I tried grep -v '^$'
in Linux and that didn't work. This file cam开发者_高级运维e from a Windows file system.
Try the following:
grep -v -e '^$' foo.txt
The -e
option allows regex patterns for matching.
The single quotes around ^$
makes it work for Cshell. Other shells will be happy with either single or double quotes.
UPDATE: This works for me for a file with blank lines or "all white space" (such as windows lines with \r\n
style line endings), whereas the above only removes files with blank lines and unix style line endings:
grep -v -e '^[[:space:]]*$' foo.txt
Keep it simple.
grep . filename.txt
Use:
$ dos2unix file
$ grep -v "^$" file
Or just simply awk:
awk 'NF' file
If you don't have dos2unix, then you can use tools like tr:
tr -d '\r' < "$file" > t ; mv t "$file"
grep -v "^[[:space:]]*$"
The -v makes it print lines that do not completely match
===Each part explained===
^ match start of line
[[:space:]] match whitespace- spaces, tabs, carriage returns, etc.
* previous match (whitespace) may exist from 0 to infinite times
$ match end of line
Running the code-
$ echo "
> hello
>
> ok" |
> grep -v "^[[:space:]]*$"
hello
ok
To understand more about how/why this works, I recommend reading up on regular expressions. http://www.regular-expressions.info/tutorial.html
If you have sequences of multiple blank lines in a row, and would like only one blank line per sequence, try
grep -v "unwantedThing" foo.txt | cat -s
cat -s
suppresses repeated empty output lines.
Your output would go from
match1
match2
to
match1
match2
The three blank lines in the original output would be compressed or "squeezed" into one blank line.
The same as the previous answers:
grep -v -e '^$' foo.txt
Here, grep -e
means the extended version of grep. '^$' means that there isn't any character between ^(Start of line) and $(end of line). '^' and '$' are regex characters.
So the command grep -v
will print all the lines that do not match this pattern (No characters between ^ and $).
This way, empty blank lines are eliminated.
I prefer using egrep
, though in my test with a genuine file with blank line your approach worked fine (though without quotation marks in my test). This worked too:
egrep -v "^(\r?\n)?$" filename.txt
Do lines in the file have whitespace characters?
If so then
grep "\S" file.txt
Otherwise
grep . file.txt
Answer obtained from: https://serverfault.com/a/688789
This code removes blank lines and lines that start with "#"
grep -v "^#" file.txt | grep -v ^[[:space:]]*$
awk 'NF' file-with-blank-lines > file-with-no-blank-lines
It's true that the use of grep -v -e '^$' can work, however it does not remove blank lines that have 1 or more spaces in them. I found the easiest and simplest answer for removing blank lines is the use of awk. The following is a modified a bit from the awk guys above:
awk 'NF' foo.txt
But since this question is for using grep I'm going to answer the following:
grep -v '^ *$' foo.txt
Note: the blank space between the ^ and *.
Or you can use the \s to represent blank space like this:
grep -v '^\s*$' foo.txt
I tried hard, but this seems to work (assuming \r
is biting you here):
printf "\r" | egrep -xv "[[:space:]]*"
Using Perl:
perl -ne 'print if /\S/'
\S
means match non-blank characters.
egrep -v "^\s\s+"
egrep already do regex, and the \s is white space.
The + duplicates current pattern.
The ^ is for the start
Use:
grep pattern filename.txt | uniq
Here is another way of removing the white lines and lines starting with the #
sign. I think this is quite useful to read configuration files.
[root@localhost ~]# cat /etc/sudoers | egrep -v '^(#|$)'
Defaults requiretty
Defaults !visiblepw
Defaults always_set_home
Defaults env_reset
Defaults env_keep = "COLORS DISPLAY HOSTNAME HISTSIZE INPUTRC KDEDIR
LS_COLORS"
root ALL=(ALL) ALL
%wheel ALL=(ALL) ALL
stack ALL=(ALL) NOPASSWD: ALL
Read lines from file exclude EMPTY Lines
grep -v '^$' folderlist.txt
folderlist.txt
folder1/test
folder2
folder3
folder4/backup
folder5/backup
Results will be:
folder1/test
folder2
folder3
folder4/backup
folder5/backup
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