bill wrote:
> 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) {
>>             // 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
Putting the code as a script did not work, I expect, because the script 
was not served by Ajax.updater.
Using Firebug I determined that the script was stripped out, which 
raises the next question.
I know that I need to set a parameter to updater so that it will not 
strip scripts.  I have read the section on
evalScripts and the need to set the function with coolFunc = function () {

The script that loads the form is invoked as:
 new Ajax.Updater('showMessageDiv',url ,  {method: 'get',evalScripts: 
true} );  // get the requested message
where url is defined as: url = 'mail/getmessage.php?n=' + message_id;
but it does not upload the script so I suspect that I have the 
evalScripts in the wrong place.

Bill Drescher
william {at} TechServSys {dot} com

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