Executing a PHP Script From A Bash Script

Question

I am trying to execute a simple PHP script from a simple BASH script. The answers on this website don't answer my issue.

Here is my BASH script

#!/bin/sh
railmove="/usr/bin/php -q /home/username/subfolder"
php "$railmove"/$shelltest.php

Here is my PHP script

#!/usr/bin/php
<html>
<head>
</head><body>

<?php
error_reporting(E_ALL);
ini_set('display_errors',1);             
require('connect_db.php');           
$timer="222222";               
$railinfo2=$mysql_link->prepare('INSERT INTO stillrunning(timer) VALUES(:timer)');
$railinfo2->execute(array(':timer'=>$timer));                          
$mysql_link=null;
?>  
</body>
</html>

I get the following error when I run my BASH script from the command line.

Could not open input file: /usr/bin/php -q /home/username/subfolder/.php

I have tried typing /usr/bin/php -q /home/username/subfolder/durable2.sh and that works fine. ie it runs

so where's $shelltest defined?

Marc B
– Marc B

2014-12-15 20:11:14 +00:00
Commented Dec 15, 2014 at 20:11 — Marc B
– Marc B, Commented Dec 15, 2014 at 20:11

Andrew · Accepted Answer · 2014-12-15 20:14:02Z

2

You already include /usr/bin/php in the $railmove string. You don't need to specify php again as a new command. Also, $shelltest doesn't appear to be defined.

So:

#!/bin/sh
railmove="/usr/bin/php -q /home/username/subfolder"
$railmove/your-php-script.php

might work.

answered Dec 15, 2014 at 20:14

Andrew

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1 Comment

user2635961 Over a year ago

Thanks. That worked. The $ in front of the shelltest was part of the problem and adding your script change also helped

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