Christian
What happens if you use passthru() instead of exec()?
Rich
-----Original Message-----
From: Christian Pichler [mailto:[EMAIL PROTECTED]]
Sent: 13 September 2002 00:08
To: '[EMAIL PROTECTED]'
Subject: [PHP-WIN] problems with exec function
Hey there,
My problem is following. I want to execute a PHP program (here: test2.php)
by another PHP program (here: test1.php) and get the return code of the
second executed PHP Program. But I always get the return code 0. Even if I
write 'return 50' in test1.php. But the rest of test1.php is working.
Because if I print the value of the output variable (test2.php), there's all
the output of test1.php.
Does anybody know why this doesn't work?
Thanks in advance!
test1.php:
==========
<?php
$i = 0;
echo "Commandline - Parameter\n\n";
print("PHP-Version: " . phpversion() . "\n\n");
print("Result: \n\n");
print("Value of argc: " . $argc . "\n\n");
print("Value(s) of argv: \n");
for ($i = 0; $i < $argc; $i++) {
print("argv[" . $i . "]: " . $argv[$i] . "\n");
}
return $argc;
?>
test2.php:
==========
<?php
$res = "";
$command = "c:\\php\\php.exe -f c:\\test1.php param1 param2 param3";
$output = array();
$retval = 0;
echo "Calling another one...\n\n";
$res = exec($command, $output, $retval);
print ("Following value was returned by exec: " . $retval . "\n");
?>
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