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  • Ajax response seems to be getting lost

    - by Ringo Blancke
    I'm using the ddslick jquery dropdown plugin in conjunction with my Rails app. In view1, I have $('#challenges_dropdown').ddslick({ <snipped> onSelected: function (data) { $.ajax({ url: "/load_data", type: "GET", data: {"id": data.selectedData.value} }); } }); I.e., I make a call to my controller to load_data. The controller receives this correctly and then at the end, makes a call to render a separate view render "data" This view contains a script snippet that needs to run in order to update some elements of my original view. For some reason, this script snippet is just not running. I'm very confused. When I use a regular link with data-remote="true", then the whole process works perfectly. However, when I make an AJAX call, it fails. What's going on?? Thanks! Ringo

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  • ajax echo returns javascript code!

    - by StevanSteve
    Hi, i have an ajax call. Php page (A) called by ajax requires some other php page (B). Page "B" is php file looking something like this <html> <head> javascript code </head> <body> PHP Code </body> Inside of "head" tags is javascript code. Now, Page "A" includes page "B", but instead of expected result, it echoes pure javascript code from included page "B"! How to prevent that?

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  • $.ajax is not working

    - by Geetha
    Hi All, In my web page there is a textbox to get the scanned barcode value. Once we scan the barcode it has to get details from the database. I am creating the change event for the textbox. Problem: $.ajax is not working. Code: var target = $('#txtBarcode'), val = target.val(); target.change(monitor()); function monitor() { var xx = $('#txtBarcode').val(); $.ajax({ type: "POST", contentType: "application/json; charset=utf-8", data: "{}", url: "HomePage.aspx/SearchProduct", dataType: "json", success: function(data) { alert("Success!!!"); } }); }

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  • jquery $.ajax call succeeds but returns nothing. (jsonp)

    - by Shawn
    $(document).ready(function() { $('#button').click(function() { try { var json = $.ajax({url : 'http://www.example.com/experimental/service.php', type : 'jsonp', success : function() {alert('success')}}); alert(json); } catch(err) { alert(err.description) } var sjson = JSON.stringify(json); $('#display').html(sjson); }) }) After a button is pressed I get an alert message that says "success" and also one that says undefined, referring to the fact that nothing was returned from the ajax call. I checked the firebug 'net' tab and indeed i get a succesful response from the server of "jsonp1272724228884( {} );" Any ideas?

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  • Jquery ajax and php die()

    - by BizMark
    Hi, I have an IE problem. I am using the jquery ajax method to call a php script. The php script just calls die(). In firefox, the error message is displayed, but in IE the success message is displayed without any data. I would prefer the error function to be called. Is there any way to fix this? I'm guessing my javascript code needs change somehow. Thanks! <?php die() ?> $.ajax({ url: "phps/php.php?id="+the_id, dataType: "json", error: function(){ alert('error'); }, success: function(data){ alert("SUCCESS"); } });

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  • Show AJAX content after images have loaded

    - by Ben4Himv
    I am developing my own lightbox kind of jquery plugin. Everything works but I want to hide the loaded content until the images have loaded in the browser from the AJAX call. I found a similar post and I am using the following script but the setTimeout function is what reveals the content and not the .load function. Am I trying to achieve the impossible? $.ajax({ url: 'meet/'+ pLoad + '.html', success: function(data) { var imageCount = $(data).filter('img').length; var imagesLoaded = 0; $(data).hide() .appendTo('#zoom_inner') .filter('img') .load( function() { ++imagesLoaded; if (imagesLoaded >= imageCount) { $('#zoom_inner').children().show(); } }); setTimeout( function() { $('#zoom_inner').children().show() }, 5000 ); } });

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  • Download content of the page using ajax jquery

    - by niao
    Greetings, how can I download some page content using ajax and jquery: I am doing something like that (2 versions inside one script): $("p").click(function() { $('#result').load('http://google.com'); $.ajax({ url='www.google.com', success: function(data) { $("result").html(data); alert('Load was performed.'); var url = 'www.wp.pl'; $('div#result').load(url); //var content = $.load(url); //alert(content); //$("#result").html("test"); } }); }); but it does not return any content

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  • How do you send an array as part of an (jquery) ajax request

    - by Ankur
    I tried to send an array as part of an ajax request like this: var query = []; // in between I add some values to 'query' $.ajax({ url: "MyServlet", data: query, dataType: "json", success: function(noOfResults) { alert(noOfResults); } }); } I wanted to see what I get back in the servlet, so I used this line: System.out.println(request.getParameterMap().toString()); Which returned {} suggesting an empty map. Firebug tells me I am getting a 400 bad request error If I send a queryString like attribute=value as the 'data' then everything works fine, so it has to do with not being able to send an array as is. What do I have to do to get that data into the servlet for further processing. I don't want to pull it out and turn it into a queryString in the JS if I can avoid it. EDIT: I used the .serializeArray() (jQuery) function before sending the data. I don't get the 400 but nothing useful is being sent through.

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  • How come $(this) is undefined after ajax call

    - by JohnnyQ
    I am doing an ajax request when an anchor tag is clicked with an ID set to href. Btw, this anchor tag is dynamically created. <a href="983" class="commentDeleteLink">delete</a> When the anchor tag is clicked the following code is executed: $('.commentDeleteLink').live('click', function(event) { event.preventDefault(); var result = confirm('Proceed?'); if ( result ) { $.ajax({ url: window.config.AJAX_REQUEST, type: "POST", data: { action : 'DELCOMMENT', comment : $('#commentText').val(), comment_id : $(this).attr('href') }, success: function(result) { alert($(this).attr('href')); //$(this).fadeOut(slow); } }); } }); When I tried to display $(this).attr('href') it says it's "undefined". What I really want to do is fadeOut the anchor tag but when I investigated the value of $(this) it is "undefined". What could be wrong with the snippet above?

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  • Use AJAX to check for a new message

    - by Luke
    Quite simply, I need to alert the end user when they have a new private message. From a combination of research and other opinion, I realise I need to use AJAX for this. The mysql query would be SELECT id FROM tbl_messages WHERE to_viewed = 1 So when someone sends a message, I want an alert to popup on the screen to inform the user without a page reload. I have absolutely no idea what I am doing, but know what I want. Really need help with this, AJAX is definitely something I want to improve as it opens up greater possibilities! Thanks

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  • jQuery ajax get

    - by Happy
    There are many class="item" blocks on the page. For each one there is different var item_link and ajax request. Ajax searches for src attribute of .message img and throws it to var src. $(".item").each(function(){ var item_link = "http://..."; $(this).prepend('<div class="src"></div>'); $.get(item_link, function(data) { var src = $('.message img', data).attr('src'); }); }); How to print var src to <div class="src"></div>? Thanks.

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  • Ajax posting to PHP

    - by JQonfused
    Hi guys, I'm testing a jQuery ajax post method on a local Apache 2.2 server with PHP 5.3 (totally new at this). Here are the files, all in the same folder. html body (jQuery library included in head): <form id="postForm" method="post"> <label for="name">Input Name</label> <input type="text" name="name" id="name" /><br /> <label for="age">Input Age</label> <input type="text" name="age" id="age" /><br /> <input type="submit" value="Submit" id="submitBtn" /> </form> <div id="resultDisplay"></div> <script src="queryRequest.js"></script> queryRequest.js $(document).ready(function(){ $('#s').focus(); $('#postForm').submit(function(){ var name = $('#name').val(); var age = $('#age').val(); var URL = "post.php"; $.ajax({ type:'POST', url: URL, datatype:'json', data:{'name': name ,'age': age}, success: function(data){ $('#resultDisplay').append("Value returned.<br />name: "+data.name+" age: "+data.age); }, error: function() { $('resultDisplay').append("ERROR!") } }); }); }); post.php <?php $name = $_POST['name']; $age = $_POST['age']; $return = array('name' => $name, 'age' => $age); echo json_encode($return); ?> After inputting the two fields and pressing 'Submit', the success method is called, text appended, but the values returned from ajax post are undefined. And then after less than a second, the text fields are emptied, and the text appended to the div is gone. Doesn't seem like it's a page refresh, though, since there's no empty page flash. What's going on here? I'm sure it's a silly mistake but Firebug isn't telling me anything.

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  • call .js file after ajax load

    - by asovgir
    I am trying to apply a .js file to a page I loaded via ajax (since ajax automatically strips the content of all the javascript). var url; var textUrl = 'local/file.js'; $('a').click(function() { url = $(this).attr('href'); $('.secondaryDiv').load(url, function() { $.getScript(textUrl, function(data, textStatus, jqxhr) { console.log(data); //data returned from getScript console.log(textStatus); //return "success" console.log(jqxhr.status); //200 }); }); return false; }); Am I approaching this the right way? I tried everything I could think of and I can't get it to work

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  • ajaxStart only called once

    - by cmroanirgo
    I'm having trouble with ajaxStart/ajaxStop. In the following code, the ajaxStart/ajaxStop code is only ever called once (the very first time). Any ideas why? var A_func = function() { $.get($(this).attr("href"), function(response) { $("#content_inner").replaceWith($("#content_inner", response)); }); return false; }; $(function() { $("A").live("click", A_func); $(document).ajaxStart(function(){$("#loading").show();alert("loading");}); $(document).ajaxStop(function(){$("#loading").hide();alert("finished");}); }); Note that the ajax requests themselves work correctly, it's just that the 'loading' icon only shows the once and then never again. My html is something like: <body> <img id="loading" width="12" height="12" src="/images/ajax-loader.gif" style="display:none" alt="loading"/> ... <div id="content"><div id="content_inner"...</div></div>... </body> I have also tried using ajaxSend/ajaxComplete and it has the same problem

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  • Form submitted via dialog opens dialog again

    - by VikingGoat
    I have a form in a jquerymobile dialog box that I am submitting via jQuery Ajax. Currently my problem is that once the form is submitted the same dialog box is opened again on top of the original dialogbox. So that my url reads before submission: url/index.php#&ui-state=dialog and then after submission: url/index.php#&ui-state=dialog#&ui-state=dialog&ui-state=dialog Has anyone ever encountered something like this before? [edit added code example] $(function(){ $("#form").submit(function(e){ e.preventDefault(); var dataString = $("#form").serialize(); errorInput = $("input[name=valOne]#valOne").val(); $.ajax({ type: "GET", url: "formHandler.php", data: dataString, dataType: "text", success: function(data){ if(data.toLowerCase().indexOf("error") >= 0){ alert(data); $(".ui-dialog").dialog("close"); $("#valOne").val(errorInput); //the reentering info so user doesn't have to }else{ $(".ui-dialog").dialog("close"); location.href="index.php"; } }, error:function (xhr, ajaxOptions, thrownError){ alert(thrownError); } }); }); });

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  • jquery checkbox and array help

    - by sea_1987
    Hi There I need to get the names and values of checkboxes that have been checked into an array name selected, I cannot for life of me get it working, below is my attempt, if someone could clrify what I am doing wrong that would be brilliant. //Location AJAX //var dataObject = new Object(); var selected = new Array(); $('#areas input.radio').change(function(){ // will trigger when the checked status changes var checked = $(this).attr("checked"); // will return "checked" or false I think. // Do whatever request you like with the checked status if(checked == true) { /*$("input:checked").each(function() { selected.push($(this).attr('name')+"="+$(this).val(); }); alert(selected)*/ getQuery = $(this).attr('name')+"="+$(this).val()+"&location_submit=Next"; $.ajax({ type:"POST", url:"/search/location", data: getQuery, success:function(data){ alert(getQuery); console.log(data); // $('body.secEmp').html(data); } }); } else { //do something to remove the content here alert("Remove"); } });

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  • Ubuntu Control Center Makes Using Ubuntu Easier

    - by Vivek
    Users who are new to Ubuntu might find it somewhat difficult to configure. Today we take a look at using Ubuntu Control Center which makes managing different aspects of the system easier. About Ubuntu Control Center A lot of utilities and software has been written to work with Ubuntu. Ubuntu Control Center is one such cool utility which makes it easy for configuring Ubuntu. The following is a brief description of Ubuntu Control Center: Ubuntu Control Center or UCC is an application inspired by Mandriva Control Center and aims to centralize and organize in a simple and intuitive form the main configuration tools for Ubuntu distribution. UCC uses all the native applications already bundled with Ubuntu, but it also utilize some third-party apps like “Hardinfo”, “Boot-up Manager”, “GuFW” and “Font-Manager”. Ubuntu Control Center Here we look at installation and use of Ubuntu Control Center in Ubuntu 10.04. First we have to satisfy some dependencies. You will need to install Font-Manager and jstest-gtk (link below)…before installing Ubuntu Control Center (UCC). Click the Install Package button. You’ll be prompted to enter in your admin password for each installation package. Installation is successful…close out of the screen. Download and install Font-Manager…again you’ll need to enter in your password to complete installation.   Once you have installed the two dependencies, you are all set to install Ubuntu Control Center (link below), double click the downloaded Ubuntu Control Center deb file to install it. Once installed you can find it under Applications \ System Tools \ UCC. Once you launch it you can start managing your system, software, hardware, and more.   You can easily control various aspects of your Ubuntu System using Ubuntu Control Center. Here we look at configuring the firewall under Network and Internet.     UCC allows easy access for configuring several aspects of your system. Once you install UCC you’ll see how easy it is to configure your Ubuntu system through an intuitive clean graphical interface. If you’re new to Ubuntu, using UCC can help you in setting up your system how you like in a user friendly way. Home Page of UCC http://code.google.com/p/ucc/ Links Download Font-Manager ManagerDownload jstest-gtkUbuntu Control Center (UCC) Similar Articles Productive Geek Tips Adding extra Repositories on UbuntuAllow Remote Control To Your Desktop On UbuntuAssign a Hotkey to Open a Terminal Window in UbuntuInstall VMware Tools on Ubuntu Edgy EftInstall Monodevelop on Ubuntu Linux TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips Xobni Plus for Outlook All My Movies 5.9 CloudBerry Online Backup 1.5 for Windows Home Server Snagit 10 How to Forecast Weather, without Gadgets Outlook Tools, one stop tweaking for any Outlook version Zoofs, find the most popular tweeted YouTube videos Video preview of new Windows Live Essentials 21 Cursor Packs for XP, Vista & 7 Map the Stars with Stellarium

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  • How to handle splitting a file under source control?

    - by sharptooth
    I have a .cpp file and .h file containing a class. Class.cpp contains the implementation and Class.h contains the definition. The class is overcomplicated so I want to separate some code and move it into a separate class. So I create NewClass.cpp and NewClass.h and move the code there. How do I handle this when the files are under SVN? I can simply "svn add" the two new files, but then they will appear as new and will have no history. I could instead "svn copy and rename" the two initial files and edit the the two old files and the two new files - then the two new files will have common history. Which approach is better from the point of version control? Should the new files share history with the old files or should they appear as new?

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  • Remote Control Adapter for PC

    - by Kairan
    Does there exist a product out there that attaches to your PC (preferably USB) that enables Infrared signals to be received (by a regular remote control) ? It seems a novel idea to be able to get a programmable remote that can be used to manipulate windows (like the Harmony remote control series) along with some programmable software (preferably Windows supported) to convert any button on a regular remote to some command on the PC. Does this exist? What is it called? This particular instance you are REQUIRED to use the crappy supplied remote, and this is not what I am looking for

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  • Can not access control panel.

    - by Amby
    I can not access Contol panel in my windows vista machine. As soon as i click the "conrol panel" item in start items, it shows up a window and then its closed automatically ( same happens if i use "control" command). Is there some program or some registry entry thats restricting it? is ther a way to control this behaviour?

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  • [Vista] Can not access control panel.

    - by Amby
    I can not access Contol panel in my windows vista machine. As soon as i click the "conrol panel" item in start items, it shows up a window and then its closed automatically ( same happens if i use "control" command). Is there some program or some registry entry thats restricting it? is ther a way to control this behaviour?

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