I am looking for a way to have fallthrough happen based on an if condition within a case condition in bash. For example:
input="foo"
VAR="1"
case $input in
foo)
if [ $VAR = "1" ]; then
# perform fallthrough
else
# do not perform fallthrough
fi
;;
*)
echo "fallthrough worked!"
;;
esac
In the above code, if the variable VAR is 1, I would like to have the case condition perform fallthrough.
You can't. The way to have a case fall through is to replace the ;; separator with ;& (or ;;&). And it's a syntax error to put that inside an if.
You could write the whole logic out as a regular conditional:
if [ "$input" != "foo" ] || [ "$VAR" = 1 ]; then
one branch ...
else # $input = "foo" && $VAR != 1
another branch...
fi
The following script turns your test "inside out" in the sense that we test $var first and then perform the fallthrough (using ;& in a case) depending on $input.
We do this because the question of whether or not to "perform the fallthrough" is really only dependent on $input if $var is 1. If it's any other value, the question of whether to do the fallthrough does not even have to be asked.
#/bin/bash
input='foo'
var='1'
case $var in
1)
case $input in
foo)
echo 'perform fallthrough'
;&
*)
echo 'fallthough worked'
esac
;;
*)
echo 'what fallthrough?'
esac
Or, without case:
if [ "$var" -eq 1 ]; then
if [ "$input" = 'foo' ]; then
echo 'perform fallthrough'
fi
echo 'fallthough worked'
else
echo 'what fallthrough?'
fi
I'd suggest restructuring your logic: put the "fallthrough" code into a function instead:
fallthrough() { echo 'fallthrough worked!'; }
for input in foo bar; do
for var in 1 2; do
echo "$input $var"
case $input in
foo)
if (( var == 1 )); then
echo "falling through"
fallthrough
else
echo "not falling through"
fi
;;
*) fallthrough;;
esac
done
done
outputs
foo 1
falling through
fallthrough worked!
foo 2
not falling through
bar 1
fallthrough worked!
bar 2
fallthrough worked!
Not something that I would do, but you could achieve something approaching with:
shopt -s extglob # for !(*)
default='*'
case $input in
(foo)
if [ "$VAR" = 1 ]; then
echo going for fallthrough
else
echo disabling fallthrough
default='!(*)'
fi ;;&
($default)
echo fallthrough
esac
Test both variables at once (bash 4.0-alpha+):
#!/bin/bash
while (($#>1)); do
input=$1 VAR=$2
echo "input=${input} VAR=${VAR}"; shift 2
if [ "$VAR" = 1 ]; then new=1; else new=0; fi
case $input$new in
foo0) echo "do not perform fallthrough" ;;
foo*) echo "perform fallthrough" ;&
*) echo "fallthrough worked!" ;;
esac
echo
done
On testing:
$ ./script foo 0 foo 1 bar baz
input=foo VAR=0
do not perform fallthrough
input=foo VAR=1
perform fallthrough
fallthrough worked!
input=bar VAR=baz
fallthrough worked!
Clean and simple.
Understand that the tested value ($new) must have only two possible values, that is why the if clause is there, to transform VAR to a Boolean value. If VAR may be made to be a Boolean, then test for 0 (not 1) in the case and remove the if.
You can make the fallthrough default but place a condition that the code only executes only if the condition is met
#!/bin/bash
input='foo'
var='1'
case $input in
foo)
echo "Do fall through"
;& #always fall through
*)
if [ $var = "1" ] #execute only if condition matches
then
echo "fallthrough took place"
fi
esac
But as ilkkachu suggested you can also use conditions rather than switch.
If you don't mind someone complaining about they don't understand your code, you could simply switch the order of the two conditionals:
input="foo"
VAR="1"
if
case $input in
foo)
[ $VAR = "1" ]
;;
esac
then
echo "fallthrough worked!"
fi
Or:
input="foo"
VAR="1"
case $input in
foo)
[ $VAR = "1" ]
;;
esac &&
echo "fallthrough worked!"
Simple and clear (at least to me). case doesn't support fallthrough itself. But you can replace *) with && after esac to make it respect to the return values of other branches.
New ;& and ;;& operators were introduced in Bash
4.0
and although they both might be useful in similar situations I think
they are of no use in your case. This is what man bash says about
these operators:
If the ;; operator is used, no subsequent matches are attempted after the first pattern match. Using ;& in place of ;; causes execution to continue with the list associated with the next set of patterns. Using ;;& in place of ;; causes the shell to test the next pattern list in the statement, if any, and execute any associated list on a successful match.
In other words, ;& is a fall through and as we know it from C and
;;& makes bash check remaining cases instead of returning from
case block entirely. You can find a nice example of ;;& in action
here: https://stackoverflow.com/a/24544780/3691891.
That being said, neither ;& nor ;;& could be used in your script
because both of them would go to *) that would be always run.
The following script works and does what you want without re-arranging the logic but consider it only as an example and never rely on it, it's too fragile. I've taken the idea from here:
#!/usr/bin/env bash
function jumpto
{
label=$1
cmd=$(sed -n "/$label:/{:a;n;p;ba};" "$0" | grep -v ':$')
cmd=$(echo "$cmd" | sed 's,;;,,')
cmd=$(echo "$cmd" | sed 's,esac,,')
eval "$cmd"
}
input="foo"
VAR="1"
case $input in
foo)
if [ $VAR = "1" ]; then
printf "perform fallthrough\n"
jumpto ft
else
printf "do not perform fallthrough\n"
fi
;;
*)
ft:
echo "fallthrough worked!"
;;
esac
