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
--
Bill Drescher
william {at} TechServSys {dot} com
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