Shell Programming and Scripting

Hi,

I am looking for a way to find a particular word in a file then print a line that precedes this line, as well as this line.

Sometimes in a log file there is only one word per line and I need to print one of the lines leading up to the single worded line.

Example - I can grep for ouch and need line 1 and the ouch line printed as one line for output.

2006-10-04 this is line one
this is line two
this is line three
ouch
this is line four

thx.

Hi - this should work...

#!/bin/sh

# p_line is previous line
# c_line is current line
# myfile is path to target file

myfile=/path/to/myfile
cat $myfile | while read c_line
do
loop=
while [ -z "$loop" ]
do
if [ `echo "$c_line" | grep -c ouch` -ne 0 ]
then
echo $p_line
echo $c_line
else
p_line=$c_line
fi
loop=1
done
done

Got perl? Try...

Code:

perl -00ne 'printf "%s %s\n",$1,$2 if $_=~/(.*)\n.*\n.*\n(ouch)/' file1

...gives...

Code:

2006-10-04 this is line one ouch

Python alternative:

Code:

#!/usr/bin/python
data = open("test.txt").readlines()
for i in range(len(data)):
 	if "ouch" in data[i] :
 		print data[i-1] , data[i]

Output:

Code:

this is line three
ouch

One line solution in awk, if you know the line number of the preceeding line

Code:

 awk 'NR==1{str=$0}$0~/ouch/{ print str,$0}' filename

Thanks all - I will give it try today.