syntax check on the command line

[kpohl@aspherio.com]

Hi!
It would be great if anyone could help me with this problem:
I want to check the syntax of my php code. Therefore I use the command
line function PHP -l, the structure is similar to the following code
(example from the php Manual.) The Problem is: I read from the standard
output with
echo stream_get_cont ents($pipes[1]);
this function is available only from PHP version 5, but I must write
for version 4.3.0. Which way could it be solved? I don't want to use a
temporary file. Please give me a hint.
Tanks, Karina


<?php
$descriptorspec = array(
0 => array("pipe", "r"), // stdin is a pipe that the child will
read from
1 => array("pipe", "w"), // stdout is a pipe that the child will
write to
2 => array("file", "/tmp/error-output.txt", "a") // stderr is a file
to write to
);

$cwd = '/tmp';
$env = array('some_opt ion' => 'aeiou');

$process = proc_open('php' , $descriptorspec , $pipes, $cwd, $env);

if (is_resource($p rocess)) {
// $pipes now looks like this:
// 0 => writeable handle connected to child stdin
// 1 => readable handle connected to child stdout
// Any error output will be appended to /tmp/error-output.txt

fwrite($pipes[0], '<?php print_r($_ENV); ?>');
fclose($pipes[0]);

echo stream_get_cont ents($pipes[1]);
fclose($pipes[1]);

// It is important that you close any pipes before calling
// proc_close in order to avoid a deadlock
$return_value = proc_close($pro cess);

echo "command returned $return_value\n ";
}

?>