This question already has an answer here:
这个问题已经有了答案:
- jQuery ajax load() java script not executing? 8 answers
- jQuery ajax load() java脚本不执行?8回答
Have the below code which loads a php page into a div, once this content is added the jquery on that php page doesnt seem to work. It only works if you open it directly (without ajax)
让下面的代码将一个php页面加载到一个div中,一旦该内容添加到php页面上的jquery似乎不能工作。只有直接打开它(不使用ajax)才有效
any ideas?
什么好主意吗?
AJAX CODE
AJAX代码
<script>
$(document).ready (function() {
$('#NavButton1').click(function(event) {
$('#content').empty();
$('#content').load('placeholder.php');
})
})
</script>
Jquery code on PHP page
PHP页面上的Jquery代码
<script>
$(document).ready(function() {
// call the tablesorter plugin
$('#myTable').tablesorter({
// sort on the first column and third column, order asc
sortList: [[0,0],[2,0]]
});
});
</script>
1 个解决方案
#1
3
document.ready
is not triggered when you load your PHP.
文档。在加载PHP时不会触发ready。
But the right way to have a script run after you load external content would be to use a callback function:
但是,在加载外部内容之后运行脚本的正确方法是使用回调函数:
$('#content').load('placeholder.php', function() {
// call the tablesorter plugin
$('#myTable').tablesorter({
// sort on the first column and third column, order asc
sortList: [[0,0],[2,0]]
});
});
#1
3
document.ready
is not triggered when you load your PHP.
文档。在加载PHP时不会触发ready。
But the right way to have a script run after you load external content would be to use a callback function:
但是,在加载外部内容之后运行脚本的正确方法是使用回调函数:
$('#content').load('placeholder.php', function() {
// call the tablesorter plugin
$('#myTable').tablesorter({
// sort on the first column and third column, order asc
sortList: [[0,0],[2,0]]
});
});