> When a form is submitted from an image (e.g. <input src="../images
> /next1.gif" name="paging" type="image" value="pageforward"> ) this
> will be passed in the URL as paging.x=SOME_NUMBER. If I try and
> access this variable as ${paging.x} in the subsequent PHP script
> I get nothing. How can I access this variable? all I want to do
> is test for it's existance ? Simple as it seems ... it doesn't
> work ?
You can't use a "." in a variable name. PHP automagically converts
it into a "_". So, look for $paging_x.
Jason
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