Block Comments in a Shell Script
Block Comments in a Shell Script
Question
Is there a simple way to comment out a block of code in a shell script?
Accepted Answer
In bash:
#!/bin/bash
echo before comment
: <<'END'
bla bla
blurfl
END
echo after comment
The '
and '
around the END
delimiter are important, otherwise things inside the block like for example $(command)
will be parsed and executed.
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There is no block comment on shell script.
Using vi
(yes, vi
) you can easily comment from line n to m
<ESC>
:10,100s/^/#/
(that reads, from line 10 to 100 substitute line start (^) with a # sign.)
and un comment with
<ESC>
:10,100s/^#//
(that reads, from line 10 to 100 substitute line start (^) followed by # with noting //.)
vi
is almost universal anywhere where there is /bin/sh
.
You can use:
if [ 1 -eq 0 ]; then
echo "The code that you want commented out goes here."
echo "This echo statement will not be called."
fi
The following should work for sh
,bash
, ksh
and zsh
.
The blocks of code to be commented can be put inside BEGINCOMMENT
and ENDCOMMENT
:
[ -z $BASH ] || shopt -s expand_aliases
alias BEGINCOMMENT="if [ ]; then"
alias ENDCOMMENT="fi"
BEGINCOMMENT
echo "This line appears in a commented block"
echo "And this one too!"
ENDCOMMENT
echo "This is outside the commented block"
Executing the above code would result in:
This is outside the commented block
In order to uncomment the code blocks thus commented, say
alias BEGINCOMMENT="if : ; then"
instead of
alias BEGINCOMMENT="if [ ]; then"
in the example above.