Passing Parameters Script

[JD]

I want to pass two parameters to a simple script running on my web server.

The script "parpass.php" looks like this:

<?php
echo "The total number of parameters passed was: $argc \n";
echo "The parameters passed were: ";
foreach (argv as $val) {
echo "$val ";
}
echo " \n ";
?>

I tried running this script by typing this in my browser window

http://www.myserver.org/parpass.php parm1 parm2

The script runs but does not report any parameters

Any help would be appreciated.

I'm new to PHP but am getting better.

Thanks to all

John

[Marian Heddesheimer]

On Sun, 27 Jun 2004 09:23:03 -0400, JD <jd******@cox.net> wrote:

try:

http://www.myserver.org/parpass.php&...t1&parm2=test2

Marian

--
Tipps und Tricks zu PHP, Coaching und Projektbetreuung
http://www.heddesheimer.de/coaching/

[Marian Heddesheimer]

On Sun, 27 Jun 2004 15:41:16 +0200, Marian Heddesheimer
<26*************@spamgourmet.com> wrote:

sorry, should be this:

http://www.myserver.org/parpass.php?...t1&parm2=test2

Marian

--
Tipps und Tricks zu PHP, Coaching und Projektbetreuung
http://www.heddesheimer.de/coaching/

[JD]

Marian Heddesheimer wrote:

On Sun, 27 Jun 2004 15:41:16 +0200, Marian Heddesheimer
<26*************@spamgourmet.com> wrote:

sorry, should be this:

http://www.myserver.org/parpass.php?...t1&parm2=test2

Marian

Thanks Mariam, This will pass the first parameter OK
but if I add the '& param2=test2' it concantenats it all into one
parameter. I've tried several combinations..

Thanks so far.

John

[Marian Heddesheimer]

On Sun, 27 Jun 2004 10:13:44 -0400, JD <jd******@cox.net> wrote:

Thanks Mariam, This will pass the first parameter OK
but if I add the '& param2=test2' it concantenats it all into one
parameter. I've tried several combinations..

check the array variable $_GET to get the variables individually

echo "<pre>";
print_r($_GET);
echo "</pre>";

Marian

--
Tipps und Tricks zu PHP, Coaching und Projektbetreuung
http://www.heddesheimer.de/coaching/