I want to grep some pattern.

[Saraswathy_1]

Hello

I want to grep for one pattern and associated time stamp from a log file..

#### <> <
> <101020> <[ServletContext(id=3078349,name=webapperer,context-path=/webapperee)] Servlet failed with Exception>
java.lang.OutOfMemoryError: unable to create new native thread
at java.lang.Thread.start(Native Method)

Iam looking for below output:-
Dec 1, 2005 2:20:28 AM PST :OutOfMemoryError : blaa.log

[Luk Vandenbussche]

Hi,

try

grep OutofMemory blaa.log

[James R. Ferguson]

Hi:

You could keep this as simple as:

# grep -i -e "Dec 1" -e "OutofMemoryError" logfile

Regards!

...JRF...

[H.Merijn Brand (procura]

# perl -nle'/<(.*?)>.*?(\w*OutOfMemoryError)/ and print "$1 : $2 : $ARGV"' blaa.log

Enjoy, Have FUN! H.Merijn

[Saraswathy_1]

Is there any way to grep for some patter and above line of the pattern.

Which will solve my problem

[H.Merijn Brand (procura]

There is, but not with plain grep.
perl and awk will do, but you cannot grep in one regex (in perl you could, but then you will have to change the input line separator, and things will get quite unreadable fast :)

In the next line, $p is the previous line:

# perl -nle'$p=~/pattern_for_previous_line/&&/<(.*?)>.*?(\w*OutOfMemoryError)/ and print "$1 : $2 : $ARGV\n(previous line is $p)\n";$p=$_' blaa.log

Enjoy, Have FUN! H.Merijn

[Sandman!]

Saraswathy,

Not sure if grep can print lines above those that match a pattern but ex can:

# ex -s input_file <> /OutOfMemoryError/-1,p
> EOF

regards!

[Arturo Galbiati]

Hi,
you can use sed:
# print the line immediately before and the line containing regexp
sed -n -e '/regexp/{x;1!p;g;p;}' -e h file

in your case:
sed -n -e '/OutOfMemoryError/{x;1!p;g;p;}' -e h blaa.log

HTH,
Art