[SOLVED] Case statement fallthrough?

Issue

In popular imperative languages, switch statements generally "fall through" to the next level once a case statement has been matched.

Example:

int a = 2;
switch(a)
{
   case 1:
      print "quick ";
   case 2: 
      print "brown ";
   case 3: 
      print "fox ";
      break;
   case 4:
      print "jumped ";
}

would print "brown fox".

However the same code in bash

A=2
case $A in
2)
  echo "QUICK"
  ;&
2)
  echo "BROWN"
  ;&
3)
  echo "FOX"
  ;&
4)
  echo "JUMPED"
  ;&
esac

only prints "BROWN"

How do I make the case statement in bash "fall through" to the remaining conditions like the first example?

(edit: Bash version 3.2.25, the ;& statement (from wiki) results in a syntax error)

running:

test.sh:

#!/bin/bash
A=2
case $A in
1)
  echo "QUICK"
  ;&
2)
  echo "BROWN"
  ;&
3)
  echo "FOX"
  ;&
esac

Gives:

./test.sh: line 6: syntax error near unexpected token ;' ./test.sh:
line 6: ;&'

Solution

The ;& and ;;& operators were introduced in bash 4.0, so if you want to stick with a five year old version of bash, you'll either have to repeat code, or use ifs.

if (( a == 1)); then echo quick; fi
if (( a > 0 && a <= 2)); then echo brown; fi 
if (( a > 0 && a <= 3)); then echo fox; fi
if (( a == 4)); then echo jumped; fi

or find some other way to achieve the actual goal.

(On a side note, don't use all uppercase variable names. You risk overwriting special shell variables or environment variables.)

Answered By - geirha

Answer Checked By - Terry (WPSolving Volunteer)