CODEWITHSUNDEEP
bashunix
creativeindian1984
creativeindian1984

Reputation: 13

check arguments in list unix

I have a list of names in a list (say name.txt) which has the set of name lists one by one.

name.txt

babu
praveen
kamal
sneha

This name will be passed as run time argument $1 in my bash script. Now I have to do a match to check if the inputted name is in my list or not.

If it's not there then I will print saying invalid name and exit. Can you help me with this?

I have tried this with 

if [[ "$1" != "babu" || "$1" != "praveen" || "$1" != "kamal" ... ]]; then
    exit
fi

but this doesn't look good professionally.

Is there any other simple and decent way to achieve this?

Upvotes: 0

Views: 173

Answers (1)

Tom Fenech
Tom Fenech

Reputation: 74595

I guess you could use grep:

if ! grep -qFxf name.txt <<<"$1"; then
    exit
fi
  • -q is "quiet mode" - the exit status indicates a match
  • -F matches fixed strings, rather than regular expressions
  • -x is a full line match, so names that are substrings won't match
  • -f means that the list of patterns are contained within a file

The string contained within the first argument is passed to grep using a here string. This is a bash feature, so if you're using another shell, you could go with something like if ! echo "$1" | grep -qFxf name.txt instead (but the bash way is preferred as it saves invoking a subprocess).

If you want to ensure that any error output from grep is suppressed, you can add 2>/dev/null to the command, as suggested in the comments. This prevents any messages sent to standard error from being displayed.

Upvotes: 2

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