CODEWITHSUNDEEP
bashconditional-statements
Oliver
Oliver

Reputation: 13

Bash if statement gives opposite response than expected

I think i'm missing something very obvious. But shouldn't the following code produce the opposite response? I thought if the statement "if s == d" is used and s is not equal to d then the if statement should return false and not run the following code. This is not what appears to happen. Can anyone explain what i've missed. I think it's something very obvious.

Thanks 

s=2
d=3
if ! [ "$s == $d" ]; then         echo "hello"; fi
if [ "$s == $d" ]; then         echo "hello"; fi
hello

Upvotes: 0

Views: 678

Answers (1)

that other guy
that other guy

Reputation: 123450

You are quoting the entire string "$s == $d" when you should be quoting the two arguments "$s" and "$d".

This means that instead of comparing $s to $d, you are checking whether "2 == 3" is a non-empty string (which it is).

This will correctly print "not equal":

s=2
d=3
if ! [ "$s" == "$d" ]; then echo "not equal"; fi
if [ "$s" == "$d" ]; then echo "equal"; fi

Upvotes: 2

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