I am using the php exec command to execute another file.
I am using the following:
exec ('php email.php');
But I would wish to pass a variable called $body to "email.php" file through the exec command. How can I pass the variable through exec command?
Pass the argument:
exec('php email.php "' . addslashes($body) . '"');
Get it in email.php:
$body = stripslashes($argv[1]);
$argv[0] is the file's name. I corrected it, thanks!Use:
exec ('php email.php firstparameter secondparameter thirdparameter');
You can also refer this : Command Line Manual
You can pass it as a parameter to email.php
exec('php email.php "'.addslashes($body).'"');
And email.php get it with
$body = stripslashes($argv[1]);
But if $body contains long text with fancy chars, it is better if you save it to a temporary file with random name.
<?php
$temp_file = uniqid().'.txt';
file_put_contents($temp_file, $body);
exec("php email.php $temp_file");
Then in email.php, get the $body from contents of $temp_file.
2 ways to pass parameters to php script from php exec command:
<?php
$fileName = '/var/www/ztest/helloworld.php 12';
$options = 'target=13';
exec ("/usr/bin/php -f {$fileName} {$options} > /var/www/ztest/log01.txt 2>&1 &");
echo "ended the calling script";
?>
see complete answer (with called script and results)
exec()? Any reason you couldn't justinclude()it?