Develop Reference

Javascript image timout

Good day, i'm having a problem with my code, can't get to show the loading image for few seconds, while POST code is getting in database and gives backinformation to show.
$("#poll_vote").click(function(){
var answer = $("input.panswer:checked").val();
var p_id = $("#p_id").val();
$("#poll_load").html("<tr><td align='center'><img src='/images/ajax/ajax4.gif'/></td></tr>");
$.ajax({
type: "POST",
data: "action=poll_vote&p_id="+p_id+"&answer="+answer+"&module="+module+"",
dataType: 'html',
url: "/ajax.php",
success: function(data)
{
$("#poll_content").html(data);
}
});
});
I would hope on your fast help, i'm begginer in java, so can't dicide it myself.

If what you want is to create a delay so the loading animation shows (I believe that is... mmm different, I'm going with that...)
what you need is to set a timeout like so:
setTimeout(function(){ alert("Hello"); }, 3000);
now in your code the function could contain the ajax call:
$("#poll_vote").click(function(){
var answer = $("input.panswer:checked").val();
var p_id = $("#p_id").val();
$("#poll_load").html("<tr><td align='center'><img src='/images/ajax/ajax4.gif'/></td></tr>");
setTimeout(function(){
$.ajax({
type: "POST",
data: "action=poll_vote&p_id="+p_id+"&answer="+answer+"&module="+module+"",
dataType: 'html',
url: "/ajax.php",
success: function(data)
{
$("#poll_content").html(data);
}
});
}, 3000);
});
or be inside the success function, which I believe is better:
$("#poll_vote").click(function(){
var answer = $("input.panswer:checked").val();
var p_id = $("#p_id").val();
$("#poll_load").html("<tr><td align='center'><img src='/images/ajax/ajax4.gif'/></td></tr>");
$.ajax({
type: "POST",
data: "action=poll_vote&p_id="+p_id+"&answer="+answer+"&module="+module+"",
dataType: 'html',
url: "/ajax.php",
success: function(data)
{
setTimeout(function(){$("#poll_content").html(data);}, 3000, data);
}
});
});
I didn't test it, so check if in the second case data can be seen inside the callback function (it should I think...)
Hope it helps.

Looped ajax only submits sometimes

I am working on a dynamic page with multiple forms that can be added and removed by the user. My jquery script goes and finds all 'form' elements and submits them with jquerys ajax method. Here is the script
$(document).ready(function () {
(function (){
var id = $(document).data('campaign_id');
$(document).on('click', '#save-button', function () {
$('form').each(function (){
var data = new FormData(this);
var form = $(this);
if(!form.parent().hasClass('hideme'))
{
$.ajax({
url: form.attr('action'),
type: 'POST',
data: data,
mimeType:"multipart/form-data",
contentType: false,
cache: false,
processData:false,
success: function(data, textStatus, jqXHR)
{
console.log('form submitted '+count);
}
});
}
});
window.location.replace('/campaign');
});
})(); //end SIAF
});//end document.ready
The problem occurs that only sometimes the form submits, I can get it to if I click the save button a few times or if I remove the window.location.redirect that runs at the end, I suspect it is something to do with the redirect occurring before the submit, but I am not sure of a solution after going through some of the documentation

You are being caught out by the asynchronous nature of Ajax. Ajax does not work in a procedural manner, unfortunately. Your success method is called as and when the Ajax request has completed, which depends on your internet connection speed and how busy the server is.
It is entirely possible, the javascript completes its each loop and the first ajax request is still sending or waiting for a response. By when the javascript is ready to do a window.location call.
Edit:
Added code to check the number of forms, and the number of ajax requests, once they have all run, it will redirect
$(document).ready(function () {
(function (){
var id = $(document).data('campaign_id');
var numForms = $('form').length;
var numAjaxRequests= 0;
$(document).on('click', '#save-button', function () {
$('form').each(function (){
var data = new FormData(this);
var form = $(this);
if(!form.parent().hasClass('hideme'))
{
$.ajax({
url: form.attr('action'),
type: 'POST',
data: data,
mimeType:"multipart/form-data",
contentType: false,
cache: false,
processData:false,
success: function(data, textStatus, jqXHR)
{
console.log('form submitted '+count);
numAjaxRequests++;
if(numAjaxRequests == numForms) {
window.location.replace('/campaign');
}
}
});
}
});
});
})(); //end SIAF
});//end document.ready

using ajax how to show the value after success without refreshing the page

i am adding the value to database by using ajax after adding i want to display the value in front end but now after success i am using window.location to show the data because of this the page getting refresh,i don't want to refresh the page to show the data ,anyone guide me how to do this.
below is my ajax
$(function() {
$(".supplierpriceexport_button").click(function() {
var pricefrom = $("#pricefrom").val();
var priceto = $("#priceto").val();
var tpm = $("#tpm").val();
var currency = $("#currency").val();
var dataString = 'pricefrom='+ pricefrom +'&priceto='+priceto+'&tpm='+tpm+'&currency='+currency;
if(pricefrom=='')
{
alert("Please Enter Some Text");
}
else
{
$("#flash").show();
$("#flash").fadeIn(400).html;
$.ajax({
type: "POST",
url: "supplierpriceexport/insert.php",
data: dataString,
cache: false,
success: function(html){
$("#display").after(html);
window.location = "?action=suppliertargetpiceexport";
$("#flash").hide();
}
});
} return false;
});
});

The code that you are using to post the data needs to return some meaningful data, JSON is useful for this, but it can be HTML or other formats.
To return your response as JSON from PHP, you can use the json_encode() function:
$return_html = '<h1>Success!</h1>';
$success = "true";
json_encode("success" => $success, "html_to_show" => $return_html);
In this piece of code, you can set your dataType or JSON and return multiple values including the HTML that you want to inject into the page (DOM):
$.ajax({
type: "POST",
url: "supplierpriceexport/insert.php",
data: dataString,
cache: false,
//Set the type of data we are expecing back
dataType: json
success: function(return_json){
// Check that the update was a success
if(return_json.success == "true")
{
// Show HTML on the page (no reload required)
$("#display").after(return_json.html_to_show);
}
else
{
// Failed to update
alert("Not a success, no update made");
}
});
You can strip out the window.location altogether, else you won't see the DOM update.

Just try to return the values that you need from the ajax function.Something like this might do.
In your insert.php
echo or return the data at the end of the function that needs to be populated into the page
$.ajax({
type: "POST",
url: "supplierpriceexport/insert.php",
data: dataString,
cache: false,
success: function(data){
//Now you have obtained the data that was was returned from the function
//if u wish to insert the value into an input field try
$('#input_field').val(data); //now the data is pupolated in the input field
}
});

Don't use window.location = "?action=suppliertargetpiceexport";
This will redirect to the page suppliertargetpiceexport
$.ajax({
type: "POST",
url: "supplierpriceexport/insert.php",
data: dataString,
cache: false,
success: function(html){
$('#your_success_element_id').html(html); // your_success_element_id is your element id where the html to be populated
$("#flash").hide();
}
});
your_success_element_id is your element id where the html to be populated

.keyup() is only working once, why?

I am using this pretty simple jquery function, but it seems to work only on the first keyup..
$('#cmentuser').keyup(function() {
var mess = document.getElementById('cmentuser').value;
var dataString = 'message='+ mess;
$.ajax({
type: "POST",
url: "atuamae.org/comentbyuser.php",
data: dataString,
success: function() {
}
});
});
any ideas on how to keep it active?

It works, also in the following form (changed mess into jQuery(this).val() and relied on jQuery when encoding the data string):
$('#cmentuser').keyup(function() {
$.ajax({
type: "POST",
url: "atuamae.org/comentbyuser.php",
data: {
'message': jQuery(this).val()
},
success: function() {
// success callback
}
});
});
Proof that it works: jsfiddle.net/xfxPR/
You may be dynamically changing some elements (eg. changing ID or assuming id does not need to be unique), or maybe unbinding the event. Just make sure the event is being attached and stays attached to the element you need.

$(document).on('keyup', '#cmentuser', function(e) {//try to find lower element then doc
var dataString = 'message='+ $(e.target).val();
$.ajax({
type: "POST",
url: "/comentbyuser.php", //no cross domain requests, no need for domain name
data: dataString,
success: function() {}
});
});

try this
$('#cmentuser').live('keyup',function() {
var mess = $(this).val();
var dataString = 'message='+ mess;
$.ajax({
type: "POST",
url: "atuamae.org/comentbyuser.php",
data: dataString,
success: function() {
}
});
});

How to call second jQuery.ajax instance on success of first and update page

I have some jQuery that is triggered on click of a link with the class 'changetag'. I'm using $.ajax() to update the database via changetag.php.
I then change the visual appearance of the link by toggling the class between on/off. The code is as follows:
$(function() {
$(".changetag").click(function(){
var element = $(this);
var I = element.attr("id");
var info = 'switch_tag=' + I;
$.ajax({
type: "POST",
url: "_js/changetag.php",
data: info,
success: function(){}
});
$("#li_"+I).toggleClass("off on");
element.toggleClass("off on");
return false;
});
});
Works perfectly. But now I want to add in a second PHP call which will pull data and update another area of the page if the above was successful.
What I'm trying to add is:
$.ajax({
url: "_js/loaddata.php",
success: function(results){
$('#listresults').empty();
$('#listresults').append(results);
}
});
But just adding it into success: function(){} doesn't seem to be working. To clarify, here is the complete code I'm testing:
$(function() {
$.ajaxSetup ({cache: false});
$(".changetag").click(function(){
var element = $(this);
var I = element.attr("id");
var info = 'switch_tag=' + I;
$.ajax({
type: "POST",
url: "_js/changetag.php",
data: info,
success: function(){
$.ajax({
url: "_js/loaddata.php",
success: function(results){
$('#listresults').empty();
$('#listresults').append(results);
}
});
}
});
$("#li_"+I).toggleClass("off on");
element.toggleClass("off on");
return false;
});
});
The PHP scripts are both called successfully and the toggle class works, but the data pulled is not written to #listresults for some reason.

Ajax calls are (by default) asynchronous. That means that this code:
$("#li_"+I).toggleClass("off on");
element.toggleClass("off on");
return false;
could be executed before the ajax call preceding it is finished. This is a common problem for programmers who are new to ajax and asynchronous code execution. Anything you want to be executed after the ajax call is done must be put into a callback, such as your success handler:
$.ajax({
type: "POST",
url: "_js/changetag.php",
data: info,
success: function(){
$("#li_"+I).toggleClass("off on");
element.toggleClass("off on");
}
});
Likewise, you could put the second ajax call in there as well:
$.ajax({
type: "POST",
url: "_js/changetag.php",
data: info,
success: function(){
$("#li_"+I).toggleClass("off on");
element.toggleClass("off on");
$.ajax({
url: "_js/loaddeals_v2.php",
success: function(results){
$('#listresults').empty();
$('#listresults').append(results);
}
});
}
});
With jQuery 1.5's Deferred Object, you can make this slicker.
function firstAjax() {
return $.ajax({
type: "POST",
url: "_js/changetag.php",
data: info,
success: function(){
$("#li_"+I).toggleClass("off on");
element.toggleClass("off on");
}
});
}
// you can simplify this second call and just use $.get()
function secondAjax() {
return $.get("_js/loaddata.php", function(results){
$('#listresults').html(results);
});
}
// do the actual ajax calls
firstAjax().success(secondAjax);
This is nice because it lets you un-nest callbacks - you can write code that executes asynchronously, but is written like synchronously-executed code.

Deprecation Notice: The jqXHR.success(), jqXHR.error(), and jqXHR.complete() callbacks are removed as of jQuery 3.0. You can use jqXHR.done(), jqXHR.fail(), and jqXHR.always() instead.
https://api.jquery.com/jQuery.ajax/#jqXHR