Using the not equal operator for string comparison

I tried to check if the PHONE_TYPE variable contains one of three valid values.

if [ "$PHONE_TYPE" != "NORTEL" ] || [ "$PHONE_TYPE" != "NEC" ] ||
   [ "$PHONE_TYPE" != "CISCO" ]
then
    echo "Phone type must be nortel,cisco or nec"
    exit
fi

The above code did not work for me, so I tried this instead:

if [ "$PHONE_TYPE" == "NORTEL" ] || [ "$PHONE_TYPE" == "NEC" ] ||
   [ "$PHONE_TYPE" == "CISCO" ]
then
    :        # do nothing
else
    echo "Phone type must be nortel,cisco or nec"
    exit
fi

Are there cleaner ways for this type of task?

Answers:

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Method 1

I guess you’re looking for:

if [ "$PHONE_TYPE" != "NORTEL" ] && [ "$PHONE_TYPE" != "NEC" ] &&
   [ "$PHONE_TYPE" != "CISCO" ]

The rules for these equivalents are called De Morgan’s laws and in your case meant:

not(A || B || C) => not(A) && not(B) && not (C)

Note the change in the boolean operator or and and.

Whereas you tried to do:

not(A || B || C) => not(A) || not(B) || not(C)

Which obviously doesn’t work.

Method 2

A much shorter way would be:

if [[ ! $PHONE_TYPE =~ ^(NORTEL|NEC|CISCO)$ ]]; then 
  echo "Phone type must be nortel, cisco or nec."
fi

• ^ – To match a starting at the beginning of line
• $ – To match end of the line
• =~ – Bash’s built-in regular expression comparison operator

Method 3

Good answers, and an invaluable lesson 😉 Only want to supplement with a note.

What type of test one choose to use is highly dependent on code, structure, surroundings etc.

An alternative could be to use a switch or case statement as in:

case "$PHONE_TYPE" in
"NORTEL"|"NEC"|"CISCO")
    echo "OK"
    ;;
*)
    echo "Phone type must be nortel,cisco or nec"
    ;;
esac

As a second note you should be careful by using upper-case variable names. This is to prevent collision between variables introduced by the system, which almost always is all upper case. Thus $phone_type instead of $PHONE_TYPE.

Though that one is safe, if you have as habit using all upper case, one day you might say IFS="boo" and you’re in a world of hurt.

It will also make it easier to spot which is what.

Not a have to but a would strongly consider.

It is also presumably a good candidate for a function. This mostly makes the code easier to read and maintain. E.g.:

valid_phone_type()
{
    case "$1" in
    "NORTEL"|"NEC")
        return 0;;
    *)
        echo "Model $1 is not supported"
        return 1;;
    esac
}

if ! valid_phone_type "$phone_type"; then
    echo "Bye."
    exit 1
fi

Method 4

You should use ANDs, not ORs.

if [ "$PHONE_TYPE" != "NORTEL" ] && [ "$PHONE_TYPE" != "NEC" ] && [ "$PHONE_TYPE" != "CISCO" ]
then

or

if [ "$PHONE_TYPE" != "NORTEL" -a "$PHONE_TYPE" != "NEC" -a "$PHONE_TYPE" != "CISCO" ]
then

Method 5

To correct an above answer (as I can’t comment yet):

PHONE_TYPE="NORTEL"
if [[ $PHONE_TYPE =~ ^(NORTEL|NEC|CISCO|SPACE TEL)$ ]]; then 
  echo "Phone type accepted."
else
  echo "Error! Phone type must be NORTEL, CISCO or NEC."
fi

Please note that you need at least bash 4 for this use of =~
It doesn’t work in bash 3.

I tested on MS Windows 7 using bash 4.3.46 (works fine) and bash 3.1.17 (didn’t work)

The LHS of the =~ should be in quotes. Above, PHONE_TYPE=”SPACE TEL” would match too.

Method 6

Just a variation proposal based on @0x80 solution:

# define phone brand list
phoneBrandList=" NORTEL NEC CISCO" ## separator is space with an extra space in first place

# test if user given phone is contained in the list
if [[ ${phoneBrandList} =~ (^|[[:space:]])"${userPhoneBrand}"($|[[:space:]]) ]]; then
    echo "found it !"
fi

Method 7

Use [[ instead

if [[ "$PHONE_TYPE" != "NORTEL" ]] || [[ "$PHONE_TYPE" != "NEC" ]] || 
   [[ "$PHONE_TYPE" != "CISCO" ]]
then
echo "Phone type must be nortel,cisco or nec"
exit 1
fi

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0