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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.

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  • 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. Commented Nov 29, 2016 at 8:56

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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
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