup. I have the database with 3 fields, id, name, email. The database is named newsletter and the table is named newsletter. Everything seems to be fine but I am getting this error Notice: Undefined index: Name in C:\wamp\www\insert.php on line 12 Notice: Undefined index: Name in C:\wamp\www\insert.php on line 13 Here is my form code. <form action="insert.php" method="post"> <input type="text" value="Name" name="Name" id="Name" class="txtfield" onblur="javascript:if(this.value==''){this.value=this.defaultValue;}" onfocus="javascript:if(this.value==this.defaultValue){this.value='';}" /> <input type="text" value="Enter Email Address" name="Email" id="Email" class="txtfield" onblur="javascript:if(this.value==''){this.value=this.defaultValue;}" onfocus="javascript:if(this.value==this.defaultValue){this.value='';}" /> <input type="submit" value="" class="button" /> </form> Here is my insert.php file. <?php $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="newsletter"; // Database name $tbl_name="newsletter"; // Table name // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // Get values from form $name=$_POST['Name']; $email=$_POST['Email']; // Insert data into mysql $sql="INSERT INTO $tbl_name(name, email)VALUES('$name', '$email')"; $result=mysql_query($sql); // if successfully insert data into database, displays message "Successful". if($result){ echo "Successful"; echo "<BR>"; echo "<a href='index.html'>Back to main page</a>"; } else { echo "ERROR"; } ?> <?php // close connection mysql_close(); ?>

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  • Passing password value through URL

    - by Steven Wright
    OK I see a lot of people asking about passing other values, URLS, random stuff through a URL, but don't find anything about sending a password to a password field. Here is my situation: I have a ton of sites I use on a daily basis with my work and oh about 90% require logins. Obviously remembering 80 bajillion logins for each site is dumb, especially when there are more than one user name I use for each site. So to make life easier, I drew up a nifty JSP app that stores all of my logins in a DB table and creates a user interface for the specific page I want to visit. Each page has a button that sends a username, password into the id parameters of the html inputs. Problem: I can get the usernames and other info to show up just dandy, but when I try and send a password to a password field, it seems that nothing gets received by the page I'm trying to hit. Is there some ninja stuff I need to be doing here or is it just not easily possible? Basically this is what I do now: http://addresshere/support?loginname=steveoooo&loginpass=passwordhere and some of my html looks like this: <form name="userform" method="post" action="index.jsp" > <input type="hidden" name="submit_login" value="y"> <table width="100%"> <tr class="main"> <td width="100" nowrap>Username:</td> <td><input type="text" name="loginname" value="" size="30" maxlength="64"></td> </tr> <tr class="main"> <td>Password: </font></td> <td><input type="password" name="loginpass" value="" size="30" maxlength="64"></td> </tr> <tr class="main"> <td><center><input type="submit" name="submit" value="Login"></center></td> </tr> </table> </form> Any suggestions?

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  • Releasing the keyboard stops shake events. Why?

    - by Moshe
    1) How do I make a UITextField resign the keyboard and hide it? The keyboard is in a dynamically created subview whose superview looks for shake events. Resigning first responder seems to break the shake event handler. 2) how do you make the view holding the keyboard transparent, like see through glass? I have seen this done before. This part has been taken care of thanks guys. As always, code samples are appreciated. I've added my own to help explain the problem. EDIT: Basically, - (void)motionBegan:(UIEventSubtype)motion withEvent:(UIEvent *)event; gets called in my main view controller to handle shaking. When a user taps on the "edit" icon (a pen, in the bottom of the screen - not the traditional UINavigationBar edit button), the main view adds a subview to itself and animates it on to the screen using a custom animation. This subview contains a UINavigationController which holds a UITableView. The UITableView, when a cell is tapped on, loads a subview into itself. This second subview is the culprit. For some reason, a UITextField in this second subview is causing problems. When a user taps on the view, the main view will not respond to shakes unless the UITextField is active (in editing mode?). Additional info: My Motion Event Handler: - (void)motionBegan:(UIEventSubtype)motion withEvent:(UIEvent *)event { NSLog(@"%@", [event description]); SystemSoundID SoundID; NSString *soundFile = [[NSBundle mainBundle] pathForResource:@"shake" ofType:@"aif"]; AudioServicesCreateSystemSoundID((CFURLRef)[NSURL fileURLWithPath:soundFile], &SoundID); AudioServicesPlayAlertSound(SoundID); [self genRandom:TRUE]; } The genRandom: Method: /* Generate random label and apply it */ -(void)genRandom:(BOOL)deviceWasShaken{ if(deviceWasShaken == TRUE){ decisionText.text = [NSString stringWithFormat: (@"%@", [shakeReplies objectAtIndex:(arc4random() % [shakeReplies count])])]; }else{ SystemSoundID SoundID; NSString *soundFile = [[NSBundle mainBundle] pathForResource:@"string" ofType:@"aif"]; AudioServicesCreateSystemSoundID((CFURLRef)[NSURL fileURLWithPath:soundFile], &SoundID); AudioServicesPlayAlertSound(SoundID); decisionText.text = [NSString stringWithFormat: (@"%@", [pokeReplies objectAtIndex:(arc4random() % [pokeReplies count])])]; } } shakeReplies and pokeReplies are both NSArrays of strings. One is used for when a certain part of the screen is poked and one is for when the device is shaken. The app will randomly choose a string from the NSArray and display onscreen. For those of you who work graphically, here is a diagram of the view hierarchy: Root View -> UINavigationController -> UITableView -> Edit View -> Problem UITextfield

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  • Assign value to HTML textbox from JSP

    - by prakash_d22
    Hello I am creating a web page to add some information about given product.I need to enter id,name,description and image as information.I need the id to be auto generated.I am using jsp and database as access.I am fetching the count(*)+1 value from database and assigning to my html text box but its showing as null.can i get some help? Code: <body> <%@page import="java.sql.*"%> <%! String no; %> <% try{ Class.forName("sun.jdbc.odbc.JdbcOdbcDriver"); Connection con = DriverManager.getConnection("jdbc:odbc:pd"); ResultSet rs = null; Statement st = con.createStatement(); String sql = ("select count(*)+1 from products"); st.executeUpdate(sql); while (rs.next()) { no=rs.getString("count(*)+1"); } rs.close(); st.close(); con.close(); } catch(Exception e){} %> <Form name='Form1' action="productcode.jsp" method="post"> <table width="1024" border="0"> <tr> <td width="10">&nbsp;</td> <td width="126">Add Product: </td> <td width="277">&nbsp;</td> <td width="583">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>Product Id:</td> <td><label> <input type="text" name="id" value="<%= no%>"/> </label></td> <td>&nbsp;</td> .... and so on

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  • It is only uploading first row's file input

    - by user1304328
    I have a application which you can access here. If you open the application please click on the "Add" button a couple of times. This will add a new row into a table below. In each table row there is an AJAX file uploader. Now the problem is that if I click on the "Upload" button in any row except the first row, then the uploading only happens in the first row so it is only uploading the first file input only. Why is it doing this and how can I get it so that when then the user clicks the "Upload" button, the file input within that row of the "Upload" button is uploaded and not the first row being uploaded? Below is the full code where it appends the file AJAX file uploaded in each table row: function insertQuestion(form) { var $tbody = $('#qandatbl > tbody'); var $tr = $("<tr class='optionAndAnswer' align='center'></tr>"); var $image = $("<td class='image'></td>"); var $fileImage = $("<form action='upload.php' method='post' enctype='multipart/form-data' target='upload_target' onsubmit='startUpload();' >" + "<p id='f1_upload_process' align='center'>Loading...<br/><img src='Images/loader.gif' /><br/></p><p id='f1_upload_form' align='center'><br/><label>" + "File: <input name='fileImage' type='file' class='fileImage' /></label><br/><label><input type='submit' name='submitBtn' class='sbtn' value='Upload' /></label>" + "</p> <iframe id='upload_target' name='upload_target' src='#' style='width:0;height:0;border:0px solid #fff;'></iframe></form>"); $image.append($fileImage); $tr.append($image); $tbody.append($tr); } function startUpload(){ document.getElementById('f1_upload_process').style.visibility = 'visible'; document.getElementById('f1_upload_form').style.visibility = 'hidden'; return true; } function stopUpload(success){ var result = ''; if (success == 1){ result = '<span class="msg">The file was uploaded successfully!<\/span><br/><br/>'; } else { result = '<span class="emsg">There was an error during file upload!<\/span><br/><br/>'; } document.getElementById('f1_upload_process').style.visibility = 'hidden'; document.getElementById('f1_upload_form').innerHTML = result + '<label>File: <input name="fileImage" type="file"/><\/label><label><input type="submit" name="submitBtn" class="sbtn" value="Upload" /><\/label>'; document.getElementById('f1_upload_form').style.visibility = 'visible'; return true; }

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  • how to upload a audio file using REST webservice in Google App Engine for Java

    - by sathya
    Am using google app engine with eclipse IDE and trying to upload a audio file. I used the File Upload in Google App Engine For Java and can able to upload the file successfully. Now am planning to use REST web service for it. I had analyzed in developers.google but i failed. Can anyone suggest me how to implement REST Web services in google app engine using Eclipse. The code google provided is shown below, // file Upload.java public class Upload extends HttpServlet { private BlobstoreService blobstoreService = BlobstoreServiceFactory.getBlobstoreService(); public void doPost(HttpServletRequest req, HttpServletResponse res) throws ServletException, IOException { Map<String, BlobKey> blobs = blobstoreService.getUploadedBlobs(req); BlobKey blobKey = blobs.get("myFile"); if (blobKey == null) { res.sendRedirect("/"); } else { res.sendRedirect("/serve?blob-key=" + blobKey.getKeyString()); }}} // file Serve.java public class Serve extends HttpServlet { private BlobstoreService blobstoreService = BlobstoreServiceFactory.getBlobstoreService(); public void doGet(HttpServletRequest req, HttpServletResponse res) throws IOException { BlobKey blobKey = new BlobKey(req.getParameter("blob-key")); blobstoreService.serve(blobKey, res); }} // file index.jsp <%@ page import="com.google.appengine.api.blobstore.BlobstoreServiceFactory" %> <%@ page import="com.google.appengine.api.blobstore.BlobstoreService" %> <% BlobstoreService blobstoreService = BlobstoreServiceFactory.getBlobstoreService(); %> <form action="<%= blobstoreService.createUploadUrl("/upload") %>" method="post" enctype="multipart/form-data"> <input type="file" name="myFile"> <input type="submit" value="Submit"> </form> // web.xml <servlet> <servlet-name>Upload</servlet-name> <servlet-class>Upload</servlet-class> </servlet> <servlet> <servlet-name>Serve</servlet-name> <servlet-class>Serve</servlet-class> </servlet> <servlet-mapping> <servlet-name>Upload</servlet-name> <url-pattern>/upload</url-pattern> </servlet-mapping> <servlet-mapping> <servlet-name>Serve</servlet-name> <url-pattern>/serve</url-pattern> </servlet-mapping> Now how to provide a rest web service for the above code. Kindly suggest me an idea.

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  • Problems with MYSQL database

    - by shinjuo
    I have a database that worked fine until I decided to add a log onto the page. here is what I have now: <body> <?php if($_SERVER['REQUEST_METHOD'] == 'POST') { require("serverInfo.php"); mysql_query("UPDATE `cardLists` SET `AmountLeft` = `AmountLeft` + ".mysql_real_escape_string($_POST['Add'])." WHERE `cardID` = '".mysql_real_escape_string($_POST['Cards'])."'"); echo "\"" .$_POST['Add'] ."\" has been added to the inventory amount for the card \"". $_POST['Cards']. "\""; mysql_query("INSERT INTO `log` (`changes`, `amount`, `cardID`, `person`, Date)VALUES('ADDED','$_POST['Add']','$_POST['Cards']', '$_POST['Person']', NOW())"); mysql_close($link); } ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <?php require("serverInfo.php"); ?> <?php $res = mysql_query("SELECT * FROM cardLists order by cardID") or die(mysql_error()); echo "<select name = 'Cards'>"; while($row=mysql_fetch_assoc($res)) { echo "<option value=\"$row[cardID]\">$row[cardID]</option>"; } echo "</select>"; ?> Amount to Add: <input type="text" name="Add" maxlength="8" /> Changes Made By: <select name="Person"> <option value="justin">Justin</option> <option value="chris">Chris</option> <option value="matt">Matt</option> <option value="dan">Dan</option> <option value="tim">Tim</option> <option value="amanda">Amanda</option> </select> <input type="submit" name ="submit" onClick= "return confirm( 'Are you sure you want to add this amount?');"> </form> <br /> <input type="button" name="main" value="Return To Main" onclick="window.location.href='index.php';" /> </body> </html> it works fine until I added the: mysql_query("INSERT INTO `log` (`changes`, `amount`, `cardID`, `person`, Date)VALUES('ADDED','$_POST['Add']','$_POST['Cards']', '$_POST['Person']', NOW())"); mysql_close($link); Can anyone see what is going on?

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  • One registry key for many products not deleted on uninstall

    - by NC1
    My company has many products, we want to create a registry key Software\$(var.Manufacturer)that will have all of our products if our customers have installed more than one (which is likely) I then want to have a secondary key for each of our products which get removed on uninstall but the main one does not. I have tried to achieve this like below but my main key gets deleted so all of my other products also get deleted from the registry. I know this is trivial but I cannot find an answer. <DirectoryRef Id="TARGETDIR"> <Component Id="Registry" Guid="*" MultiInstance="yes" Permanent="yes"> <RegistryKey Root="HKLM" Key="Software\$(var.Manufacturer)" ForceCreateOnInstall="yes"> <RegistryValue Type="string" Name="Default" Value="true" KeyPath="yes"/> </RegistryKey> </Component> </DirectoryRef> <DirectoryRef Id="TARGETDIR"> <Component Id="RegistryEntries" Guid="*" MultiInstance="yes" > <RegistryKey Root="HKLM" Key="Software\$(var.Manufacturer)\[PRODUCTNAME]" Action="createAndRemoveOnUninstall"> <RegistryValue Type="string" Name="Installed" Value="true" KeyPath="yes"/> <RegistryValue Type="string" Name="ProductName" Value="[PRODUCTNAME]"/> </RegistryKey> </Component> </DirectoryRef> EDIT: I have got my registry keys to stay using the following code. However they only all delete wen all products are deleted, not one by one as they need to. <DirectoryRef Id="TARGETDIR"> <Component Id="Registry" Guid="FF75CA48-27DE-430E-B78F-A1DC9468D699" Permanent="yes" Shared="yes" Win64="$(var.Win64)"> <RegistryKey Root="HKLM" Key="Software\$(var.Manufacturer)" ForceCreateOnInstall="yes"> <RegistryValue Type="string" Name="Default" Value="true" KeyPath="yes"/> </RegistryKey> </Component> </DirectoryRef> <DirectoryRef Id="TARGETDIR"> <Component Id="RegistryEntries" Guid="D94FA576-970F-4503-B6C6-BA6FBEF8A60A" Win64="$(var.Win64)" > <RegistryKey Root="HKLM" Key="Software\$(var.Manufacturer)\[PRODUCTNAME]" ForceDeleteOnUninstall="yes"> <RegistryValue Type="string" Name="Installed" Value="true" KeyPath="yes"/> <RegistryValue Type="string" Name="ProductName" Value="[PRODUCTNAME]"/> </RegistryKey> </Component> </DirectoryRef>

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  • PHP: Remove Simple Session with Get-Method

    - by elmaso
    Hello, I want to Remove the Sessions from this php code, actually if someone searches i get this url search.php?searchquery=test but if I reload the page, the results are cleaned. how can I remove the Sessions to get the Results still, if someone reloads the page? this are the codes: search.php <?php session_start(); ?> <form method="get" action="querygoogle.php"> <label for="searchquery"><span class="caption">Search this site</span> <input type="text" size="20" maxlength="255" title="Enter your keywords and click the search button" name="searchquery" /></label> <input type="submit" value="Search" /> </form> <?php if(!empty($_SESSION['googleresults'])) { echo $_SESSION['googleresults']; unset($_SESSION['googleresults']); } ?> querygoogle.php <?php session_start(); $url = 'http://www.example.com'; $handle = fopen($url, 'rb'); $body = ''; while (!feof($handle)) { $body .= fread($handle, 8192); } fclose($handle); $json = json_decode($body); foreach($json->responseData->results as $searchresult) { if($searchresult->GsearchResultClass == 'GwebSearch') { $formattedresults .= ' <div class="searchresult"> <h3><a href="' . $searchresult->unescapedUrl . '">' . $searchresult->titleNoFormatting . '</a></h3> <p class="resultdesc">' . $searchresult->content . '</p> <p class="resulturl">' . $searchresult->visibleUrl . '</p> </div>'; } } $_SESSION['googleresults'] = $formattedresults; header("Location: search.php?searchquery=" . $_GET['searchquery']); exit; ?> thank you for your help!!

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  • Help with IF THEN breaking when comparing results from MYSQL query.

    - by roydukkey
    I'm have a problem with an invite system. The if statement seems to break. It shows the message "Fail" but the UPDATE statement still executes. Why do both the THEN and the ELSE excute? $dbConn = new dbConn(); // Check if POST user_username and user_hash are matching and valid; both are hidden for fields $sql = "SELECT user_username " . "FROM table_users " . "WHERE user_id=".mysql_real_escape_string($_POST["user_id"])." " . "AND user_hash='".mysql_real_escape_string($_POST["user_hash"])."' " . "AND user_enabled=0;"; $objUser = $dbConn->query($sql); // If result contains 1 or more rows if( mysql_num_rows($objUser) != NULL ){ $objUser = mysql_fetch_assoc($objUser); $ssnUser->login( $objUser["user_username"] ); $sql = "UPDATE table_users SET " . "user_enabled=1, " . "user_first_name='".mysql_real_escape_string($_POST["user_first_name"])."', " . "user_last_name='".mysql_real_escape_string($_POST["user_last_name"])."', " . "user_password='".mysql_real_escape_string( md5($_POST["user_password"]) )."' " . "WHERE user_id=".mysql_real_escape_string($_POST["user_id"]).";"; $dbConn->query($sql); echo "Success"; header( "Refresh: 5; url=/account/?action=domains" ); } else { echo "Fail"; } This dbConn Class is as follows: class dbConn{ var $username = "xxxx_admin"; var $password = "xxxxxxxx"; var $server = "localhost"; var $database = "xxxx"; var $objConn; function __construct(){ $conn = mysql_connect( $this->server, $this->username, $this->password, true ); if( !$conn ){ die("Could not connect: ".mysql_error() ); } else { $this->objConn = $conn; } unset($conn); } function __destruct(){ mysql_close( $this->objConn ); unset( $this ); } function query( $query, $db = false ){ mysql_select_db( $db != false ? $db : $this->database, $this->objConn ); $result = mysql_query( $query ); unset($query,$db); return $result; } }

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  • Trying to run multiple HTTP requests in parallel, but being limited by Windows (registry)

    - by Nailuj
    I'm developing an application (winforms C# .NET 4.0) where I access a lookup functionality from a 3rd party through a simple HTTP request. I call an url with a parameter, and in return I get a small string with the result of the lookup. Simple enough. The challenge is however, that I have to do lots of these lookups (a couple of thousands), and I would like to limit the time needed. Therefore I would like to run requests in parallel (say 10-20). I use a ThreadPool to do this, and the short version of my code looks like this: public void startAsyncLookup(Action<LookupResult> returnLookupResult) { this.returnLookupResult = returnLookupResult; foreach (string number in numbersToLookup) { ThreadPool.QueueUserWorkItem(lookupNumber, number); } } public void lookupNumber(Object threadContext) { string numberToLookup = (string)threadContext; string url = @"http://some.url.com/?number=" + numberToLookup; WebClient webClient = new WebClient(); Stream responseData = webClient.OpenRead(url); LookupResult lookupResult = parseLookupResult(responseData); returnLookupResult(lookupResult); } I fill up numbersToLookup (a List<String>) from another place, call startAsyncLookup and provide it with a call-back function returnLookupResult to return each result. This works, but I found that I'm not getting the throughput I want. Initially I thought it might be the 3rd party having a poor system on their end, but I excluded this by trying to run the same code from two different machines at the same time. Each of the two took as long as one did alone, so I could rule out that one. A colleague then tipped me that this might be a limitation in Windows. I googled a bit, and found amongst others this post saying that by default Windows limits the number of simultaneous request to the same web server to 4 for HTTP 1.0 and to 2 for HTTP 1.1 (for HTTP 1.1 this is actually according to the specification (RFC2068)). The same post referred to above also provided a way to increase these limits. By adding two registry values to [HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Internet Settings] (MaxConnectionsPerServer and MaxConnectionsPer1_0Server), I could control this myself. So, I tried this (sat both to 20), restarted my computer, and tried to run my program again. Sadly though, it didn't seem to help any. I also kept an eye on the Resource Monitor (see screen shot) while running my batch lookup, and I noticed that my application (the one with the title blacked out) still only was using two TCP connections. So, the question is, why isn't this working? Is the post I linked to using the wrong registry values? Is this perhaps not possible to "hack" in Windows any longer (I'm on Windows 7)? Any ideas would be highly appreciated :) And just in case anyone should wonder, I have also tried with different settings for MaxThreads on ThreadPool (everyting from 10 to 100), and this didn't seem to affect my throughput at all, so the problem shouldn't be there either.

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  • Rails, Edit page update in a window

    - by Mike
    I have my code working so that I have a table of businesses. There's a pencil icon you can click on the edit the business information. The edit information comes up in a partial inside of a modal pop up box. The only problem is that once they make the changes they want and click update, it sends them to the 'show' page for that business. What I want to happen is have the pop up box close and have it update the information. This is my update function in my controller. def update @business = Business.find(params[:id]) respond_to do |format| if @business.update_attributes(params[:business]) flash[:notice] = 'Business was successfully updated.' format.html { redirect_to(business_url(@business)) } format.js else format.html { render :action => "edit" } format.xml { render :xml => @business.errors, :status => :unprocessable_entity } end end end I tried following railscast 43 and i created an .rjs file but I couldn't get that to work at all. My update was still taking me to the show page. Any help would be appreciated. EDIT: Added some more code. <% form_for(@business) do |f| %> <%= f.error_messages %> <p> <%= f.label :name %><br /> <%= f.text_field :name %> </p> ... <%= f.label :business_category %><br /> <%= f.select :business_category_id, @business_categories_map, :selected => @business.business_category_id %> </p> <p> <%= f.label :description %><br /> <%= f.text_area :description %> </p> <p> <%= f.submit 'Update' %> </p> <% end %> This is my form inside of my edit page which is being called through the index in a pop up by: <div id="popupEdit<%=h business.id %>" class="popupContact"> <a class="popupClose<%=h business.id %>" id="popupClose">x</a> <% if business.business_category_id %> <% @business = business %> <%= render "business/edit" %> <% end %> </div>

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  • UIVIewController not released when view is dismissed

    - by Nelson Ko
    I have a main view, mainWindow, which presents a couple of buttons. Both buttons create a new UIViewController (mapViewController), but one will start a game and the other will resume it. Both buttons are linked via StoryBoard to the same View. They are segued to modal views as I'm not using the NavigationController. So in a typical game, if a person starts a game, but then goes back to the main menu, he triggers: [self dismissViewControllerAnimated:YES completion:nil ]; to return to the main menu. I would assume the view controller is released at this point. The user resumes the game with the second button by opening another instance of mapViewController. What is happening, tho, is some touch events will trigger methods on the original instance (and write status updates to them - therefore invisible to the current view). When I put a breakpoint in the mapViewController code, I can see the instance will be one or the other (one of which should be released). I have tried putting a delegate to the mainWindow clearing the view: [self.delegate clearMapView]; where in the mainWindow - (void) clearMapView{ gameWindow = nil; } I have also tried self.view=nil; in the mapViewController. The mapViewController code contains MVC code, where the model is static. I wonder if this may prevent ARC from releasing the view. The model.m contains: static CanShieldModel *sharedInstance; + (CanShieldModel *) sharedModel { @synchronized(self) { if (!sharedInstance) sharedInstance = [[CanShieldModel alloc] init]; return sharedInstance; } return sharedInstance; } Another post which may have a lead, but so far not successful, is UIViewController not being released when popped I have in ViewDidLoad: // checks to see if app goes inactive - saves. [[NSNotificationCenter defaultCenter] addObserver:self selector:@selector(resignActive) name:UIApplicationWillResignActiveNotification object:nil]; with the corresponding in ViewDidUnload: [[NSNotificationCenter defaultCenter] removeObserver:self name:UIApplicationWillResignActiveNotification object:nil]; Does anyone have any suggestions? EDIT: - (void) prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender{ NSString *identifier = segue.identifier; if ([identifier isEqualToString: @"Start Game"]){ gameWindow = (ViewController *)[segue destinationViewController]; gameWindow.newgame=-1; gameWindow.delegate = self; } else if ([identifier isEqualToString: @"Resume Game"]){ gameWindow = (ViewController *)[segue destinationViewController]; gameWindow.newgame=0; gameWindow.delegate = self;

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  • JSP to Bean to Java class Validation

    - by littlevahn
    I have a rather simple form in JSP that looks like this: <form action="response.jsp" method="POST"> <label>First Name:</label><input type="text" name="firstName" /><br> <label>Last Name:</label><input type="text" name="lastName" /><br> <label>Email:</label><input type="text" name="email" /><br> <label>Re-enter Email:</label><input type="text" name="emailRe" /><br> <label>Address:</label><input type="text" name="address" /><br> <label>Address 2:</label><input type="text" name="address2" /><br> <label>City:</label><input type="text" name="city" /><br> <label>Country:</label> <select name="country"> <option value="0">--Country--</option> <option value="1">United States</option> <option value="2">Canada</option> <option value="3">Mexico</option> </select><br> <label>Phone:</label><input type="text" name="phone" /><br> <label>Alt Phone:</label><input type="text" name="phoneAlt" /><br> <input type="submit" value="submit" /> </form> But when I try and access the value of the select box in my Java class I get null. Ive tried reading it in as a String and an Array of strings neither though seems to be grabbing the right value. The response.jsp looks like this: <%@ page language="java" %> <%@ page import="java.util.*" %> <%@page contentType="text/html" pageEncoding="UTF-8"%> <%! %> <jsp:useBean id="formHandler" class="validation.RegHandler" scope="request"> <jsp:setProperty name="formHandler" property="*" /> </jsp:useBean> <% if (formHandler.validate()) { %> <jsp:forward page="success.jsp"/> <% } else { %> <jsp:forward page="retryReg.jsp"/> <% } %> I already have Java script validation in place but I wanted to make sure I covered validation and checking for non-JS users. The RegHandler just uses the name field to refer to the value in the form. Any Idea how I could access the select box's value?

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  • get renamed file names of multiple upload form [js array] in codeigniter

    - by artmania
    Hi friends, I use codeigniter. I have a multiple image upload form. The code below is working well for uploading, but I also need to save file names to database. How can I get the names in here? I spent hours & hours :/ but could not sort it :/ Appreciate helps!!! uploadform.php echo form_open_multipart('gallery/upload'); <input type="file" name="photo" size="50" /> <input type="file" name="thumb" size="50" /> <input type="submit" value="Upload" /> </form> I have a controller between form view and model load model (of course : )) but didnt post here because of no need. gallery_model.php function multiple_upload($upload_dir = 'uploads/', $config = array()) { /* Upload */ $CI =& get_instance(); $files = array(); if(empty($config)) { $config['upload_path'] = realpath($upload_dir); $config['allowed_types'] = 'gif|jpg|jpeg|jpe|png'; $config['max_size'] = '2048'; } $CI->load->library('upload', $config); $errors = FALSE; foreach($_FILES as $key => $value) { if( ! empty($value['name'])) { if( ! $CI->upload->do_upload($key)) { $data['upload_message'] = $CI->upload->display_errors(ERR_OPEN, ERR_CLOSE); // ERR_OPEN and ERR_CLOSE are error delimiters defined in a config file $CI->load->vars($data); $errors = TRUE; } else { // Build a file array from all uploaded files $files[] = $CI->upload->data(); } } } // There was errors, we have to delete the uploaded files if($errors) { foreach($files as $key => $file) { @unlink($file['full_path']); } } elseif(empty($files) AND empty($data['upload_message'])) { $CI->lang->load('upload'); $data['upload_message'] = ERR_OPEN.$CI->lang->line('upload_no_file_selected').ERR_CLOSE; $CI->load->vars($data); } else { return $files; } /* ------------------------------- Insert to database */ // problem is here, i need file names to add db. // if there is already same names file at the folder, it rename file itself. so in such case, I need renamed file name :/ } }

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  • jQuery slider onChange auto submit form

    - by bikey77
    I'm using a jQuery slider as a price range selector in a form. I'd like to have the form submit automatically when one of the values has been changed. I used a couple of examples I found on SO but they didn't work with my code. <form action="itemlist.php" method="post" enctype="application/x-www-form-urlencoded" name="priceform" id="priceform" target="_self"> <div id="slider-holder"> Prices: From <span id="pricefromlabel">100 &#8364;</span> To <span id="pricetolabel">500 &#8364;</span> <input type="hidden" id="pricefrom" name="pricefrom" value="100" /> <input type="hidden" id="priceto" name="priceto" value="500" /> <div id="slider-range"></div> <input name="Search" type="submit" value="Search" /> </div> </form> This is the code that displays the values of the slider and updates 2 hidden form fields I use to store the prices in order to submit: <script> $(function() { $("#slider-range" ).slider({ range: true, min: 0, max: 1000, values: [ <?=$minprice?>, <?=$maxprice?> ], start: function (event, ui) { event.stopPropagation(); }, slide: function( event, ui ) { $( "#pricefrom" ).val(ui.values[0]); $( "#priceto" ).val(ui.values[1]); $( "#pricefromlabel" ).html(ui.values[0] + ' &euro;'); $( "#pricetolabel" ).html(ui.values[1] + ' &euro;'); } }); return false; }); </script> I've tried adding this code as well as an data-autosubmit="true" attribute to the div but no result. $(function() { $('[data-autosubmit="true"]').change(function() { parentForm = $(this).('#priceform'); clearTimeout(submitTimeout); submitTimeout = setTimeout(function() { parentForm.submit() }, 100); }); I've also tried adding a $.post() event to the slider but I'm not very good with jQuery so I'm probably doing it wrong. Any help will be appreciated.

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  • expand div with focus-jquery

    - by Joel
    Hi guys, I'm revisiting this after a few weeks, because I could never get it to work before, and hoping to now. Please look at this website-notice the newsletter signup form at the top right. http://www.rattletree.com I am wanting it to look exactly this way for now, but when the user clicks in the box to enter their email address, the containing div will expand (or simply appear) above the email field to also include a "name" and "city" field. I'm using jquery liberally in the sight, so that is at my disposal. This form is in the header so any id info, etc can't be withing the body tag... This is what I have so far: <div class="outeremailcontainer"> <div id="emailcontainer"> <?php include('verify.php'); ?> <form action="index_success.php" method="post" id="sendEmail" class="email"> <h3 class="register2">Newsletter Signup:</h3> <ul class="forms email"> <li class="email"><label for="emailFrom">Email: </label> <input type="text" name="emailFrom" class="info" id="emailFrom" value="<?= $_POST['emailFrom']; ?>" /> <?php if(isset($emailFromError)) echo '<span class="error">'.$emailFromError.'</span>'; ?> </li> <li class="buttons email"> <button type="submit" id="submit">Send</button> <input type="hidden" name="submitted" id="submitted" value="true" /> </li> </ul> </form> <div class="clearing"> </div> </div> css: p.emailbox{ text-align:center; margin:0; } p.emailbox:first-letter { font-size: 120%; font-weight: bold; } .outeremailcontainer { height:60px; width: 275px; background-image:url(/images/feather_email2.jpg); /*background-color:#fff;*/ text-align:center; /* margin:-50px 281px 0 auto ; */ float:right; position:relative; z-index:1; } form.email{ position:relative; } #emailcontainer { margin:0; padding: 0 auto; z-index:1000; display:block; position:relative; } Thanks for any help! Joel

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  • how to set objectForKey to UISwitch to access switch for save selected option

    - by Rkm
    SystemUIViewcontroller has button event. Tap on Button fire InfoTableviewController. InfoTableview has UISwitch. How to set objectForKey to Info to access UIswitch ViewDidload of System... - (void)viewDidLoad { [super viewDidLoad]; self.Infoarray = [NSMutableArray array]; Info *info = [[Info alloc]initWithNibName:@"Info" bundle:nil]; [self.Infoarray addObject:[NSDictionary dictionaryWithObjectsAndKeys:info, @"viewController", nil]]; } SystemUIviewcontroller button event... -(IBAction) Info_Button_Clicked:(id) sender { Info *info = [[Info alloc]initWithNibName:@"Info" bundle:[NSBundle mainBundle]]; [self.navigationController pushViewController:info animated:YES]; [info release]; } Here for Info TableviewController... - (void)viewDidLoad { [super viewDidLoad]; self.Soundarray = [NSArray arrayWithObjects: [NSDictionary dictionaryWithObjectsAndKeys: @"Sounds", @"labelKey", self.SoundsswitchCtl, @"viewKey", nil],nil]; } // Customize the appearance of table view cells. - (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { UITableViewCell *cell = nil; if ([indexPath section] == 0) { static NSString *kDisplayCell_ID = @"DisplayCellID"; cell = [self.tableView dequeueReusableCellWithIdentifier: kDisplayCell_ID]; if (cell == nil) { cell = [[[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier: kDisplayCell_ID] autorelease]; cell.selectionStyle = UITableViewCellSelectionStyleNone; } else { // the cell is being recycled, remove old embedded controls UIView *viewToRemove = nil; viewToRemove = [cell.contentView viewWithTag:1]; if (viewToRemove) [viewToRemove removeFromSuperview]; } cell.textLabel.text = [[self.Soundarray objectAtIndex: indexPath.row] valueForKey:@"labelKey"]; UIControl *control = [[self.Soundarray objectAtIndex: indexPath.row] valueForKey:@"viewKey"]; [cell.contentView addSubview:control]; return cell; } } - (UISwitch *)SoundsswitchCtl { if (SoundsswitchCtl == nil) { CGRect frame = CGRectMake(198.0, 12.0, 94.0, 27.0); SoundsswitchCtl = [[UISwitch alloc] initWithFrame:frame]; [SoundsswitchCtl addTarget:self action:@selector(switch_Sounds:) forControlEvents:UIControlEventValueChanged]; // in case the parent view draws with a custom color or gradient, use a transparent color SoundsswitchCtl.backgroundColor = [UIColor clearColor]; [SoundsswitchCtl setAccessibilityLabel:NSLocalizedString(@"StandardSwitch", @"")]; SoundsswitchCtl.tag = 1; // tag this view for later so we can remove it from recycled table cells } return SoundsswitchCtl; }

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  • How can I create a Searchstring for a Google AJAX Search API?

    - by elmaso
    Hello, i have this code to get the search resutls from the api: querygoogle.php: <?php session_start(); // Here's the Google AJAX Search API url for curl. It uses Google Search's site:www.yourdomain.com syntax to search in a specific site. I used $_SERVER['HTTP_HOST'] to find my domain automatically. Change $_POST['searchquery'] to your posted search query $url = 'http://ajax.googleapis.com/ajax/services/search/web?rsz=large&v=1.0&start=20&q=' . urlencode('' . $_POST['searchquery']); // use fopen and fread to pull Google's search results $handle = fopen($url, 'rb'); $body = ''; while (!feof($handle)) { $body .= fread($handle, 8192); } fclose($handle); // now $body is the JSON encoded results. We need to decode them. $json = json_decode($body); // now $json is an object of Google's search results and we need to iterate through it. foreach($json->responseData->results as $searchresult) { if($searchresult->GsearchResultClass == 'GwebSearch') { $formattedresults .= ' <div class="searchresult"> <h3><a href="' . $searchresult->unescapedUrl . '">' . $searchresult->titleNoFormatting . '</a></h3> <p class="resultdesc">' . $searchresult->content . '</p> <p class="resulturl">' . $searchresult->visibleUrl . '</p> </div>'; } } $_SESSION['googleresults'] = $formattedresults; header('Location: ' . $_SERVER['HTTP_REFERER']); exit; ?> search.php <?php session_start(); ?> <form method="post" action="querygoogle.php"> <label for="searchquery"><span class="caption">Search this site</span> <input type="text" size="20" maxlength="255" title="Enter your keywords and click the search button" name="searchquery" /></label> <input type="submit" value="Search" /> </form> <?php if(!empty($_SESSION['googleresults'])) { echo $_SESSION['googleresults']; unset($_SESSION['googleresults']); } ?> but with this code, I cant add a searchstring.. how can i add a search string like search.php?search=keyword ? thanks

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