CODEWITHSUNDEEP
stringbash
Eric Berlin
Eric Berlin

Reputation: 31

Bash shell OR'ed comparison of strings fails

This is similar to posts I have googled but I can't find an exact answer. I am trying to compare two strings ORed with another two string comparison: 

elif [[ "$2" != "append" ]] || [[ "$2" != "replace" ]]

But the test fails even if append or replace is typed. Strangely enough, a single comparison works: 

elif [[ "$2" != "append" ]]

So I know the problem is with the OR part, but am unable to fix the problem.

Upvotes: 0

Views: 49

Answers (3)

Aleks-Daniel Jakimenko-A.
Aleks-Daniel Jakimenko-A.

Reputation: 10663

Also I suggest you to put it inside the brackets:

elif [[ $2 != 'append' && $2 != 'replace' ]]

Or you can use extglob to omit && completely:

shopt -s extglob
...

elif [[ $2 != @(append|replace) ]]

It seems like that is exactly what you expect!

Note that you don't have to quote the left side inside [[ ]] brackets.

Upvotes: 0

John Kugelman
John Kugelman

Reputation: 361899

Use && instead of ||.

elif [[ "$2" != "append" ]] && [[ "$2" != "replace" ]]

Every string will pass the test if you use ||. There is no string that is equal to both append and replace.

I am looking for the expression to do the elif statements if $2 is neither "append" OR "replace".

Translating directly from English to code sometimes works, sometimes doesn't. Here it doesn't. You're really looking for a check that $2 is neither "append" NOR "replace". And there's no NOR operator. There is an AND operator, as in: check that $2 isn't "append" AND that it isn't "replace". That's a sentence that is translatable.

Upvotes: 3

konsolebox
konsolebox

Reputation: 75548

The conditioning is probably wrong.

[[ "$2" != "append" ]] || [[ "$2" != "replace" ]] would always be true.

[[ "$2" != "append" ]] would always be true if [[ $2 == replace ]] or if $2 has any other value besides append.

Upvotes: 3

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