[jQuery] Re: ajax append problem

Hello Michael,

thanks for this.
I got a error: > html.find is not a function
Any idea?

<script type='text/javascript' src='http://ajax.googleapis.com/ajax/
libs/jquery/1.3.2/jquery.min.js'></script>
<script type="text/javascript">
$(document).ready(function(){
$.ajax({
url: "leistung.html",
cache: false,
success: function(html) {
var loadCont = html.find('#cont');
$(".content").append(loadCont);
}
});
});
</script>

On Apr 27, 5:07 pm, Michael Lawson <mjlaw...@us.ibm.com> wrote:
> instead of  $(".content").append(html);
> do
> var contentToAppend = html.find("#cont");
> $(".content").append(contentToAppend);
>
> cheers
>
> Michael Lawson
> Content Tools Developer, Global Solutions, ibm.com
> Phone:  1-828-355-5544
> E-mail:  mjlaw...@us.ibm.com
>
> 'Examine my teachings critically, as a gold assayer would test gold. If you
> find they make sense, conform to your experience, and don't harm yourself
> or others, only then should you accept them.'
>
>   From:       "clicforw...@googlemail.com" <clicforw...@googlemail.com>                                                          
>
>   To:         "jQuery (English)" <jquery-en@googlegroups.com>                                                                    
>
>   Date:       04/27/2009 11:03 AM                                                                                                
>
>   Subject:    [jQuery] ajax append problem                                                                                        
>
> Hello,
>
> im using this script to load a external site into my DOM.
> It works fine but i need just a part of this test.html.
>
> For example: I want to append just the div "#cont" from test.html
> How can i do that?
>
> <script type="text/javascript">
> $.ajax({
>   url: "test.html",
>   cache: false,
>   success: function(html){
>     $(".content").append(html);
>   }});
>
> </script>
>
> Thanks for Help!
>
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> < 1KViewDownload
>
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