Using argument to call PHP in shell script

Question

I have the following line in a shell script:

/usr/local/bin/php /home/script_to_run.php;

This works fine with our setup, until I add an argument like so:

/usr/local/bin/php /home/script_to_run.php?needed_variable=1;

At which point, I get the "Could not open input file" error.

Ideas on how to get this to work?

Thanks!

user142162 · Accepted Answer · 2013-09-03 21:22:12Z

2

There is no query string in the command line; you need to use arguments instead:

/usr/local/bin/php /home/script_to_run.php 1;

You then access the value with the $argv variable:

$value = $argv[1];

For more advanced command line argument parsing, take a look at getopt.

answered Sep 3, 2013 at 21:22

user142162

Sign up to request clarification or add additional context in comments.

1 Answer 1