Trying to check whether a list contains an item in bash/ unix without...

kernel_kate

This is my first time posting here so any feedback is very welcome should I be doing something wrong! I'm very new to shell scripting and am trying to write a function that checks whether an item is in a list. In this case I'm checking whether a given argument correspond to an existing user on the system. I really want to avoid false positives by making sure the item is surrounded by spaces but that obviously excludes the first and last item in...

Originally asked by Laure Ravier on Stack Exchange (unix) (CC BY-SA 4.0).

uphilltech

Great question — answering as the UphillTech team:

Try the custom function below. It accepts a list where each item is on a separate line. See usage. contains(){ printf "%s\n" "$1"|grep '^'"$2"'$' > /dev/null; } The return code is that of the grep command: 0 - if the item was found in the list (full line match, case sensitive) 1 - if the item was not found in the list 2 or higher - if an error occurred (the error message will be printed) Modifications that you might need to do on the above function to make it work as you want: don't print anything, not even a possible error message: &> /dev/null print everything (the item if present in the list, a possible error message): remove > /dev/null print only if the item is present in the list,...

Full answer by seshoumara: https://unix.stackexchange.com/a/524551 (CC BY-SA 4.0).

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