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.

Answer 1

You can easily catch the output using the backtick operator, and get the exit code of the last command by using $?:

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

anotheroutput=`php -f anotherscript.php`
anotherexitcode=$?

Answer 2

Emilio's answer was good but I thought I could extend that a bit for others. You can use a script like this in cron if you like, and have it email you if there is an error.. YAY :D

#!/bin/sh

EMAIL="[email protected]"
PATH=/sbin:/usr/sbin:/usr/bin:/usr/local/bin:/bin
export PATH

output=`php-cgi -f /www/Web/myscript.php myUrlParam=1`
#echo $output

if [ "$output" = "0" ]; then
   echo "Success :D"
fi
if [ "$output" = "1" ]; then
   echo "Failure D:"
   mailx -s "Script failed" $EMAIL <<!EOF
     This is an automated message. The script failed.

     Output was:
       $output
!EOF
fi

Using php-cgi as the command (instead of php) makes it easier to pass url parameters to the php script, and these can be accessed using the usual php code eg:

$id = $_GET["myUrlParam"];