UNIX for Dummies Questions & Answers

I have a file that contains the following:
Mon Dec 3 15:52:57 PST 2o007: FAILED TO PROCESSED FILE 200712030790881200.TXT - exit code=107
Tue Dec 4 09:08:57 PST 2007: FAILED TO PROCESSED FILE 200712030790879200a.TXT - exit code=107

This file also has a lot more stuff since it is a log file.
I am currently using the following but there must be a shorter way of doing it.

Code:

LOGDIR=/path/to/log/directory

grep FAILED $LOGDIR/processor.log | grep `date +20%y%m%d%y` | grep 107 > tempfile

I am trying the following but it does not work.

Code:

grep -E "FAILED[\w]*`date +20%y%m%d%y`[\w]*107

Also, how can I just grep this log file and get just the ones that have today's date and a.TXT on the line.

TIA
Mike

Use %Y instead of 20%y for safety.

Code:

egrep "FAILED.*`date +%Y%m%d%y`.*107$" $LOGDIR/processor.log > tempfile

Code:

egrep "^`date`.*\.TXT"

Thanks that worked great. I am pretty new to reg expressions. I do understand how you did it.

Mike