sh if or loop

[jdb91]

This should be easy but I've forgotten how to do it and it's driving me crazy.

All I want to do is this:



#!/bin/sh

if [ $FOO != "A" || "B" || "C" ]; then
exit 1
fi


But it's saying the syntax is wrong, so I tried



if [ $FOO != "A" ] || [ $FOO != "B" ]; then
exit 1
fi


But that exits if FOO = A because when it hits the B check, they don't match.

I'm sure this is simple, but I can't remember how to do it.

Cheers.

[Some Penguin]

(x != A) && (x != B) <=> !((x == A) || (x == B))

What you're doing there, '(x != A) || (x != B)' will always be true if A != B because one of the two clauses must be true.

[jdb91]

(x != A) && (x != B) <=> !((x == A) || (x == B))

What you're doing there, '(x != A) || (x != B)' will always be true if A != B because one of the two clauses must be true.

Ah of course, stupid me, I've done that before!

Many thanks.

[geirha]

On a side note, use lowercase variable names, and quote parameter expansions; "$foo", not $foo.

I'd use case for that particular case



case "$foo" in
A|B|C) echo "Great!";;
*) echo "Not great"; exit 1;;
esac

[bashologist]

As Some Penguin said case probably is best for that situation, but to correct your if syntax you didn't quote the $FOO which might lead to the syntax error in practice. You don't need quotes if you used double [[.

So rather than this.


if [ $FOO != "A" ] || [ $FOO != "B" ]; then
exit 1
fi


You maybe could've done this.


if [[ $FOO != "A" || $FOO != "B" ]]; then
exit 1
fi


I may be wrong but I'm just pointing that out.

[geirha]

@bashologist That's good bash syntax, yes, but the OP is writing an sh script, not bash. Also, the logic is still wrong...

http://en.wikipedia.org/wiki/De_Morgan%27s_laws