I have defined the following variable:
myVar=true
now I'd like to run something along the lines of this:
if [ myVar ]
then
echo "true"
else
echo "false"
fi
The above code does work, but if I try to set
myVar=false
it will still output true. What might be the problem?
edit: I know I can do something of the form
if [ "$myVar" = "true" ]; then ...
but it is kinda awkward.
Thanks
This question is related to
bash
bash doesn't know boolean variables, nor does test (which is what gets called when you use [).
A solution would be:
if $myVar ; then ... ; fi
because true and false are commands that return 0 or 1 respectively which is what if expects.
Note that the values are "swapped". The command after if must return 0 on success while 0 means "false" in most programming languages.
SECURITY WARNING: This works because BASH expands the variable, then tries to execute the result as a command! Make sure the variable can't contain malicious code like rm -rf /
Note that the if $myVar; then ... ;fi construct has a security problem you might want to avoid with
case $myvar in
(true) echo "is true";;
(false) echo "is false";;
(rm -rf*) echo "I just dodged a bullet";;
esac
You might also want to rethink why if [ "$myvar" = "true" ] appears awkward to you. It's a shell string comparison that beats possibly forking a process just to obtain an exit status. A fork is a heavy and expensive operation, while a string comparison is dead cheap. Think a few CPU cycles versus several thousand. My case solution is also handled without forks.
Source: Stackoverflow.com