May 31, 2013
Calling ajax once
User2310422’s Question:
I have created a button on my website when it is clicked, I sent data on some php file using ajax and return the results. The code I am using is below,
My Goal :
- When that button is clicked for the first time. I want to send the data on some
phpfile usingajaxand return the results, and - When it is clicked for the second time, I just want to hide the content, and
- When it is clicked for the third time, I just want to show the container without calling the ajax again.
jQuery:
$(function() {
$('#click_me').click(function(){
var container = $('#container').css('display');
var id = $('#id').html();
if(container == 'none'){
$.ajax({
type: 'POST',
data: {id: id},
url: "ajax/get_items.php",
}).done(function(data) {
$('#container').html(data);
}).success(function(){
$('#container').show('fast');
});
}else if(container == 'block'){
$('#container').hide('fast');
}
});
});
Html :
<input type="button" id="click_me" value="Click Me"/>
<div id="container"></div>
You can do this by defining a simple variable counting the clicks.
$(function() {
var clickCount = 1; //Start with first click
$('#click_me').click(function(){
switch(clickCount) {
case 1: //Code for the first click
// I am just pasting your code, if may have to change this
var container = $('#container').css('display');
var id = $('#id').html();
if(container == 'none'){
$.ajax({
type: 'POST',
data: {id: id},
url: "ajax/get_items.php",
}).done(function(data) {
$('#container').html(data);
}).success(function(){
$('#container').show('fast');
});
}else if(container == 'block'){
$('#container').hide('fast');
}
break;
case 2:
//code for second click
break;
case 3:
//Code for the third click
break;
});
clickCount++; //Add to the click.
});
