Reputation: 36899
I have the following scenario.
I have a index.php page with the following JQuery code included
jQuery(document).ready(function(){
jQuery('#sIMG img').click(function() {
var currentSRC = jQuery(this).attr('src');
var altSRC = jQuery(this).attr('title');
var imgID = jQuery(this).attr('id');
var cat = jQuery(this).attr('name');
/*Fade, Callback, swap the alt and src, fade in */
jQuery('#main').fadeOut('fast',function() {
jQuery('#main').load("detail.php?id="+imgID+"&category="+cat);
jQuery('#main').fadeIn('fast');
});
});
});
Now I have two div tags called #main and #right in the index.php page. When I click on a menu item right changes to a bunch of images, if I click on one of those images the above code should take effect and load into the main div, but it's just not working. the images are located within a div called sIMG. Any help will be appreciated
Upvotes: 4
Views: 9848
Reputation: 187110
Try using live
jQuery('#sIMG img').live("click",function(){
});
As of jQuery 1.7, the .live() method is deprecated. Use .on() to attach event handlers.
jQuery('#sIMG img').on("click",function(){
});
Upvotes: 12
Reputation: 9530
One of the reasons that the jquery click don't work is that you have dupplicates id's in the form.
Upvotes: 0
Reputation: 37526
I think what you're doing is setting "click" on the array that is return there. Try this:
jQuery('#sIMG img').each(function() {
jQuery(this).click(function() {
});
});
Upvotes: 1