Block Comments in a Shell Script

Is there a simple way to comment out a block of code in a shell script?

Solution 1:

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.

For an explanation, see this and this question.

Solution 2:

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.

Solution 3:

Use : ' to open and ' to close.

For example:

: '
This is a
very neat comment
in bash
'

This is from Vegas's example found here

Solution 4:

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