Question

Hii i have this code :

            $handle = popen("php -q nah.php -p=". '$part' . " 2>&1", "r");
            while (!feof($handle))
            {
                 $read = fread($handle, 2096);
                    echo $read;
            }
            pclose($handle);

And the the nah.php looks like this:

echo "yaya1\n";
$options = getopt("p:");
$part = $options["p"];
echo $part;
print_r($options);

The output of this comes out to be:

yaya1 Array (
[p] => )

While the $part in the main file has content.
What can be the error in it ?
How can i fix the error ?
Thanks
And all this is done in php-cli

Answer 1

You’ll want to replace the line that runs popen with this:

$handle = popen("php -q nah.php -p=". escapeshellarg($part) . " 2>&1", "r");

This uses the escapeshellarg() ^[docs] function in order to wrap the $part variable in quotes (and escape any quotes inside it), so that it can be used as a shell argument safely.

The error was that you had $part within single quotes, so it was being sent to the shell directly as $part instead of the value being replaced by PHP. The shell then tried to look for a shell variable called $part to replace it with, but since there was none, it simply replaced it with a blank.

This code would also have worked:

$handle = popen("php -q nah.php -p=". $part . " 2>&1", "r");

However, that would not have wrapped the variable in single quotes, so if there were any spaces in it, only the first word would have been considered the value of argument p, all the other ones would have been considered different arguments.

It’s generally good practice to use escapeshellarg() when sending arguments to a shell command since it takes care of quoting the argument for you.