I am trying to submit a form using .ajax() jquery function
function submitForm()
{
$("#quiz").ajaxForm({
target: '#result',
type: 'post',
beforeSubmit: before_submit,
success: showResponse
}).submit();
}
the problem is form gets submitted in second attempt i.e. when i click twice.
$(document).ready(function(){
$("form#submit").submit(function() {
var fname = $('#fname').attr('value');
var lname = $('#lname').attr('value');
$.ajax({
type: "POST",
url: "ajax.php",
data: "fname="+ fname +"& lname="+ lname,
success: function(){
$('form#submit').hide(function(){$('div.success').fadeIn();});
}
});
return false;
});
});
Are you Initializing it in the $(document).ready(function()) ? I hope not.
First initialize it in the document ready minus the submit() method and call submit() in the submitForrm()
The things inside ajaxform() must be initialized before calling them that is why most jquery plugins or widgets are initialized in the document ready section.
That is why when you call it for the first time it is initialized and on the second time it is actually executed.
Related
i want to refresh a particular div on ajax success, im using the below code but the whole page getting refreshed.
<script type="text/javascript">
$('#post_submit').click(function() {
var form_data = {
csrfsecurity: $("input[name=csrfsecurity]").val(),
post_text: $('#post_text').val()
};
$.ajax({
url: "<?php echo site_url('/post_status'); ?>",
type: 'POST',
data: form_data,
success: function(response){
$(".home_user_feeds").html("markUpCreatedUsingResponseFromServer");
}
return false;
});
return false;
});
</script>
you have an extra return false which is inside the $.ajax block which most probably causes an error so your form isn't submitted via ajax. If you remove that, you shouldn't have any issues.
Use the submit event of the form and remove the return false from the ajax callback:
$('#myFormId').on('submit', function() {
var form_data = {
csrfsecurity: $("input[name=csrfsecurity]").val(),
post_text: $('#post_text').val()
};
$.ajax({
url: "<?php echo site_url('/post_status'); ?>",
type: 'POST',
data: form_data,
success: function(response){
$(".home_user_feeds").html("markUpCreatedUsingResponseFromServer");
}
});
return false;
});
Remove the return false from inside your $.ajax function. Its a syntax error. The $.ajax function only expects a json object as an argument. "return false" cannot be part of a json object. You should keep the JavaScript console open during testing at all times - Press Ctrl-Shift-J in Chrome and select console to see any JS errors.
Also suggest you use <input type=button> instead of <input type=submit> or <button></button>
Below is an Ajax POST variable I use to return some information to an ASP MVC3 View. However, I cannot get the .dialg() pop-up function to work. Right now you click on the icon that calls GetProgramDetails(pgmname), and nothing happens. First time using Ajax, so any suggestions would be appreciated. Thx!
<script src="http://code.jquery.com/jquery-1.8.3.js" type="text/javascript"></script>
<script src="http://code.jquery.com/ui/1.9.2/jquery-ui.js" type="text/javascript"></script>
<script type="text/javascript">
function GetProgramDetails(pgmname) {
var request = $.ajax({
type: 'POST',
url: '/BatchPrograms/PopDetails',
data: { programName: pgmname },
dataType: 'html'
});
request.done(function (data) {
$('#data').dialog();
});
</script>
EDIT
I've updated the request.done function to include a simple alert to see if the code was being called. After stepping through with Chrome's debugger, I saw that the code inside was completely skipped over.
request.done(function (data) {
alert("HERE!");
$('#programExplanation').html(data);
});
SECOND EDIT
Here is the controller code the ajax is returning a value from:
[HttpPost]
public string PopDetails(string programName)
{
BatchPrograms batchprograms = db.BatchPrograms.Find(programName);
if (batchprograms == null) return string.Empty;
StringBuilder s = new StringBuilder();
s.Append(batchprograms.ProgramName + " - " + batchprograms.ShortDescription);
s.Append("<br />Job Names: " + batchprograms.PrdJobName + ", " + batchprograms.QuaJobName );
s.Append("<br /> " + batchprograms.Description);
return s.ToString();
}
You need to use the success method to handle the callback, like so:
var request = $.ajax({
type: 'POST',
url: '/BatchPrograms/PopDetails',
data: { programName: pgmname },
dataType: 'html'
}).success(function(data){ $('#data').dialog()} );
This will launch the dialog for you, but if you want to get the response data to work with it, you can have GetProgramDetails take a second parameter which is a callback for after the data is loaded like so:
function GetProgramDetails(pgmname, callback) {
var request = $.ajax({
type: 'POST',
url: '/BatchPrograms/PopDetails',
data: { programName: pgmname },
dataType: 'html'
}).success(callback);
}
This way after the response is received you can handle what to do with the data in your implementation of the callback, in this case it seems like you will be setting data in the dialog and launching the dialog.
I have a .js class named Widget.js
In widget.js class I am initiating a errors.ascx control class that has a JS script function "GetErrors()" defined in it.
Now, when I call GetErrors from my widgets.js class it works perfectly fine.
I have to populate a few controls in widgets.js using the output from GetErrors() function.
But the issue is that at times the GetErrors() takes a lot of time to execute and the control runs over to my widgets class. and the controls are populated without any data in them.
So the bottom line is that I need to know the exact usage of the OnSuccess function of Jquery.
this is my errors.ascx code
var WidgetInstance = function () {
this.GetErrors = function () {
$.ajax({
url: '/Management/GetLoggedOnUsersByMinutes/',
type: 'GET',
cache: false,
success: function (result) {
result = (typeof (result) == "object") ? result : $.parseJSON(result);
loggedOnUsers = result;
}
});
},.....
The code for the Widgets.js file is
function CreateWidgetInstance() {
widgetInstance = new WidgetInstance();
widgetInstance.GetErrors();
}
now I want that The control should move from
widgetInstance.GetErrors();
only when it has produced the results.
any Help???
You can use jQuery Deferreds. $.ajax() actually returns a promise. So you can do the following:
var WidgetInstance = function () {
this.GetErrors = function () {
return $.ajax({
url: '/Management/GetLoggedOnUsersByMinutes/',
type: 'GET',
cache: false
});
},.....
Then you can process the results like so...
widgetInstance.GetErrors().done(function(result){
//process the resulting data from the request here
});
Hi Simply use async:false in your AJAX call.. It will block the control till the response reaches the client end...
var WidgetInstance = function () {
this.GetErrors = function () {
$.ajax({
url: '/Management/GetLoggedOnUsersByMinutes/',
type: 'GET',
cache: false,
async: false,
success: function (result) {
result = (typeof (result) == "object") ? result : $.parseJSON(result);
loggedOnUsers = result;
}
});
},.....
I did a simple solution for this..
I just called my populating functions in the onSuccess event of the GetErrors() of my control and everything worked perfectly..
Was hoping to use the popup and I am pretty sure I am trying to use it incorrectly. Any ideas on how this should work? Can you use the popup in this manner?
<script>
function onSuccess(data, status)
{
data = $.trim(data);
$("#notification").text(data);
}
function onError(data, status)
{
data = $.trim(data);
//$("#notification").text(data);
$("#notification").popup(data); }
$(document).ready(function() {
$("#submit").click(function(){
var formData = $("#callAjaxForm").serialize();
$.ajax({
type: "POST",
url: "sendmsg.php",
cache: false,
data: formData,
success: onSuccess,
error: onError
});
return false;
});
});
</script>
I'm assuming you are trying to use the JQM popup widget, first your missing the closing } from your onError function. Second to use the popup widget you can first set the data
$("#myPopupContent").text(data)
Then to display you use the open method
$("#myPopup").popup("open")
Hey. I need some help with jQuery Ajax calls. In javascript I have to generste ajax calls to the controller, which retrieves a value from the model. I am then checking the value that is returned and making further ajax calls if necessary, say if the value reaches a particular threshold I can stop the ajax calls.
This requires ajax calls that need to be processes one after the other. I tried using async:false, but it freezes up the browser and any jQuery changes i make at the frontend are not reflected. Is there any way around this??
Thanks in advance.
You should make the next ajax call after the first one has finished like this for example:
function getResult(value) {
$.ajax({
url: 'server/url',
data: { value: value },
success: function(data) {
getResult(data.newValue);
}
});
}
I used array of steps and callback function to continue executing where async started. Works perfect for me.
var tasks = [];
for(i=0;i<20;i++){
tasks.push(i); //can be replaced with list of steps, url and so on
}
var current = 0;
function doAjax(callback) {
//check to make sure there are more requests to make
if (current < tasks.length -1 ) {
var uploadURL ="http://localhost/someSequentialToDo";
//and
var myData = tasks[current];
current++;
//make the AJAX request with the given data
$.ajax({
type: 'GET',
url : uploadURL,
data: {index: current},
dataType : 'json',
success : function (serverResponse) {
doAjax(callback);
}
});
}
else
{
callback();
console.log("this is end");
}
}
function sth(){
var datum = Date();
doAjax( function(){
console.log(datum); //displays time when ajax started
console.log(Date()); //when ajax finished
});
}
console.log("start");
sth();
In the success callback function, just make another $.ajax request if necessary. (Setting async: false causes the browser to run the request as the same thread as everything else; that's why it freezes up.)
Use a callback function, there are two: success and error.
From the jQuery ajax page:
$.ajax({
url: "test.html",
context: document.body,
success: function(){
// Do processing, call function for next ajax
}
});
A (very) simplified example:
function doAjax() {
// get url and parameters
var myurl = /* somethingsomething */;
$.ajax({
url: myurl,
context: document.body,
success: function(data){
if(data < threshold) {
doAjax();
}
}
});
}
Try using $.when() (available since 1.5) you can have a single callback that triggers once all calls are made, its cleaner and much more elegant. It ends up looking something like this:
$.when($.ajax("/page1.php"), $.ajax("/page2.php")).done(function(a1, a2){
// a1 and a2 are arguments resolved for the page1 and page2 ajax requests, respectively
var jqXHR = a1[2]; /* arguments are [ "success", statusText, jqXHR ] */
alert( jqXHR.responseText )
});