Reputation:
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.
Upvotes: 2
Views: 174
Reputation: 21492
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
Upvotes: 2