Setting an argument with bash
bash, linux, unix
Solution
The problem is that quotes are not processed after variable substitution. So it looks like you're trying to define a macro named `'_transaction_color`.
Try using an array:
INSTALL_OPT=(-Uvh --define '_transaction_color 3')
then:
"$INSTALL_CMD" "${INSTALL_OPT[@]}" myPackage.rpm
It's important to put `${INSTALL_OPT[@]}` inside double quotes to get the requoting.
Problem
I frequently run a simple bash command: ``` rpm -Uvh --define "_transaction_color 3" myPackage.rpm ``` which works properly. But now I'm trying to script it into a bash file, and make it more flexible: ``` #!/bin/bash INSTALL_CMD=rpm INSTALL_OPT="-Uvh --define '_transaction_color 3'" ${INSTALL_CMD} ${INSTALL_OPT} myPackage.rpm ``` However, this keeps generating the error: ``` error: Macro % has illegal name (%define) ``` The error is coming from how `--define` and the quoted `_transaction_color` is handled. I've tried a variety of escaping, different phrasing, even making `INSTALL_OPT` an array, handled with `${INSTALL_OPT[@]}`. So far, my attempts have not worked. Clearly, what I want is extremely simple. I'm just not sure how to accomplish it. How can I get bash to handle my `--define` argument properly?