Bash case menu wont call to function: Command not found

I have tried to create a bash script which has a case menu, that calls a function when chosen:

#!/bin/bash # Menu PS3='Vælg en funktion: ' options=("1: fileSplitter" "2: rowGrepper" "3: grepCounter?" "4: exit") select opt in "${options[@]}" do case $opt in "1: fileSplitter") fileSplitter ;; "2: rowGrepper") rowGrepper ;; "3: grepCounter?") grepCounter ;; "4: exit") break ;; *) echo "$REPLY er ikke tilgængeligt.." ;; esac done # Functions function fileSplitter() { echo "fileSplitter function" } function rowGrepper() { echo "rowGrepper function" } function grepCounter() { echo "grepCounter" } 

But when choosing option one, two or three that calls the specified functions, i get the error: fileSplitter: command not found

1 Answer

Thank you, the problem was caused by the order of menu and functions. It works perfectly when stating the functons before the menu:

#!/bin/bash #Functions: function fileSplitter() { echo "fileSplitter function" } function rowGrepper() { echo "rowGrepper function" } function grepCounter() { echo "grepCounter" } # Menu: PS3='Vælg en funktion: ' options=("1: fileSplitter" "2: rowGrepper" "3: grepCounter?" "4: exit") select opt in "${options[@]}" do case $opt in "1: fileSplitter") fileSplitter ;; "2: rowGrepper") rowGrepper ;; "3: grepCounter?") grepCounter ;; "4: exit") break ;; *) echo "$REPLY er ikke tilgængeligt.." ;; esac done 

Result:

./csv_helper.sh 1) 1: fileSplitter 2) 2: rowGrepper 3) 3: grepCounter? 4) 4: exit Vælg en funktion: 1 fileSplitter function