unable to set variable in case statement bash

问题

I'm trying to set a variable based on a bunch of input conditions. Here's a small sample of the code:

#!/bin/bash
INSTANCE_SIZE=""
case "$1" in
   "micro")
     $INSTANCE_SIZE="t1.micro"
     ;;
   "small")
     $INSTANCE_SIZE="m1.small"

     ;;
esac
echo $INSTANCE_SIZE

When I run the script with the -ex switch and specify the proper argument:

+ case "$1" in
+ =m1.small
./provision: line 19: =m1.small: command not found

回答1:

You need to remove the $ sign in the assignments - INSTANCE_SIZE="m1.small". With the dollar sign, $INSTANCE_SIZE gets substituted with its value and no assignment takes place - bash rather tries to execute the command that resulted from the interpolation.

来源：https://stackoverflow.com/questions/6675879/unable-to-set-variable-in-case-statement-bash