Bash-Script: How to insert Variables into Bash-Script?

Question

The following bash script is working completely fine:

#!/bin/bash echo '!PaSsWoRd!' | openconnect --csd-wrapper=/home/user/.cisco/csd-wrapper.sh --authenticate --user=abcde123 --authgroup="tunnel My Company" --passwd-on-stdin vpn.mycompany.com 

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However, I want to replace the previous input parameters with variables like that:

#!/bin/bash WRAPPER=/home/user/.cisco/csd-wrapper.sh USER=abcde123 PASSWD=!PaSsWoRd! AUTHGROUP=tunnel My Company DOMAIN=vpn.mycompany.com echo '$PASSWD' | openconnect --csd-wrapper=$WRAPPER --authenticate --user=$USER --authgroup="$AUTHGROUP" --passwd-on-stdin $DOMAIN 

Unfortunately this attempt does not work anymore. I think I have to put in some quote chars or similar. Do you know what is wrong with the bash script below?

Answer 1

The lack of quotes around the assignment to a variable containing ! is the problem here. The shell tries to interpret the ! character to run the history expansion first before assigning it to the variable. The quoting should prevent the shell from interpreting the contents within as '..' as special and keep it as-is.

The assignment should have been written as

PASSWD='!PaSsWoRd!' 

and pass the variable with quoted expansion

echo "$PASSWD" | openconnect --csd-wrapper="$WRAPPER" --authenticate --user="$USER" --authgroup="$AUTHGROUP" --passwd-on-stdin "$DOMAIN" 

Or turn of the history expansion in the script temporarily by including the line, set +H at the top of the script. Subsequently do set -H at the end to disable it. This is not recommended and much recommended to use the proper quoted approach above.

Answer 2

The same (correct) logic that you are applying when using strings in your first example has to be used when assigning strings to variables, too.

In AUTHGROUP=tunnel My Company the three words are split. And PASSWD=!PaSsWoRd! works in a script, but bear in mind that the ! would be interpreted in a shell, triggering history expansion.

Note that single quotes will preserve the literal value of each enclosed character, while double quotes will allow for exceptions: $, `, \ and ! (see QUOTING in man bash). Then, PASSWD='!PaSsWoRd!' will work, PASSWD="!PaSsWoRd!" won't (but, again, ! is problematic only in interactive shells, not in scripts).

At the same time, you will need double quotes when using variables: echo '$PASSWD' will echo the literal $PASSWD, not the variable's value.