I'm not familiar with posting to a data base. One thing I noticed was that you have a mismatch in the script. If you want it to output to flashfile, you must have one and only one element in the form with that name (right now I see two, the text input and a select). And the script is currently set to output to:
Code:
var radio_concat = {
form : 'flashform',
buttons : ['color', 'effect', 'font'],
output : 'swf_file',
extension : '.swf',
////////////// Stop Editing //////////////
func : function(e){
e = e? e : window.event; . . .
You will also need to configure the name of the form into the script, it is currently set (as you can see) as 'flashform' in the script.
Anyways, I can see now that folks without javascript will never even see this part of the form, so we can forget about the doing the work over again on POST that is done by the script. You can just grab the value of of the text input. But, as I say, it must be configured in the script and be the only element in the form that has that name.
To more fully understand the script, perhaps you should try it out using only the code and supporting markup from my demo as separate page, just to see what it does.
But the important thing I think you need to do is make sure that the various items here:
Code:
form : 'flashform',
buttons : ['color', 'effect', 'font'],
output : 'swf_file',
extension : '.swf',
are configured properly for your form, and that you have no unintended duplicate named elements (other than your radio groups configured in the script as buttons : - and those names should only be repeated in each of their own given radio groups) in the form that could conflict.
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