Hi all,
I am struck with the below requirement. I need to grep a particular pattern in a file and then print next n lines of it for further processing.
I have used the below code
grep -A 3 "pattern" filename
But it is throwing error as below.
grep: illegal option -- A
Can anyone help me with some alternates for this.
Thanks.
That's an option to GNU grep, which not all systems have. What is your system?
Anyway, this should do:
awk 'PAT ~ $0 { N=LINES } (N--) > 0' LINES=5 PAT="qwerty" inputfile
Use nawk on Solaris.
Hi Corona,
Thanks for the reply. My system is Solaris . But i need this for AIX too
I am getting error when used with awk and no output when used with nawk as below
$ awk 'PAT ~ $0 { N=LINES } (N--) > 0' LINES=5 PAT="ISEND" sqloutput1.log
awk: syntax error near line 1
awk: bailing out near line 1
$ nawk 'PAT ~ $0 { N=LINES } (N--) > 0' LINES=5 PAT="ISEND" sqloutput1.log
$
Hello ssk250,
Could you please use /usr/xpg4/bin/awk , and /usr/xpg6/bin/awk in place of awk this should work.
Thanks,
R. Singh
Sorry, there was a mistake in my code.
awk '$0 ~ PAT { N=LINES } (N--) > 0' LINES=5 PAT="ISEND" sqloutput1.log
(N--) > 0 can overflow with large files. Better (N && N--) !
Hi All,
Thanks All. The below script worked for me
/usr/xpg4/bin/awk '$0 ~ PAT { N=LINES } (N--) > 0' LINES=5 PAT="ISEND" sqloutput1.log
Just want to know... In a similar way how the lines before this pattern can be printed.
Perhaps something like:
/usr/xpg4/bin/awk '
function pprev( i) {
for(i = (NR > LINES) ? NR - LINES + 1 : 1; i <= NR; i++)
print l[i % LINES]
print ""
}
{ l[NR % LINES] = $0
}
$0 ~ PAT {
pprev()
}' LINES=3 PAT='9' file
If file contains 100 lines numbered from 100 to 001:
100
099
098
098
...
003
002
001
the output produced is:
100
099
100
099
098
099
098
097
098
097
096
097
096
095
096
095
094
095
094
093
094
093
092
093
092
091
092
091
090
091
090
089
081
080
079
071
070
069
061
060
059
051
050
049
041
040
039
031
030
029
021
020
019
011
010
009
The following shell command emulates the ggrep -E -A4 , and works on all Unix:
PATH=/usr/xpg4/bin:/bin:/usr/bin awk '$0~PAT {a=A+1} (a && a--)' A=4 PAT="ISEND" sqloutput1.log
Omit the A=x to have the normal grep -E behavior.
Hi.
Some general alternatives, as opposed to single-purpose solutions:
Print lines above, below pattern-matched line (context, window); "only" string matching pattern
1) GNU grep -A -B; ggrep on some Solaris
2) peg, a perl script, does not use "-o" like GNU grep;
allows -A, -B
3) ack, a perl script, can use "-o" like GNU grep; allows -A,
-B
4) cgrep does not use "-o" like GNU grep; allows -nline,
+nline for context like GNU grep -A, -B
5) bool, prints context in bytes
6) xtcgrep, extension of CPAN grep; allow -C context; allows -o;
complicated calling sequence; http://freecode.com/projects/greppm
Best wishes ... cheers, drl
The probably easiest way is to use good old sed instead of all the fancy tools:
sed -n '/<search-pattern>/ {;N;...N;p;}' /path/to/file
Put in x-1 "N"s to display x lines of text after the matched pattern.
I hope this helps.
bakunin
Thanks ... ggrep is working on Solaris.
Problem solved. But the previous sed solution needs a fix; (at least on Solaris) it does not print at the end of a file, because the N command terminates the script. Therefore, each N is to be preceded by a $p .
The following works with a number for the trailing lines (here: 3).
sed -e '/^<search-pattern>/!d;:L' -e '$p;N;s/\n/&/3;t' -e 'bL' file
---------- Post updated at 10:19 AM ---------- Previous update was at 10:14 AM ----------
Hmm interesting, GNU sed prints at the end of the file, so the $p would print twice...
Which sed is correct?
You objection is correct and it was my bad not to think my solution through to the end. GNU-sed - as it is so often is the case - gets it wrong (again).
bakunin
An alternative would be to use: $!N instead of N , which should give equal results with GNU sed..