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Closed 10 years ago.
Possible Duplicate:
Use grep to report back only line numbers
I only want to see the line number. I don't need to see the remaining output.
Pipe your grep -n output, which normally looks something like:
11: stuff that matched
43: more stuff that matched
through sed to strip out the matching parts:
grep -n pattern file | sed -e 's/:.*//g'
11
43
grep -n or --line-number option will do this for you. You can find this information in the grep help file, which you can find by using grep --help or grep --help | less to read it more carefully. Also consider using the manual page: man grep
You could use awk too.
grep -n word file | awk -F: '{ print $1 }'
As #Barmar pointed out you could just use an awk one-liner as such:
awk '/regex/ { print NR }' file
Since you don't have awk you could also use cut:
grep -n word file | cut -d: -f1
Related
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.
This question already has answers here:
Can grep show only words that match search pattern?
(15 answers)
Closed 8 years ago.
I have a bash file that has below line along with other lines.
var BUILD_VERSION = '2014.17.10_23';
I just want to extract 2014.17.10_23 and this value may change so something like grep for 2014* . However when I do that I get the whole line returned instead of the value 2014.17.10_23.
What would be the best way to achieve this?
Thanks
Using awk:
awk -F= '/BUILD_VERSION/{print $2}' input | tr -d "[' ;]"
And with sed:
sed -n "/BUILD_VERSION/s/.*'\([^']*\)'.*/\1/p" input
grep 'BUILD_VERSION' <your file> | sed -e 's/var BUILD_VERSION = //g'
Would get you '2014.17.10_23'; tweak the sed expression (or pipe it through a few more) to get rid of quotes.
It would be a 1 liner regex in Perl...
Here is another awk solution:
awk -F' = ' '/BUILD_VERSION/ {gsub(/\x27|;/,""); print $NF}'
You can use this awk
awk -F\' '/BUILD_VERSION/ {print $2}' file
2014.17.10_23
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 "$MYNUMDOWN"p file.txt #display below keyword
The plus point of the script is you can change the keyword in VAL variable as you like and execute to get the needed output.
This question already has answers here:
sed unknown option to `s' in bash script [duplicate]
(4 answers)
Closed 2 years ago.
I am trying
grep searchterm myfile.csv | sed 's/replaceme/withthis/g'
and getting
unknown option to `s'
What am I doing wrong?
Edit:
As per the comments the code is actually correct. My full code resembled something like the following
grep searchterm myfile.csv | sed 's/replaceme/withthis/g'
# my comment
And it appears that for some reason my comment was being fed as input into sed. Very strange.
use the --expression option
grep searchterm myfile.csv | sed --expression='s/replaceme/withthis/g'
use "-e" to specify the sed-expression
cat input.txt | sed -e 's/foo/bar/g'
To make sed catch from stdin , instead of from a file, you should use -e.
Like this:
curl -k -u admin:admin https://$HOSTNAME:9070/api/tm/3.8/status/$HOSTNAME/statistics/traffic_ips/trafc_ip/ | sed -e 's/["{}]//g' |sed -e 's/[]]//g' |sed -e 's/[\[]//g' |awk 'BEGIN{FS=":"} {print $4}'
If you are trying to do an in-place update of text within a file, this is much easier to reason about in my mind.
grep -Rl text_to_find directory_to_search 2>/dev/null | while read line; do sed -i 's/text_to_find/replacement_text/g' $line; done
Open the file using vi myfile.csv
Press Escape
Type :%s/replaceme/withthis/
Type :wq and press Enter
Now you will have the new pattern in your file.