How to build a conditional assignment in bash?

Question

I'm looking a way to build conditional assignments in bash:

In Java it looks like this:

int variable= (condition) ? 1 : 0;

Thanks in advance

Answer 1

As per Jonathan's comment:

variable=$(( 1 == 1 ? 1 : 0 ))  

EDIT:

I revised the original answer which just echo'd the value of the condition operator, it didn't actually show any assignment.

Answer 2

If you want a way to define defaults in a shell script, use code like this:

: ${VAR:="default"}

Yes, the line begins with ':'. I use this in shell scripts so I can override variables in ENV, or use the default.

This is related because this is my most common use case for that kind of logic. ;]

Answer 3

another way

case "$variable" in
  condition ) result=1;;
  *) result=0;;
esac

Answer 4

myvar="default" && [[ <some_condition_is_true> ]]  && myvar="non-default"

real examples:

DELIM="" && [[ "$APP_ENV_RESOLVED" != "" ]] && DELIM=$INNER_DELIM

The condition can be "(( ... ))" as well:

filepath=/proc/drbd && (( $# > 0 )) && filepath=$1