I have an sh script that contains the line $PHP_COMMAND -r 'echo get_include_path();' I can not edit this script, but I need the eventual command line to be (equivalent to) php -d include_path='/path/with spaces/dir' -r 'echo get_include_path();' How can I achieve this? Below is a script that demonstrates the problem. #!/bin/sh # shell script quoting problem demonstration # I need to be able to set a shell variable with a command with # some options, like so PHP_COMMAND="php -d 'include_path=/path/with spaces/dir'" # then use PHP_COMMAND to run something in another script, like this: $PHP_COMMAND -r 'echo get_include_path();' # the above fails when executed. However, if you copy/paste the output # from this line and run it in the CLI, it works! echo "$PHP_COMMAND -r 'echo get_include_path();'" php -d include_path='/path/with spaces/dir' -r 'echo get_include_path();' # what's going on? # this is also interesting echo "\n--------------------" # this works great, but only works if include_path doesn't need quoting PHP_COMMAND="php -d include_path=/path/to/dir" echo "$PHP_COMMAND -r 'echo get_include_path();'" $PHP_COMMAND -r 'echo get_include_path();' echo "\n--------------------" # this one doesn't when run in the sh script, but again if you copy/paste # the output it does work as expected. PHP_COMMAND="php -d 'include_path=/path/to/dir'" echo "$PHP_COMMAND -r 'echo get_include_path();'" $PHP_COMMAND -r 'echo get_include_path();' Script also available online: http://gist.github.com/276500