[linux] Delete empty lines using sed

I am trying to delete empty lines using sed:

sed '/^$/d'

but I have no luck with it.

For example, I have these lines:

xxxxxx


yyyyyy


zzzzzz

and I want it to be like:

xxxxxx
yyyyyy
zzzzzz

What should be the code for this?

Another option without sed, awk, perl, etc

strings $file > $output

strings - print the strings of printable characters in files.

My bash-specific answer is to recommend using perl substitution operator with the global pattern g flag for this, as follows:

$ perl -pe s'/^\n|^[\ ]*\n//g' $file
xxxxxx
yyyyyy
zzzzzz

This answer illustrates accounting for whether or not the empty lines have spaces in them ([\ ]*), as well as using | to separate multiple search terms/fields. Tested on macOS High Sierra and CentOS 6/7.

FYI, the OP's original code sed '/^$/d' $file works just fine in bash Terminal on macOS High Sierra and CentOS 6/7 Linux at a high-performance supercomputing cluster.

With help from the accepted answer here and the accepted answer above, I have used:

$ sed 's/^ *//; s/ *$//; /^$/d; /^\s*$/d' file.txt > output.txt

`s/^ *//`  => left trim
`s/ *$//`  => right trim
`/^$/d`    => remove empty line
`/^\s*$/d` => delete lines which may contain white space

This covers all the bases and works perfectly for my needs. Kudos to the original posters @Kent and @kev

You can do something like that using "grep", too:

egrep -v "^$" file.txt

For me with FreeBSD 10.1 with sed worked only this solution:

sed -e '/^[     ]*$/d' "testfile"

inside [] there are space and tab symbols.

test file contains:

fffffff next 1 tabline ffffffffffff

ffffffff next 1 Space line ffffffffffff

ffffffff empty 1 lines ffffffffffff

============ EOF =============

You can say:

sed -n '/ / p' filename    #there is a space between '//'

sed '/^$/d' should be fine, are you expecting to modify the file in place? If so you should use the -i flag.

Maybe those lines are not empty, so if that's the case, look at this question Remove empty lines from txtfiles, remove spaces from start and end of line I believe that's what you're trying to achieve.

I believe this is the easiest and fastest one:

cat file.txt | grep .

If you need to ignore all white-space lines as well then try this:

cat file.txt | grep '\S'

Example:

s="\
\
a\
 b\
\
Below is TAB:\
    \
Below is space:\
 \
c\
\
"; echo "$s" | grep . | wc -l; echo "$s" | grep '\S' | wc -l

outputs

7
5

sed

• '/^[[:space:]]*$/d'
• '/^\s*$/d'
• '/^$/d'
• -n '/^\s*$/!p'

grep

• .
• -v '^$'
• -v '^\s*$'
• -v '^[[:space:]]*$'

awk

• /./
• 'NF'
• 'length'
• '/^[ \t]*$/ {next;} {print}'
• '!/^[ \t]*$/'

You are most likely seeing the unexpected behavior because your text file was created on Windows, so the end of line sequence is \r\n. You can use dos2unix to convert it to a UNIX style text file before running sed or use

sed -r "/^\r?$/d"

to remove blank lines whether or not the carriage return is there.

This works in awk as well.

awk '!/^$/' file
xxxxxx
yyyyyy
zzzzzz

I am missing the awk solution:

awk 'NF' file

Which would return:

xxxxxx
yyyyyy
zzzzzz

How does this work? Since NF stands for "number of fields", those lines being empty have 0 fiedls, so that awk evaluates 0 to False and no line is printed; however, if there is at least one field, the evaluation is True and makes awk perform its default action: print the current line.