UNIX for Advanced & Expert Users

hi there,

i have to read lines [ time-stamps ] from the file, where the line is just above the pattern am looking for
typically this looks like this

<time-stamp>|-----
<pattern am searching >......
<time-stamp>|.....
<some garbage >
....


the log file is big [2M lines ]
wc -l ~/log/ompe.log.20081203
2808766

after getting the list of line numbers here the pattern is
am trying to read time-stamps from the perticualr line using

for line_no in `cat /tmp/jms1`
do
prv_line_no=$(expr $line_no - 1 )
sed -n "${prv_line_no}p" $1 >> /tmp/jms_times
# head -${prv_line_no} $1 | tail -1 >> /tmp/jms_times1
done

sed option is very slow , on the other hand using head/tail option is fast inatially but as the line number is increases the retreval slows down

is there any fast way to retreive a perticular line number

try going by the line number with
grep -n "pattern" file | cut -d ":" -f1

your line no will be echo `grep -n "pattern" file | cut -d ":" -f1` - 1 |bc

then you can get the line printed with sed without having extracted the two lines and the head|tailing them ..

Hi,

a sed one liner is enough, you don't need the for-loop and all
the other processes. This should speed up your search.

Code:

sed -n '/pattern/{g;1!p;};h' file

It will search for pattern but print the line immediately before
the line containg pattern, but not the line with pattern itself.

HTH Chris

thanks you very much every one , for the help

hi chris,
i tried your command , its not giving the output

can you explain me the command " {g:1!p;};h "

$ sed -n '/JMS message rec/{g;1!p;};h' ompeJMSServerB.log.20081202004203 | head -2


$
can i execute this command in any shell ?

Sorry,

i have overlooked your answer:

-n -- print only lines when asked
'/pattern/ -- if you match pattern
{g;1!p} -- then execute the following group of command
g -- overwrite the pattern space with the hold space
1!p -- print the pattern space if it is not the first lien
;h' -- this is executed on all lines and means save every line in the
hold space.

file -- the file to proceed.

What it does: a) it saves every line in the hold space, b) if your pattern
is matched, the pattern space -- the line with the pattern you searched --
is replaced by the line previously saved in the hold space -- that is the
line preceeding your pattern. Thus the pattern you searched is replaced by
the line preceeding it.

This sed-command should be the usable on all unixes.

If you have problems, please give concrete examples of what you are
searching for and the data in your file.

hi chris,

am trying to use your command but getting the following error !

tcenh253!omsrec:~/infraBin/logs/tmp [8]> sed -n '/JMS message rec/{g;1!p;};h' ompeJMSServerB.log.20081207004201 | wc
p: Event not found.

when i try to remove shell-interpretation of "!" like below all am getting is blank/empty lines

tcenh253!omsrec:~/infraBin/logs/tmp [11]> sed -n '/JMS message rec/{g;1\!p;};h' ompeJMSServerB.log.20081207004201


tcenh253!omsrec:~/infraBin/logs/tmp [12]>grep "JMS message rec" ompeJMSServerB.log.20081207004201 |wc
552 2760 28704


typically the logfile looks like below

-------------------------------------------
at amdocs.epi.appserver.ClientConnectionCls.doAs(ClientConnectionCls.java:175)
at amdocs.epi.jms.JmsQueueConsumerCls$ListenerThread.run(JmsQueueConsumerCls.java:711)
at java.lang.Thread.run(Thread.java:534)

|D |02:00:48 |13 |AL_0 |
------------ JMS message received -----------------
Message Object: weblogic.jms.common.TextMessageImpl
Replay Queue: OMS_Ics_iq@EAIJMSServer_mserver2
Message Text:
------------------------------------