How to comment out particular lines in a shell script

Can anyone suggest how to comment particular lines in the shell script other than #?

Suppose I want to comment five lines. Instead of adding # to each line, is there any other way to comment the five lines?

You can comment section of a script using a conditional.

For example, the following script:

DEBUG=false
if ${DEBUG}; then
echo 1
echo 2
echo 3
echo 4
echo 5
fi
echo 6
echo 7

would output:

6
7

In order to uncomment the section of the code, you simply need to comment the variable:

#DEBUG=false

(Doing so would print the numbers 1 through 7.)

Yes (although it's a nasty hack). You can use a heredoc thus:

#!/bin/sh

# do valuable stuff here
touch /tmp/a

# now comment out all the stuff below up to the EOF
echo <<EOF
...
...
...
EOF

What's this doing ? A heredoc feeds all the following input up to the terminator (in this case, EOF) into the nominated command. So you can surround the code you wish to comment out with

echo <<EOF
...
EOF

and it'll take all the code contained between the two EOFs and feed them to echo (echo doesn't read from stdin so it all gets thrown away).

Note that with the above you can put anything in the heredoc. It doesn't have to be valid shell code (i.e. it doesn't have to parse properly).

This is very nasty, and I offer it only as a point of interest. You can't do the equivalent of C's /* ... */

for single line comment add # at starting of a line
for multiple line comments add ' (single quote) from where you want to start & add ' (again single quote) at the point where you want to end the comment line.

You have to rely on '#' but to make the task easier in vi you can perform the following (press escape first):

:10,20 s/^/#

with 10 and 20 being the start and end line numbers of the lines you want to comment out

and to undo when you are complete:

:10,20 s/^#//