如何在bash中为php命令传递变量

Question

On bash funcion I need pass varible por a external command (php)

panel=$(php -r '$ini_array = parse_ini_file("/root/.name/name.ini");echo $ini_array['panel'];')

Work properly.

But I need pass path on variable and key of array also. Try several form but all fails. '', "", ...

php -r '$ini_array = parse_ini_file(`$sPath`);echo $ini_array["$sKey"];'

It's possible?

Answer 1

Pass the variable through the environment, and retrieve it from the environment on the PHP side.

sPath=$sPath sKey=$sKey \
  php -r '$ini_array = parse_ini_file($_ENV["sPath"]);echo $ini_array[$_ENV["sKey"]];'

The initial sPath=$sPath and sKey=$sKey are not necessary if you have used export to put your shell variables in the environment already.