shell scripting help

[Archana1]

Hi,

Iam mid of Shell scripting and Iam looking for command which should NOT capture
"WARN [00000003]" entry from below given log.

2010-04-02T01:57:18,814 WARN [00000003] sas - SAS Object Spawner Daemon III has completed initi
2010-04-02T01:57:18,815 WARN [00000004] sas - SAH021999I Server SAS Object Spawner Daemon III (

[James R. Ferguson]

Hi:

# grep -v "WARN [00000003]" file

...will output everything from 'file' that does *not* match.

You might use this in cases like this:

# grep "WARN" file|grep -v "WARN [00000003]"

...which would look for all instances of "WARN" in 'file', but skip those matching "WARN [00000003]".

Regards!

...JRF...
Regards!

...JRF...

[Kapil Jha]

you can use
egrep -v "WARN [00000003]|second_word|3rd_word"

although its lil heavy but quite handy.

BR,
Kapil+

[Dennis Handly]

>Kapil: egrep -v "WARN [00000003]|second_word|3rd_word"
>although it's lil heavy but quite handy.

Rather than use the egrep hammer, it may be easier to use multiple -e options:
grep -v -e "WARN [00000003]" -e second_word -e 3rd_word

Also since "[" is special to grep, it needs to be quoted or use fgrep instead:
grep -v -e 'WARN \[00000003]'

[Steven Schweda]

It might help if you specified exactly what
you do wish to extract from the file, as well
as what you wish not to extract from it.