pass quoted string argument in sh script

Question

I simply don't understand

command-line (works):

$ someprog -flag 'key1=val 1,key2=val 2,key3=val 3'

sh direct (works):

#!/bin/sh
someprog -flag 'key1=val 1,key2=val 2,key3=val 3'

sh variable (fails):

#!/bin/sh
OPTSTR="'key1=val 1,key2=val 2,key3=val 3'"
someprog -flag "$OPTSTR"      #fails
someprog -flag ${OPTSTR}      #fails
someprog -flag $OPTSTR        #fails
eval "someprog -flag $OPTSTR" #works

Answer 1

In the first two examples you quote your argument. Those quotes are only seen by your shell, which treats the whole string as a single argument instead of splitting it on the spaces. The quotes itself are not seen by the program.

When typing OPTSTR="'key1=val 1,key2=val 2,key3=val 3'" the content of your variable contains the single quotes, and will therefore later be given to your program. Your program sees the quotes, does not expect them and therefore fails.

Simply use:

OPTSTR="key1=val 1,key2=val 2,key3=val 3"
someprog -flag "$OPTSTR"

The difference between single and double quotes is, whether you want your shell to replace variables and such or not.