I have a basic string equality check in a Bash script, but the output is not as expected.
To reproduce, copy the code below into an executable file (called 'deploy' in my examples below).
#!/bin/bash
echo $1
if [[ "$1" -eq "--help" ]] || [[ "$1" -eq "-h" ]]; then
echo "hello"
fi
If I run the script like so:
./deploy -h
The output is:
-h
hello
If I run the script like so:
./deploy --help
The output is:
-help
Why does the conditional statement not resolve to true?
-eq compares integers. Use == or = to compare strings.
if [[ "$1" == "--help" ]] || [[ "$1" == "-h" ]]; then
echo "hello"
fi
You could omit the quotes. Variable expansions on the left-hand side of == are safe when using double brackets.
You can also use || inside the brackets. It's not possible to do that with single brackets, but double brackets are a syntactical feature with special parsing rules that allow it.
if [[ $1 == --help || $1 == -h ]]; then
echo "hello"
fi
If it gets more complicated you might also consider a case block.
case $1 in
-h|--help)
echo "hello";;
esac
If
-eqis for numerical comparison, how come./deploy -hworked as expected?
Arithmetic evaluation normally prints an error message when given illegal expressions, but as it happens the two strings you're asking it to evaluate are syntactically valid.
-h negates the value of the undefined variable $h. The result is 0.--help is decrements the undefined variable $help. The result is -1.Try an invalid string and you'll get an error.
$ ./deploy 'foo bar'
bash: [[: foo bar: syntax error in expression (error token is "bar")
$ ./deploy @
bash: [[: @: syntax error in expression (error token is "@")