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script for get lines with specific date

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# 1  
Old 12-20-2011
script for get lines with specific date
I'm a newbie in AIX, i want to make a script for grep any lines with date bellow 20
Code:
PRINT0089-88615        data1            3072   Mon Dec 19 17:53:49 WITA 2011
PRINT0089-88616        data1            4096   Mon Dec 19 17:53:49 WITA 2011
PRINT0089-88618        data1            5120   Mon Dec 19 17:54:19 WITA 2011
PRINT0272-88668        data1            3072   Mon Dec 19 20:36:42 WITA 2011
PRINT0331-88700        data1            2048   Tue Dec 20 01:03:10 WITA 2011
PRINT0331-88701        data1           77824   Tue Dec 20 01:03:11 WITA 2011
PRINT0152-88713        data1          654336   Tue Dec 20 01:11:53 WITA 2011
PRINT0089-88794        data1            3072   Tue Dec 20 07:20:05 WITA 2011
PRINT0089-88831        data1            3072   Tue Dec 20 07:48:36 WITA 2011
PRINT0089-88890        data1            3072   Tue Dec 20 08:03:07 WITA 2011
PRINT0089-88900        data1            3072   Tue Dec 20 08:07:38 WITA 2011
PRINT0331-90127        data1          100352   Wed Dec 21 01:03:14 WITA 2011
PRINT0089-90130        data1            3072   Wed Dec 21 01:08:48 WITA 2011
PRINT0089-90207        data1            3072   Wed Dec 21 07:46:07 WITA 2011
PRINT0089-90238        data1            3072   Wed Dec 21 07:56:57 WITA 2011
PRINT0089-90265        data1            3072   Wed Dec 21 08:03:58 WITA 2011

please help me!
# 2  
Old 12-20-2011
For anything from 1900-1999 try:

Code:
grep "WITA 19[0-9][0-9]$" logfile

# 3  
Old 12-20-2011
"Below 20" is ambiguous, currently the log file has only data for december, so the below code will do.
Code:
awk '$6<=20' logfile

But is that all you want?

--ahamed
# 4  
Old 12-20-2011
Quote:
Originally Posted by Chubler_XL
For anything from 1900-1999 try:

Code:
grep "WITA 19[0-9][0-9]$" logfile

I write the code
Code:
file=/print/printqeueu
file1=/print/printqeueu1
awk '$6 < 20 {print}' $file > $file1

but i must manual edit on script every change of date, for example 20.
how to get the date 3 days before today ?
# 5  
Old 12-20-2011
Code:
awk '$6<=(date-diff)' date=$(date +%d) diff=3 logfile

--ahamed
This User Gave Thanks to ahamed101 For This Post:
michlix
# 6  
Old 12-20-2011
As ahamed101 says currently your fine, but what happens on the 1st and 2nd (especially in Jan)?

Awk probably needs to be smarter and deal with a passed date.
# 7  
Old 12-20-2011
Chubler XL
For this time, my problem has been solved. But i don't know if in 1st or 2nd Jan. I should find out, i can not test the script because the server is for production.
Thank's for your comment...
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