Bash脚本：将带空格的命令和arguments存储在变量中，然后执行

Question

Been banging my head against the wall for a couple hours so time to call in the experts. Writing a small script to run some reports on one of my office's systems and I was asked to take care of a Bash script for it. The program called "auto_rep" takes various options such as "-t" to run one task (to generate one type of report) and a "-1" to exit after one task. The options are separated by spaces when running the command from command-line. The command works directly from command line but I cannot get it to work from a script...

Below is the snippet of code causing me issues:

cmd=$(auto_rep -t createfin1report -1)
echo "running ${cmd} command..."
echo
eval $cmd

The problem is when I run the script, only the "auto_rep" part of the command (from $cmd variable) is run; basically running the program without any options. And it creates tons of reports without the "-t createfin1report -1" part of the command (yikes.). Glad I only tried it on our test system.

Anyone have any tips to help me out? Is my approach way off? BTW - had tried just storing the command in a non-array (cmd="auto_rep -t createfin1report -1") and that was causing me other headache with a "command not found" errors:)...

Thanks in advance!

Answer 1

Save output to an array, then executing this array.

declare -a cmd
cmd=( $(auto_rep -t createfin1report -1) )
echo Executing: "${cmd[@]}"
"${cmd[@]}"

Please make sure the output is a valid command, and spaces have been correctly placed in double-quotes.