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Using variable inside awk


 
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# 15  
Old 08-06-2009
you can tell shell to make date as a whole string and not separate it.
Code:
grep "`date +"%m/%d/%Y %H" -d "1 hour ago"`" file

# 16  
Old 08-06-2009
Panyam, Its working perfectly. Thanks a lot!
# 17  
Old 08-06-2009
proper computation of hour is needed so it can handle 00:00:00 (12AM) minus one hour.
wouldnt be posible with 00-1 unless you have different log for the day ofcourse
# 18  
Old 08-06-2009
Ryandegreat,
It is giving error message as follows.

Code:
ksh: : cannot execute
grep: can't open %H -d 1
grep: can't open hour
grep: can't open ago

Panyam,
Will the code work for all the timings? what about 1 am?
# 19  
Old 08-06-2009
it works on me can you check your script again..
Code:
-bash-3.2$ cat file
(08/06/2009 05:48:45.992)(:)
(08/06/2009 05:48:46.641)(:)
(08/06/2009 15:00:00
(08/06/2009 15:59:00
-bash-3.2$  grep "`date +"%m/%d/%Y %H" -d "1 hour ago"`" file
(08/06/2009 15:00:00
(08/06/2009 15:59:00
-bash-3.2$ date +"%m/%d/%Y %H"
08/06/2009 16
-bash-3.2$



---------- Post updated at 10:04 PM ---------- Previous update was at 10:03 PM ----------

dont forget the qoute and backticks
# 20  
Old 08-06-2009
i tried the same example that you gave. Got error again. I don't think -d option supported on my SunOS box.
# 21  
Old 08-06-2009
i have an old SunOS
Code:
# uname -a
SunOS vaiconp01 5.10 Generic_118833-36 sun4u sparc SUNW,Sun-Fire-V240
# date +"%m/%d/%Y %H" -d "1 hour ago"
08/06/2009 17
#

another option would be to look for a timezone which is late by an hour(could not think of anything else)

Code:
# HOURAGO=`TZ=GMT-2 date +"%m/%d/%Y %H"`
# echo $HOURAGO
08/06/2009 16
# date +"%m/%d/%Y %H"
08/06/2009 17

at least you get the idea

---------- Post updated at 10:37 PM ---------- Previous update was at 10:36 PM ----------

just play around with GMT+-number

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