[RESOLVED] php activation of html form?

u-sit

When testing my IPN listener code in PayPal’s sandbox I wish to use pre-set customer selections, hence I set
$sndbxRGSTRTN = true; and $presetclient = true;

I tried w/o success ...

<form>
<!--  lines of inputs -> 
<input type="image"  src="http://images.paypal.com/images/x-click-but01.gif"
  name="submit" alt="Make payments with PayPal - it&#1267; fast, free and secure!" />
   <?PHP
   If($sndbxRGSTRTN && $presetclient=true && !isset($_POST["submit"]))   $_POST['submit']='submit';
   ?>
</form>
        <?php   // End of Buy Now Button
	echo "L.327 \$POST=".var_export($POST,$result=true)."<p>";
	//(more lines of php)
	?>

The echo returns; ‘L.327 $POST=NULL’
Apparently I don’t understand how to do this.

Any suggestions appreciated.
u-sit

bradgrafelman

Where do you define the variable $POST?

u-sit

I'm trying to use $_POST, which I thought is built-in to http. I presume that <form> inserts go into this variable and its contents should be available on exiting the form. Is this not correct? I don't recall ever defining it.

u-sit

bradgrafelman

u-sit;10995897 wrote:
I'm trying to use $_POST

Then why did you write $POST instead?

u-sit

OOPS! I see it now. Made the correction and got
L.327 $POST=array ( 'submit' => 'submit', )
which, tells me that none of the pre-set variables in the form got loaded into $_POST.
Gotta work on that now.

Thanks for your insight.
u-sit

u-sit

Another oversight -- must really be getting tired.
Correcting the http var to $POST got the contents of $POST but the echo statement still had $POST=. So the echoed contents did reveal missing data even with the erroneous var name.

u-sit