When loading execute both operations together. updateChatAJAx(); updateChatAJAx1(); Then setInterval for both actions var flag=0; $(function(){ setInterval(function () { if(flag==0) { updateChatAJAx(); } else { updateChatAJAx1(); } }, 2000); }); Change the flag to choose operation. flag=0 -> updateChatAJAx(); and flag=1 -> updateChatAJAx1(); solved two set interval function in single PHP file is not working
Your code… $(“#content”).load(“content.html #about”); is like the example in the “Script Execution” section of the .load() documentation. When calling .load() using a URL without a suffixed selector expression, the content is passed to .html() prior to scripts being removed. This executes the script blocks before they are discarded. If .load() is called with a selector … Read more
var pages = [“page1.php”,”page2.php”,”folder/page3.php”] var requests = []; function multipleAjax(string) { /*where string is “?value=somevalue&bool=true” or something*/ for (i=0; i<pages.length;i++) { requests[i] = getXMLHttpRequest(); if (requests[i] != null) { requests[i].open(“Post”, “https://stackoverflow.com/” + pages[i] + string); requests[i].onreadystatechange = function () { if (requests[i].readyState == 4) { /*code to handle responseText returned*/ }; }; }; requests[i].send(); }; … Read more
Write a function and trigger it on blur of your email field. Pass your email as a parameter of your function and check it into your database. This is your email field: <input type=”text” class=”form-control” name=”email” id=”email” value=”” /> This is your JavaScript code: $(document).ready(function() { $(“#email”).blur(function(){ var email = $(“#email”).val(); if(email != “”){ $.ajax({ … Read more
First result on Google: http://www.datatables.net/ DataTables is a plug-in for the jQuery Javascript library. It is a highly flexible tool, based upon the foundations of progressive enhancement, and will add advanced interaction controls to any HTML table. Please try searching for your answer first. 1 solved It’s possible to create ajax table with pagination in … Read more
Make sure that jQuery is properly loaded. Use firebug as a Firefox debugger or the Chrome console. Are you accessing a different domain? If so you will need to use a json callback. $.ajax({ url: “http://ihound.com.au/livechat/php/app.php/widget-init.js”, jsonpCallback: “jsonpcallback”, jsonp: false, dataType: “jsonp” }).done(function(data){ console.log(data); // array of objects }); 4 solved Jquery getScript showing ReferenceError: … Read more
//perhaps, you probably do it like this : var result = {}; $.each([{id: 1, add_1: “123”, add_2: “add1”}, {id: 2, add_1: “456”, add_2: “add2”} ], function(i, item ){ for(var pro in item ){ result[pro] = result[pro] || []; result[pro].push(item[pro]); } }) console.log(result); 2 solved How i change array in js get by key [closed]
SEE (and please read) DOC $( document ).ajaxSend(myFunction); 1 solved Call a javascript function on each ajax request [closed]
After submit run a loop and build your array. For example: var arr = []; $(form).find(‘input[name=”product”]’).each(function(index,element){ arr.push({ ‘label’: $(element).parent().find(‘label’).text(), ‘value’:$(element).val() }); }); solved Form ajax group label and value on submit
Heres my suggestion for you. Hopefully this makes sense and works for you. Please try to understand what is happening rather than just copy and hope it works. Your form, on submit, gets sent to the JS file. This grabs the form values, uses JSON to send them to your php file which processes everything … Read more
These links may help. Submit a form with Ajax and jQuery: http://net.tutsplus.com/tutorials/javascript-ajax/submit-a-form-without-page-refresh-using-jquery/ Validate a form: http://jqueryvalidation.org solved Submit a Form with Validations using AJAX and PHP [closed]
nice question! They way you could do something like that would be.. to try and check what is your url – in this case is: header(“Location: ../Home.php?signup=success”); … and basically see if it contains the keyword “signup=success”. To do this, you need $url=”http://” . $_SERVER[‘SERVER_NAME’] . $_SERVER[‘REQUEST_URI’]; if (strpos($url,’signup=success’) !== false) { echo ‘Thank you … Read more
You can’t declare a variable within an object literal. Move pid outside: var pid = … $.ajax({ … }); solved Chrome SyntaxError: Unexpected Identifier [closed]
Use this code instead: $(document).on(“click”, “#btnaddd”, function() { alert(“click”); }); If it still does not work then you might have used the same ID btnadd for two different elements and JQUERY only matches the first one. 2 solved function to fire on button [closed]
If you want to convert your list of dictionaries to a list of lists on the server side you can do it like so: >>> data = [{‘tahun’: ‘2010’, ‘apel’: 100, ‘pisang’: 200, ‘anggur’: 300, ‘nanas’: 400, ‘melon’: 500}, {‘tahun’: ‘2011’, ‘apel’: 145, ‘pisang’: 167, ‘anggur’: 210, ‘nanas’: 110, ‘melon’: 78}] >>> [[x[‘tahun’], x[‘apel’]] for … Read more