How to pass shell variables into a PHP function call as arguments?

Question

I'm calling a PHP script from command line:

/usr/bin/php -r "require '/path/to/php/dummy.php'; echo getFunction(\"DummyParameter\")";

The call works like a charm, and the PHP function returns the correct value. I want to replace DummyParameter with a $ShellVariable. When I replace DummyParameter, it is supposed to be passed as "$ShellVariable" string. Here is what I have tried:

/usr/bin/php -r "require '/path/to/php/dummy.php'; echo getFunction(\"$ShellVariable\")";

Is there any way to pass the shell variable into the PHP script as a function argument?

Thanks in advance.

You probably can—but getting proper escaping will possibly be very difficult to get right and will require rather messy code. Is it a proof of concept? Because it's trivial if you just write the code in a file. — Álvaro González
– Álvaro González, Commented Nov 29, 2016 at 8:56

Ruslan Osmanov · Accepted Answer · 2016-11-29 07:09:39Z

2

Pass the CLI arguments and use the $argv array, e.g.:

php -r 'printf("%s\n", $argv[1]);' "$(date)"

Sample Output

Tue Nov 29 14:09:07 +07 2016

answered Nov 29, 2016 at 7:09

Ruslan Osmanov

21.8k8 gold badges52 silver badges62 bronze badges

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

Collectives™ on Stack Overflow

How to pass shell variables into a PHP function call as arguments?

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related