I want to use "awk" or "sed" to print all the lines that start with comm= from the file filex, Note that each line contains "comm=somthing"
for example : comm=rm , comm=ll, comm=ls ....
How can i achieve that ?
For lines that start with comm=
sed -n '/^comm=/p' filex
awk '/^comm=/' filex
If comm= is anywhere in the line then
sed -n '/comm=/p' filex
awk '/comm=/' filex
You could use grep also :
grep comm= filex
this will display all the lines containing comm=.
Here's an approach using grep:
grep -o '\<comm=[[:alnum:]]*\>'
This treats a word as consisting of alphanumeric characters; extend the character class as needed.
If grep is ok to use, you could give a try to:
grep -E "^comm=" file
Related
How do I delete all lines in a text file which do not start with the characters #, & or *? I'm looking for a solution using sed or grep.
Deleting lines:
With grep
From http://lowfatlinux.com/linux-grep.html :
The grep command selects and prints lines from a file (or a bunch of files) that match a pattern.
I think you can do something like this:
grep -v '^[\#\&\*]' yourFile.txt > output.txt
You can also use sed to do the same thing (check http://lowfatlinux.com/linux-sed.html ):
sed '^[\#\&\*]/d' yourFile.txt > output.txt
It's up to you to decide
Filtering lines:
My mistake, I understood you wanted to delete the lines. But if you want to "delete" all other lines (or filter the lines starting with the specified characters), then grep is the way to go:
grep '^[\#\&\*]' yourFile.txt > output.txt
sed -n '/^[#&*].*/p' input.txt > output.txt
this should work.
sed -ni '/^[#&*].*/p' input.txt
this one will edit the input file directly, be careful +
egrep '^(&|#|\*)' input.txt > output.txt
I would like to print the contents of a file, but all lines starting with # I want to ignore. I was trying some stuff with grep and awk, but it kept printing the whole file, or just printed the lines starting with #. I you could give me a push in the right way, or a grep/awk command that would print anyline in the file that does not start with #.
Use the -v option of grep to negate the condition:
grep -v '^#' file
You can use the ! operator:
awk '!/^ *#/ { print; }'
This negates the result of the match. I also included lines that start with spaces and then #, but you can tailor the regex how you like.
You could use grep to exclude all lines that begin with # using the -v option
grep -v '^#' filename
If you're a fan of sed:
sed '/^#/d' filename
This would also leave out lines with whitespace before the # :
awk '$1!~/^#/' file
or
grep -v '^[[:blank:]]*#' file
Here is the grep PCRE way,
grep -P '^(?!#)' file
I am trying to extract text between pattern1 (fixed) and pattern2 (this can be p2-1/p2-2).
can you please tell me how to achieve this in a single command?
A file starts with start and ends with either end or close
File1:
======
junktest
data
start
stackoverflow
sed
close
File2:
======
data2
start
stackoverflow
end
I can extract text from File1 with
sed -n "/start/,/close/p"
And from File2 with
sed -n "/start/,/end/p"
I need a single sed command to achieve both..
something like:
sed -n "/start/, /close or end /p"
Both GNU sed and BSD sed:
sed -nE '/start/,/close|end/p' file
This awk looks better
awk '/start/,/end|close/' file
sed -n -E "/Word1/,/Word2-1/p" | sed -n -E "/Word1/,/Word2-2/p"
Easy with awk:
$ awk '/start/{p=1}p{print}/end|close/{p=0}' file
In a very large file I need to find the position (line number) of a string, then extract the 2 lines above and below that string.
To do this right now - I launch vi, find the string, note it's line number, exit vi, then use sed to extract the lines surrounding that string.
Is there a way to streamline this process... ideally without having to run vi at all.
Maybe using grep like this:
grep -n -2 your_searched_for_string your_large_text_file
Will give you almost what you expect
-n : tells grep to print the line number
-2 : print 2 additional lines (and the wanted string, of course)
You can do
grep -C 2 yourSearch yourFile
To send it in a file, do
grep -C 2 yourSearch yourFile > result.txt
Use grep -n string file to find the line number without opening the file.
you can use cat -n to display the line numbers and then use awk to get the line number after a grep in order to extract line number:
cat -n FILE | grep WORD | awk '{print $1;}'
although grep already does what you mention if you give -C 2 (above/below 2 lines):
grep -C 2 WORD FILE
You can do it with grep -A and -B options, like this:
grep -B 2 -A 2 "searchstring" | sed 3d
grep will find the line and show two lines of context before and after, later remove the third one with sed.
If you want to automate this, simple you can do a Shell Script. You may try the following:
#!/bin/bash
VAL="your_search_keyword"
NUM1=`grep -n "$VAL" file.txt | cut -f1 -d ':'`
echo $NUM1 #show the line number of the matched keyword
MYNUMUP=$["NUM1"-1] #get above keyword
MYNUMDOWN=$["NUM1"+1] #get below keyword
sed -n "$MYNUMUP"p file.txt #display above keyword
sed -n "$MYNUMDOWN"p file.txt #display below keyword
The plus point of the script is you can change the keyword in VAL variable as you like and execute to get the needed output.
I tried grep -v '^$' in Linux and that didn't work. This file came 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