Retrieve exit status from php script inside of shell script

Question

I have a bash shell script that calls a few PHP scripts like this.

#!/bin/bash

php -f somescript.php

php -f anotherscript.php

I want to compose an error log and/or an activity report based on the results of those scripts.

Is there any way I can get the exit status of the php script in the shell script?

I could either use integer exit statuses or string messages.

Emilio P. · Accepted Answer · 2014-01-14 16:10:15Z

up vote 2 down vote accepted

You can easily catch the output using the backtick operator:

#!/bin/bash
output=`php -f somescript.php`

anotheroutput=`php -f anotherscript.php`

answered Jan 14 '14 at 16:10

Emilio P.
513

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leepowers · Answer 2 · 2013-09-06 00:16:44Z

The $output parameter of the exec command can be used to get the output of another PHP program:

callee.php

<?php
echo "my return string\n";
echo "another return value\n";
exit(20);

caller.php

<?php
exec("php callee.php", $output, $return_var);
print_r(array($output, $return_var));

Running caller.php will output the following:

Array
(
    [0] => Array
        (
            [0] => my return string
            [1] => another return value
        )

    [1] => 20
)

Note the exit status must be number in the range of 0-254. See exit for more info on return status codes.

ahh thanks but is there any way to get the output when the php scripts are called from a bash script? That's what I meant to ask. I suppose I could rewrite the bash script in PHP to use the functionality you've described, which is useful. — Buttle Butkus, Sep 6 '13 at 0:40

2 Answers 2