Hey there,
You're using it partially in the wrong way.
1st:
function showUser(str)
{
$('#content1').load(select.php?q=+str);
}
No syntax mistake in the code above!
But the bad part is this:
$().ajaxSend(function(r,s){
$(#content1).load(Loading data, please wait...);
});
Which will
Hello Isaak,
Thank you for your kind reply.
I tried this code:
function showUser(str)
{
$('#content1').load(select.php?q=+str);
$('#content1').ajaxSend(function(e,r,s)
{
$(this).html('Loading data, please wait...');
}
}
but it caused the AJAX request to stop working. When I remove
Place this part of code outside your function and inside the jQuery
on-DOM-ready function:
$(function()
{
$('#content1').ajaxSend(function(e,r,s)
{
$(this).html('Loading data, please wait...');
}
});
Note that your functions must not be inside the on-DOM-ready function. So
your
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