ID's must be unique. Your animation will definitely choke on 2 the same
ghost2008 wrote:
Hey Charlie,
this is pseudo, but they are exactly the same...
Is this a problem?
Thanks and regards
That looks like it's posting twice ?
Any particular reason to do that ?
L
ghost2008 wrote:
Hey guys,
this is amazing. Thanks for your answers.
I implemented the following:
$.ajax({
type: "POST",
url: "some.php",
data: "name=test",
Hey Charlie,
this is pseudo, but they are exactly the same...
Is this a problem?
Thanks and regards
do the 2 divs really have same ID? the one being loaded and one loading
into?
I'm assuming this is pseudo code but just checking
ghost2008 wrote:
Hey guys,
this is amazing. Thanks for your answers.
I implemented the following:
$.ajax({
type: "POST",
url: "some.php",
data:
Hey guys,
this is amazing. Thanks for your answers.
I implemented the following:
$.ajax({
type: "POST",
url: "some.php",
data: "name=test",
success: function(data){
$("#div").hide("slow");
You should then try to parse the 'data' and then get it from it.But in my
opinion, you should add some additional post params to the post request, say
such as {'data': ''#div'}
and then in some.php you can conditionally only return the contents of the
div you want instead of the complete 'data'
Th
$.load() lets you get specific elements only like this $.load
("some.php #div)
I have not done this with a full $.ajax but since they are using same
jQuery, in your $.ajax try
url: some.php #div
ghost2008 wrote:
Hello there,
I am not sure, if this works. I make a function call with t
Hi,
Just implement the falg so that specific data will show into the div.
For Example:
$.ajax({
type: "POST",
url: "some.php?ajx=1",
data: "name=test",
success: function(data){
$("#div").html(data);
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