Replace content inside div instead of show/hide

JavaScript
1

Can anyone tell me how I can change the code below so that instead of all the fade in fade out stuff the function will actually replace whatever is inside a div called Myholder with the response this script pulls in?

Here’s the code I need to change.

<script type="text/javascript">
$(document).ready(function() {
$('#wait_1').hide();
$('#drop_1').change(function(){
$('#wait_1').show();
$('#result_1').hide();
$.get("func.php", {
func: "drop_1",
drop_var: $('#drop_1').val()
}, function(response){
$('#result_1').fadeOut();
setTimeout("finishAjax('result_1', '"+escape(response)+"')", 400);
});
return false;
});
});

function finishAjax(id, response) {
$('#wait_1').hide();
$('#'+id).html(unescape(response));
$('#'+id).fadeIn();
}
</script>

drop_1 is the name of a drop down which when changed runs this script which brings back a second dropdown with data relating to the first one.

If I need to post more info to get some help with this please let me know.

Thanks in advance.

2

Currently it is a div called result1 that is affected:


$('#result_1').fadeOut();
setTimeout("finishAjax('result_1', '"+escape(response)+"')", 400);

If you don’t want the fadeout/fadein parts occurring, you can remove them so that you just have:


}, function(response){
    setTimeout("finishAjax('myHolder', '"+escape(response)+"')", 400);
});
...
function finishAjax(id, response) {
    $('#wait_1').hide();
    $('#'+id).html(unescape(response));
}
3

thanks for the help, works perfectly now.