jQuery获得load（）函数的新结果

Question

I have this code in index.html

$(document).ready(function() {

$('#generate').click(function() {

//$("#results").empty();
$("#results").html("");

$("#results").load("generate.php"); 

});

});

Also, I have this code in generate.php :

<?php

$img1 =  "<img src='logo_0.png' width='350px' height='350x' /><br />";
$img2 =  "<img src='logo_1.png' width='350px' height='350x' /><br />";

echo $img1.$img2;

?>

If I replace these images with different images (manually by copy & paste but keeping their names), I still get same images,

I assumed that happens because the dom objects are already created, so I tried

$("#results").empty();

and

$("#results").html("");

but images don't change,

Thank you

Answer 1

I was going to add a comment but didn't have enough reputation. What about adding the current date/time to the end of the URL to make it unique with the goal of preventing the browser from loading from cache?

You could also use an $.ajax() function to control cache or turn if off for all requests. See this answer for more details.