T.J. Crowder wrote:
> Hi,
> 'Tis indeed very easy.  Say you have a form wrapped in a div:
>     <div id='formwrapper'><form>
>     ....
>     </form></div>
> You can post it like so and take the result (which is presumed to be
> an HTML snippet in this case) and use that to update the container:
>     new Ajax.Updater('formwrapper', someurl, {
>         parameters:  $('formwrapper').down('form').serialize(true),
>         onFailure: function(response) {
>             // ...show a failure message...
>         }
>     });

That seemed so easy when I read the post, but. . .

How do I trigger the submission ?
I tried onsubmit = {the code above} and it just submitted regularly and 
replaced the whole page with the output, not just the div.

I tried putting the code above as a script after the form, but still in 
the div and got the same result.


Bill Drescher
william {at} TechServSys {dot} com

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