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Extracting lines in file based on time

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# 1  
Old 07-14-2009
Extracting lines in file based on time
Hi,

anyone has any ideas on how do we extract lines from a file with format similiar to this: (based on current time)

Jun 18 00:16:50 .......... ............. ............
Jun 18 00:17:59 .......... ............. ............
Jun 18 01:17:20 .......... ............. ............
Jun 18 01:18:51 .......... ............. ............

for example: the time now is 01:20:00 and the result will return me records no later than 15mins ago.
Jun 18 01:17:20 .......... ............. ............
Jun 18 01:18:51 .......... ............. ............

by the way, i'm using shell script
# 2  
Old 07-14-2009
If you can share the script that you have done, we can extend it for additional requirement...
# 3  
Old 07-14-2009
Currently what i've done is only to retrieve records no later than an hour.
want to extend the feature to include retrieving of records within 15mins of current time. I don't think my current method can be applicable to 15mins intervals..

Code:
 
str=`TZ=GMT-7 date +%b" "%d" "%H:` # this sets a time earlier by one hour.
 
#echo str will give you something like "Jun 18 01:"
 

grep "$str" test.log > new.log  # test.log refers to the file format that i've mentioned earlier.

# 4  
Old 07-15-2009
Anyone has any idea on how to go about doing it?
# 5  
Old 07-17-2009
any help?
# 6  
Old 07-17-2009
Assuming Jun 18 does not change, and only time stame HH:MI:SS is considered
Try:
Code:
#!/bin/ksh
#set -x

_minPrev=15
_currTimeInSec=$(echo `date "+%T"` | awk -F":" '{print $1*3600+$2*60+$3}')
(( _prevTimeInSec = ${_currTimeInSec} - (0+60*${_minPrev}+0) ))

while read line
do
   _timeInFile=$(echo $line | awk '{FS="[: ]+"}{print $3*3600+$4*60+$5}')

   if [[ ${_timeInFile} -ge ${_prevTimeInSec} && ${_timeInFile} -le ${_currTimeInSec} ]]; then
      echo $line
   fi
done < input


I think extending it for MM and DD should not be an issue for you.
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