Menu
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
279
The format is to type the name, the equals sign = , and the value. Note there isn't a space before or after the equals sign. Giving a variable a value is often referred to as assigning a value to the variable.
Variables in Bash Scripts are untyped and declared on definition. Bash also supports some basic type declaration using the declare option, and bash variables are case sensitive.
A variable in bash can contain a number, a character, a string of characters. You have no need to declare a variable, just assigning a value to its reference will create it.
The case statement starts with the case keyword followed by the $variable and the in keyword. The statement ends with the case keyword backwards - esac . The script compares the input $variable against the patterns in each clause until it finds a match.
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.
79
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
