UNIX for Dummies Questions & Answers

Hi,

Is it possible to display a specific number of lines starting from a line having a particular text using grep command?

e.g. I have a text file with the contents below:

AAA
BBB
CCC
DDD
EEE
FFF

I want to display 3 lines starting with the line having "BBB" to get the result below:

BBB
CCC
DDD


Steve

grep -A 2

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DeepakS,

Thanks for your reply but I'm using HP-UX B.11.11 and that command doesn't work.

I know that it can be done using "SED" with the command below but I'm looking for a shorter method.

Code:

sed -n '/BBB/,$p' FILENAME | sed -n '1,3p'

I guess it depends on what you're trying to do. If you just need to save a few keystrokes throw this into a script:

Code:

#! /bin/bash

LINES=$(($1+1))

sed -n "/$3/,${LINES}p" $2

Parameter 1 is the number of lines to display, parameter 2 is the filename, and parameter 3 is the search string. Not as neat as GNU but it works.

In addition to my original question I want to find multiple occurrences of the string.

e.g. I have a file called 20071225.log with the contents below:

SUCCESS
1234
2433
SUCCESS
4321
6523
ERROR
5678
8423
SUCCESS
7890
5324
ERROR
3456
3214

I want to display 3 lines below all the occurrences of the line having the string "ERROR" to get the output below:

ERROR
5678
8423
ERROR
3456
3214


I can get the above result with the shell below but I would like a shorter method possibly in one line of command :

Code:

STRING=$1
FILE=$2

DONE=N

while [ $DONE = N ]
do
  sed -n "/$STRING/,\$p" $FILE | sed -n '1,3p'
  sed -n "/$STRING/,\$p" $FILE | sed '1,3d' > $$.txt
  mv $$.txt $FILE
  [[ `cat $FILE | wc -l` -eq 0 ]] && DONE=Y
done

Code:

sed -n '/ERROR/ { N; N; /./p }' foo

This will match ERROR followed by two lines.

Code:

awk 'f&&f--; /BBB/{f=3} ' file