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.
Related
I have a text file with layout as:
tableName1|counterVariable1
tableName2|counterVariable2
I want to replace the counterVariable1 with some other variable say counterVariableNew.
How can I accomplish this?
I have tried various SED/AWK approaches, closest one is mentioned below:
cat $fileName | grep -w $tableName | sed -i 's/$tableName\|counterVariable/$tableName\|counterVariableNew'
But all the 3 commands are not merging properly, please help!
Your script is an example of [ useless use of cat ]. But the key point here is to escape the pipe delimiter which has a special meaning(it stands for OR) when used with awk FS. So below script should do
# cat 42000479
tableName1|counterVariable1
tableName2|counterVariable2
tableName3|counterVariable2
# awk -F\| '$1=="tableName2"{$2="counterVariableNew"}1' 42000479
tableName1|counterVariable1
tableName2 counterVariableNew
tableName3|counterVariable2
An alternate way of doing the same stuff is below
# awk -v FS='|' '$1=="tableName2"{$2="counterVariableNew"}1' 42000479
Stuff inside the single quote will not be expanded.
awk -F'|' -v OFS='|' '/tableName1/ {$2="counterVariableNew"}1' file
tableName1|counterVariableNew
tableName2|counterVariable2
This will search for A (tableName1) and replace B (counterVariable1) to counterVariableNew.
Or by using sed :
sed -r '/tableName1/ s/(^.*\|)(.*)/\1counterVariableNew/g' file
tableName1|counterVariableNew
tableName2|counterVariable2
For word bounded search: Enclose the pattern inside \< and \> .
sed -r '/\<tableName1\>/ s/(^.*\|)(.*)/\1counterVariableNew/g' file
awk -F'|' -v OFS='|' '/\<tableName1\>/ {$2="counterVariableNew"}1' file
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 need to get a list of matches with grep including filename and line number but without the match string
I know that grep -Hl will give only file names and grep -Hno will give filename with only matching string. But those not ideal for me. I need to get a list without match but with line no. For this grep -Hln doesn't work. I tried with grep -Hn 'pattern' | cut -d " " -f 1 But it doesn't cut the filename and line no properly.
awk can do that in single command:
awk '/pattern/ {print FILENAME ":" NR}' *.txt
You were pointing it well with cut, only that you need the : field separator. Also, I think you need the first and second group. Hence, use:
grep -Hn 'pattern' files* | cut -d: -f1,2
Sample
$ grep -Hn a a*
a:3:are
a:10:bar
a:11:that
a23:1:hiya
$ grep -Hn a a* | cut -d: -f1,2
a:3
a:10
a:11
a23:1
I guess you want this, just line numbers:
grep -nh PATTERN /path/to/file | cut -d: -f1
example output:
12
23
234
...
Unfortunately you'll need to use cut here. There is no way to do it with pure grep.
Try
grep -RHn Studio 'pattern' | awk -F: '{print $1 , ":", $2}'
I have a product which has a command called db2level whose output is given below
I need to extract 8.1.1.64 out of it, so far i came up with,
db2level | grep "DB2 v" | awk '{print$5}'
which gave me an output v8.1.1.64",
Please help me to fetch 8.1.1.64. Thanks
grep is enough to do that:
db2level| grep -oP '(?<="DB2 v)[\d.]+(?=", )'
Just with awk:
db2level | awk -F '"' '$2 ~ /^DB2 v/ {print substr($2,6)}'
db2level | grep "DB2 v" | awk '{print$5}' | sed 's/[^0-9\.]//g'
remove all but numbers and dot
sed is your friend for general extraction tasks:
db2level | sed -n -e 's/.*tokens are "DB2 v\([0-9.]*\)".*/\1/p'
The sed line does print no lines (the -n) but those where a replacement with the given regexp can happen. The .* at the beginning and the end of the line ensure that the whole line is matched.
Try grep with -o option:
db2level | grep -E -o "[0-9]+\.[0-9]+\.[0-9]\+[0-9]+"
Another sed solution
db2level | sed -n -e '/v[0-9]/{s/.*DB2 v//;s/".*//;p}'
This one desn't rely on the number being in a particular format, just in a particular place in the output.
db2level | grep -o "v[0-9.]*" | tr -d v
Try s.th. like db2level | grep "DB2 v" | cut -d'"' -f2 | cut -d'v' -f2
cut splits the input in parts, seperated by delimiter -d and outputs field number -f
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