Passing GET Variable from one Bash/PHP Script to another

Question

I am not sure the proper name for it, but I am executing PHP code within a Bash script on my Linux server. I have two of these Bash files and want to be able to pass a GET variable from one file to the next.

Here is a simplified version of the 1st file:

#!/usr/bin/php -q
<?php

require("bash2.sh?id=1");

Here is a simplified version of the 2nd file:

#!/usr/bin/php -q
<?php

echo $_GET['id'];

Currently, when I execute the 1st file on a Crontab, I get an error that says :

PHP Warning: require(bash2.sh?id=1): failed to open stream: No such file or directory in /home/bash/bash1.sh on line 2

If I remove the ?id=1 from the require(), it executes without an error.

I'm afraid this won't work... require searches for a exact name of the script, you can't pass variables to it since it's not a URL, it's a path. The only ways I know to pass variables between scripts are session_start() and header() but I'm afraid these won't work without a web server. — Naryl
– Naryl, Commented Nov 27, 2012 at 16:51

Itay Moav -Malimovka · Accepted Answer · 2013-07-04 12:19:41Z

2

You r thinking web... What u put in the require is the actual file name the PHP engine will look for using the OS. i.e. it looks for a file called bash2.sh?id=1 which u obviously do not have.

Either u call another script from withing, say with system('./bash2.sh 2'); Or, include, and use the method below to pass data.

file1

<?php
$id = 1;
require("bash2.sh");

file2

<?php
echo $id;

If u use the first example ( system('./bash2.sh 2');) Then in bash2.sh you will access the variable in the following way:

<?php
echo $argv[1]; //argv[0] is the script name

edited Jul 4, 2013 at 12:19

answered Nov 27, 2012 at 16:50

Itay Moav -Malimovka

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1 Comment

Chris Over a year ago

Should be $argv[1] not argv[1]

rekire · Accepted Answer · 2012-11-27 16:50:11Z

2

You cannot add a parameter to a static file on your harddrive. But you can define a global variable which is accessable by the reqired script.

<?php
$id=1
require("bash2.php");

and for your bash2.php:

<?php
echo $id;

answered Nov 27, 2012 at 16:50

rekire

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Comments

pregmatch · Accepted Answer · 2012-11-27 17:24:32Z

1

No dude you should use arguments. When you execute php script (I am guessing in cron job), you add arguments like some.php variable1 variable2 ..... etc`,

and then in php you get that varibale with $argv[0], $argv[1] .... etc.

That is the way from bash scripts.

answered Nov 27, 2012 at 17:24

pregmatch

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Comments

hanleyhansen · Accepted Answer · 2012-11-27 17:18:57Z

0

Try something like this:

<?php

    $id = $_GET['id'];
    exec("./bash2.sh $id");

?>

And then in the bash script you'll be able to access the first parameter passed as $1.

More info here and here

Hope this helps!

edited Nov 27, 2012 at 17:18

answered Nov 27, 2012 at 16:50

hanleyhansen

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