grep one line + one line more is possible??

[Carme Torca]

Hi,

I would like to grep one word in one file, and I have to take this line and the line after... If its possible with grep??... if not is it possibe with another command??

Exemple

If I have this text in a file:

Completed: ALTER DATABASE DATAFILE '/oracle/BUGD02/oradata1/B
Wed Nov 19 18:29:44 2003

Thread 1 advanced to log sequence 2
Current log# 2 seq# 2 mem# 0: /oracle/BUGD02/oradata1/BUGD02/logBUGD02_2.dbl
Wed Nov 19 18:31:07 2003
Thread 1 advanced to log sequence 3
Current log# 3 seq# 3 mem# 0: /oracle/BUGD02/oradata1/BUGD02/logBUGD02_3.dbl
(.....)

Wed Nov 19 18:35:07 2003
Completed: CREATE TABLESPACE RBS DATAFILE '/oracle/BUGD02/ora
Wed Nov 19 18:35:07 2003

I would like to take only the line with Completed and the line with de date.


Thanks!
Carmen.

[Naveej.K.A]

hi,

may look a bit complicated, but is should work

awk '/(^Completed)|(^Mon) | (^Tue) | (^Wed) | (^Thu) | (^Fri) | (^Sat) | (^Sun)/ {print $0}' filename

with best wishes
Naveej

[Mark Grant]

If you installed GNU grep you could use grep with the "-C 1" option which would print one line of context i.e the line after.

You can get GNU grep here

http://hpux.connect.org.uk/hppd/hpux/Gnu/grep-2.5.1/

[Carme Torca]

Hi,

This only shows lines whith Completed and Mon.... ¿?

Completed: ALTER DATABASE CLOSE NORMAL
Mon Apr 26 00:10:55 2004
Completed: ALTER DATABASE DISMOUNT
Mon Apr 26 03:53:58 2004
Mon Apr 26 03:54:02 2004
Mon Apr 26 03:54:07 2004

But I only need lines with 1) Completed and 2) the date....

If there are more lines with Completed or dates in the middle I would like ignore it.
If it possible with awk??

Thanks!
Carme

[Mark Grant · Accepted solution]

I'm sure someone could do this better but here is a perl grep that shows the next line too

perl -ne '/text to match/ || next;print $_;$_=<>;print' datafile

[Nicolas Dumeige]

Hello Carme,

I use a script to get all errors in the alert*.log and the "lines around", i.e. 2 before and 5 after without getting duplicated lines - for what it's worth - :

grep -n '^ORA' $PATH/$ALERT_FILE | while read line
do
expr $line - 2 | read before
expr $line + 5 | read after

if [ $last -lt $before ] ; then
printf "%d,%dp\n" $before $after >> $SED_SCRIPT
else
printf "%d,%dp\n" `expr $last + 1` $after >> $SED_SCRIPT
fi

last=$after

expr $line + 1 | read after
printf "%d,%dp\n" $line $after >> $SED_SCRIPT
done
sed -nf $SED_SCRIPT $ALERTE_FILE


Change it to suit your need (^ORA --> ^Completed ; ....)

Cheers

Nicolas

[Bharat Katkar]

Carmen,
what about this one:

# cat filename | egrep 'Completed|Mon|Tues|Wed|Thur|Fri|Sat|Sun'

Regards,

[Mark Grant]

Carmen,

I did just notice that my perl thing above will hang if it matches something in the last line of the file because it tries to read the next line.

This change stops that happening

perl -ne '/text to match/ || next;print $_;if(not eof){$_=<>;print}' datafile

[Bharat Katkar]

Carmen,
See if this one is useful..

#!/usr/bin/ksh
list=`cat file1 | cut -d " " -f 1`
echo " Enter file name to search patterns \n"
read VAR2
myfunct()
{
while true
do
VAR1=$1
if [ $# -gt 0 ]
then
for VAR1 in Mon Wed Tues Fri Sat Sun Completed
do
cat $VAR2 | grep "$VAR1"
done
shift
else
exit
fi
done
}
myfunct $list

Regards,

[Simon Page_1]

Hi Carmen

Try this: sed -n '/^Completed: /{N;p;}'

Regards
Simon

[Alan Mason]

Its Unix - You can always do it in one line!
(though sometimes the lines can be quite long!!)

cat logfile | sed -n -e '/^Completed/{N;p;}'

Regards
Alan Mason