bash: Variable substitution - If variable is set use it else print word and exit.

indigowinged

Take into account the main readme for this section. README first.

Syntax is:
${name?word}

Use the variable $name is set, otherwise print word and exit.

In a shell script.

Brazil1=SugarCane
Amazon=${Brazil1?NoDeforestation}
echo "Healthy Planet"

With Brazil1 set to SugarCane, run the script.

$ ./healthyplanet
Healthy Planet

Now comment out Brazil1 from the script and run it again. The script exited before the echo statement in the script, because Brazil1 was not set.

$ ./healthplanet
./healthplanet: line 4: Brazil1: NoDeforestation
$ echo $?
1