UNIX for Dummies Questions & Answers

Hello, im trying to grep the first 3 lines(starting with ERROR) from an output report and put it in a separate file... any help would be appreciated. Below is the message contained in the report

ERROR >Invalid basic data (CMOFCT-5070)
Cusip: 05950XAL0 CMO#508039 Tra# 11 rec# 175
Firts Int Pay Date: 1061110
Bond type flag: V
Floater type Flag: 3
Floater Reset Start Date: 0
Floater Reset Frequency: 0

thanks

Give this a try,

grep -n "^ERROR" /tmp/err | head -3

INPUT:
OUTPUT
Thanks
Nagarajan Ganesan.

Hi thanks for the help.. unfortunately it didnt work.. maybe im explainin it wrong... i am running a file and it spits out an output file r.(filename) and in that output file it has several messages... im lookin to take out the first 3 lines of messages that start with ERROR and put it into another file named geek... thanks

geekeezoid,
See if this would work for you:
sed -n '/ERROR/,$p' input_file | head -3

Hi, thanks... i think it took out the first 3 lines of the first ERROR message.. but there are several ERROR messages in the r. output file... any way to take the first 3 lines of all the ERROR messages? thanks

Geekeezoid,
Try this one:

Code:

mTotal=0
while read EachLine
do
 if [[ $mTotal -gt 0 ]]; then
   echo $EachLine
   mTotal=$mTotal+1
   if [[ $mTotal -gt 2 ]]; then
     mTotal=0
   fi
 else
   mCnt=`echo $EachLine | egrep -c "ERROR"`
   if [[ $mCnt -ne 0 ]]; then
     echo $EachLine
     mTotal=1
   fi
 fi
done < input_error_file

Code:

awk '/^ERROR/ { end = NR+3 } NR < end { print }' report > new_file