I have a file text.txt which contains the below words.
1. moon,one
2. sun,two
3. well,three
4. doll,four
if i grep this file using sun
grep -i sun text.txt
I will get the output
sun,two
But, my requirement is I need to grep with the word which is starting with sun not exactly sun.
grep -i sunlight text.txt
Here I need the same output for grep -i sun text.txt.
You don't need awk or gawk, nor sed. Just do
grep -o 'sun.*'
Other more complex / elegant solutions may be available depending on the system you are using.
What you are looking for are regular expressions.
In your case, it would be
grep -i 'sun.*' text.txt
Try using -o, as showed in the documentation.
The -o make grep return only the matched part. You can also use regular expressions.
grep -io sun text.txt
Is this what you're looking for?
awk -F ',' '/^[SsuUnN]/ {print $0}' test.txt
or if you want to search the pattern "sun" in general from the input_file, then use this:
awk -F ',' 'BEGIN{IGNORECASE=1} /sun/ {print $0}' test.txt
Related
I need to grep some pattern and further i need to print some output within that. Currently I am using the below command which is working fine. But I like to eliminate using multiple pipe and want to use single awk command to achieve the same output. Is there a way to do it using awk?
root#Server1 # cat file
Jenny:Mon,Tue,Wed:Morning
David:Thu,Fri,Sat:Evening
root#Server1 # awk '/Jenny/ {print $0}' file | awk -F ":" '{ print $2 }' | awk -F "," '{ print $1 }'
Mon
I want to get this output using single awk command. Any help?
You can try something like:
awk -F: '/Jenny/ {split($2,a,","); print a[1]}' file
Try this
awk -F'[:,]+' '/Jenny/{print $2}' file.txt
It is using muliple -F value inside the [ ]
The + means one or more since it is treated as a regex.
For this particular job, I find grep to be slightly more robust.
Unless your company has a policy not to hire people named Eve.
(Try it out if you don't understand.)
grep -oP '^[^:]*Jenny[^:]*:\K[^,:]+' file
Or to do a whole-word match:
grep -oP '^[^:]*\bJenny\b[^:]*:\K[^,:]+' file
Or when you are confident that "Jenny" is the full name:
grep -oP '^Jenny:\K[^,:]+' file
Output:
Mon
Explanation:
The stuff up until \K speaks for itself: it selects the line(s) with the desired name.
[^,:]+ captures the day of week (in this case Mon).
\K cuts off everything preceding Mon.
-o cuts off anything following Mon.
I am trying to find the number an even occured in my log file.
Command:
grep -Eo "2016-08-30" applciationLog.log* -c
Output:
applciationLog.log.1:0
applciationLog.log.2:0
applciationLog.log.3:0
applciationLog.log.4:0
applciationLog.log.5:7684
applciationLog.log.6:9142
applciationLog.log.7:8699
applciationLog.log.8:0
What I actually need is sum of all these values 7684 + 9142 + 8699 = 25525. Any suggestion I can do it? Anything I can append to the grep to enable it.
Any help or pointers are welcome and appreciated.
If you want to keep your grep command, pipe its output to awk, the quick and dirty way is down here:
grep -Eo "aaa" -c aaa.txt bbb.txt -c | awk 'BEGIN {cnt=0;FS=":"}; {cnt+=$2;}; END {print cnt;}'
Or use use awk regex directly:
awk 'BEGIN {cnt=0}; {if(/aaa/) {cnt+=1;}}; END {print cnt;}' aaa.txt bbb.txt
As addition to the already given answer by ghoti:
You can avoid awk -F: by using grep -h:
grep -c -h -F "2016-08-30" applicationLog.log* | awk '{n+=$0} END {print n}'
This means no filenames and only the counts are printed by grep and we can use the first field for the addition in awk.
See if this works for you:
grep -Eo "2016-08-30" applciationLog.log* -c | awk -F':' 'BEGIN {sum = 0;} {sum += $2;} END {print sum;}'
We use awk to split each line up with a delimeter of :, sum up the numbers for each line, and print the result at the end.
The grep command doesn't do arithmetic, it just finds lines that match regular expressions.
To count the output you already have, I'd use awk.
grep -c -F "2016-08-30" applciationLog.log* | awk -F: '{n+=$2} END {print n}'
Note that your grep options didn't make sense -- -E tells the command to use Extended regular expressions, but you're just looking for a fixed string (the date). So I swapped in the -F option instead. And -o tells grep to print the matched text, which you've overridden with -c, so I dropped it.
An alternative using for-loop and arithmetic expansion could be:
x=0
for i in $(grep -hc "2016-08-30" applciationLog.log*);do
x=$((x+i))
done
echo "$x"
An easy alternative is to merge all the files before grep sees them:
cat applciationLog.log* | grep -Eo "2016-08-30" -c
In my directory have have hundreds of files, each file contains lot of text along with a lines similar to this-
Job_1-Run.log:[08/27/20 01:28:40] Total Jobs Cancelled for Job_1_set0 = 10
I do
grep '^Total Jobs Cancelled' ./*
to get that above line.
Then I do a pipe
| awk 'BEGIN {cnt=0;FS="="}; {cnt+=$2;}; END {print cnt;}'
so my final command is-
grep '^Total Jobs Cancelled' ./* | awk 'BEGIN {cnt=0;FS="="}; {cnt+=$2;};END {print cnt;}'
and result is the sum. e.g. -
900
I am using Cmder # https://cmder.net/
Thanks to the answer by #alagner, #john above
I am trying to read/grep a particular word or content that is before a period (.).
e.g. file1 has abinaya.ashok and I want to grep whatever is before the period (.) without hardcoding anything.
if I try
grep \.\ file1
it gives abinaya.ashok.
I've tried: grep\*\.\ file1
it doesn't give anything.Can we find it using grep commands or should we do it only using awk command? Any thoughts?
Using GNU grep for PCRE regex (for non-greedy and positive look-ahead), you can do:
echo 'abinaya.ashok' | grep -oP '.*?(?=\.)'
abinaya
Using awk:
echo 'abinaya.ashok' | awk -F\. '{print $1}'
abinaya
Check the following simple examples.
Including the dot:
$ echo abinaya.ashok | grep -o '.*[.]'
abinaya.
Without the dot:
$ echo abinaya.ashok | grep -o '^[^.]\+'
abinaya
Hope I understand you correctly:
sed -n 's/\..*//p' file1 | grep whatever
sed expression will print only part before dot (lines without dot are not printed).
Now use grep to search what you need.
I want to grep a part of string that has numbers and dots(.) in it.
For example:
/home/xar/11.1.0/hez
/uaa/14.0.2.5/grd/pc
What i want is only this part of line:
14.0.2.5
11.1.0
I realized that cut command is not enough for this problem.
You can use this grep:
$ grep -o "[0-9.]*" file
11.1.0
14.0.2.5
-o is for "print just the matched part".
"[0-9.]*" matches any combination of numbers and dots.
sed version:
sed -n 's!.*/\([0-9.]*\)/.*!\1!p' input
You can use awk, but here grep is the correct tool.
awk -F/ '{for (i=1;i<=NF;i++) if ($i~/[0-9.]+/) print $i}' file
11.1.0
14.0.2.5
I am in need of trimming some text with grep, I have tried various other methods and havn't had much luck, so for example:
C:\Users\Admin\Documents\report2011.docx: My Report 2011
C:\Users\Admin\Documents\newposter.docx: Dinner Party Poster 08
How would it be possible to trim the text file, so to trim the ":" and all characters after it.
E.g. so the output would be like:
C:\Users\Admin\Documents\report2011.docx
C:\Users\Admin\Documents\newposter.docx
use awk?
awk -F: '{print $1':'$2}' inputFile > outFile
you can use grep
(note that -o returns only the matching text)
grep -oe "^C:[^:]" inputFile > outFile
That is pretty simple to do with grep -o:
$ grep -o '^C:[^:]*' input
C:\Users\Admin\Documents\report2011.docx
C:\Users\Admin\Documents\newposter.docx
If you can have other drives just replace C by .:
$ grep -o '^.:[^:]*' input
If a line can start with something different than a drive name, you can consider both the occurrence a drive name in the beginning of the line and the case where there is no such drive name:
$ grep -o '^\(.:\|\)[^:]*' input
cat inputFile | cut -f1,2 -d":"
The -d specifies your delimiter, in this case ":". The -f1,2 means you want the first and second fields.
The first part doesn't necessarily have to be cat inputFile, it's just whatever it takes to get the text that you referred to. The key part being cut -f1,2 -d":"
Your text looks like output of grep. If what you're asking is how to print filenames matching a pattern, use GNU grep option --files-with-matches
You can use this as well for your example
grep -E -o "^C\S+"| tr -d ":"
egrep -o "^C\S+"| tr -d ":"
\S here is non-space character match