I have file(file.txt) contains the following
aa=testing
bb=hello
cc=hi
Expected result
the value of aa is testing
How to use grep to find the value of aa?
Use a positive lookbehind in grep:
grep -Po "(?<=aa=).*" file.txt
Output
testing
grep -oP 'aa=\K.*' file.txt
Output:
testing
See: http://www.charlestonsw.com/perl-regular-expression-k-trick/
awk -F= '/^aa=/ { print $2 }' file
sed -n '/^aa=/s|^.*=||p' file
sed -n 's|^aa=||p' file
Output:
testing
Related
I have a bunch of strings that I have to fetch the 'port_num' from -
"76 : client=new; tags=circ, LINK; port_num=switch01; far_port=Gi1/0"
The word might be in a different place in the string and it might be a different length, but it always says 'port_num=' before it and ';' after it...
I only want this bit- 'switch01'
Currently I use-
| grep -Eo 'port_num=.+' | cut -d"=" -f2 | cut -d";" -f1'
But there has got to be a better way
You can try grep -oP '(?<=port_num=).+(?=;)', if you run this:
echo "76 : client=new; tags=circ, LINK; port_num=switch01; far_port=Gi1/0" \
| grep -oP '(?<=port_num=).+(?=;)'
result will be:
switch01
Updated answer: grep -oP '(?<=port_num=)[^;]+(?=;)'
This is what I would use:
... | grep -E 'port_num=.+' | sed 's/^.*port_num=\([^;]*\).*$/\1/'
This works with or without the -o on grep, and the availability of -P will depend on the version of grep you have. (e.g., my grep does not have it). I'm not saying the other answers that rely on -P aren't any good -- they look fine to me. But grep -P will be less portable.
IMHO, piping grep with sed allows each utility to do what it specializes in -- grep is for selecting lines, sed is for modifying lines.
This can be done in a simple sed command:
s="76 : client=new; tags=circ, LINK; port_num=switch01; far_port=Gi1/0"
sed 's/.*port_num=\([^;]*\);.*/\1/' <<< "$s"
switch01
... | grep -Po 'port_num.+(?=;)'
This uses grep's Perl Compatible Regular Expression (PCRE) syntax. The (?=;) is a look-ahead assertion which looks for a match with ";" but doesn't include it in the matched output.
This produces:
port_num=switch01
As #Vladimir Kovpak noted, if you want to exclude the "port_num=" string from this output, add a look-behind assertion:
... | grep -Po '(?<=port_num).+(?=;)'
I would like some advice on how to exclude a word in a line using grep but still keep the line?
So I have tried:
grep -v '1.942134' results.tbl | egrep '*.fits' results.tbl
to try to list all the string with extension .fits but exclude "1.942134" in the sentence but it still returns the full lines.
Any advice?
Or you can use awk
awk '/\.fits/ && !/1\.942134/` results.tbl
PS you should escape the . in both sed and awk or else it will mean just any character.
You should pipe to sed. Sed has lots of abilities, some of them more complicated than others, but one of its best is regexp substitutions.
grep '\.fits$' | sed 's/1.942134//'
below text is in a file,
Pseudo name=Apple
Code=42B
state=fault
Pseudo name=Prance
Code=43B
state=good
need to grep for 42B from the above file so that the output only should display
Pseudo name=Apple
Code=42B
state=fault
perl -00ne "print if /Code=42B/i"
use before and after modifiers
grep -B 1 -A 1 42B file.txt
Unfortunately, only on AIX:
grep -p Code=42B file.txt
Basically I have a very large text file and each line contains
tag=yyyyy;id=xxxxx;db_ref=zzzzz;
What I want is to grep out the id, but the id can change in length and form, I was wondering if its possible to use grep -o and then grep for "id=" then extract everything that comes after it until the semicolon?
You could do:
$ grep -o 'id=[^;]*' file
And if you don't want to inlcude the id= part you can using positive look-behind:
$ grep -Po '(?<=id=)[^;]*' file
try :
grep -Po "(?<=id=)[^;]*" file
Via grep:
grep -o 'id=[^;]*'
Via awk:
awk -F';' '{ print $2}' testlog
id=xxxxx
edit: see sudo_O's answer for the look-behind. it's more to the point of your question, IMO.
You could try this awk. It should also work if there are multiple id= entries per line and it would not give a false positive for ...;pid=blabla;...
awk '/^id=/' RS=\; file
Try the following:
grep -oP 'id=\K[^;]*' file
perl -lne 'print $1 if(/id=([^\;]*);/)' your_file
tested:
> echo "tag=yyyyy;id=xxxxx;db_ref=zzzzz; "|perl -lne 'print $1 if(/id=([^\;]*);/)'
xxxxx
>
I have a file that possibly contains bad formatting (in this case, the occurrence of the pattern \\backslash). I would like to use grep to return only the line numbers where this occurs (as in, the match was here, go to line # x and fix it).
However, there doesn't seem to be a way to print the line number (grep -n) and not the match or line itself.
I can use another regex to extract the line numbers, but I want to make sure grep cannot do it by itself. grep -no comes closest, I think, but still displays the match.
try:
grep -n "text to find" file.ext | cut -f1 -d:
If you're open to using AWK:
awk '/textstring/ {print FNR}' textfile
In this case, FNR is the line number. AWK is a great tool when you're looking at grep|cut, or any time you're looking to take grep output and manipulate it.
All of these answers require grep to generate the entire matching lines, then pipe it to another program. If your lines are very long, it might be more efficient to use just sed to output the line numbers:
sed -n '/pattern/=' filename
Bash version
lineno=$(grep -n "pattern" filename)
lineno=${lineno%%:*}
I recommend the answers with sed and awk for just getting the line number, rather than using grep to get the entire matching line and then removing that from the output with cut or another tool. For completeness, you can also use Perl:
perl -nE 'say $. if /pattern/' filename
or Ruby:
ruby -ne 'puts $. if /pattern/' filename
using only grep:
grep -n "text to find" file.ext | grep -Po '^[^:]+'
You're going to want the second field after the colon, not the first.
grep -n "text to find" file.txt | cut -f2 -d:
To count the number of lines matched the pattern:
grep -n "Pattern" in_file.ext | wc -l
To extract matched pattern
sed -n '/pattern/p' file.est
To display line numbers on which pattern was matched
grep -n "pattern" file.ext | cut -f1 -d: