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 fin
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
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
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