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Shell Programming and Scripting

Script Advice please?

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Top Forums Shell Programming and Scripting Script Advice please?
# 8  
Old 06-18-2007
Edit: Quick question:
What is this part doing?
Code:
mFirstPart='^... .. '

It is a regular expression as follows:
1) Begining of the line.
2) Three characters.
3) One space.
4) Two characters.
5) One space.
This is done to make sure the "egrep" is using the first date and not the second one:
Code:
Jun 18 14:17:56 routername 36806: Jun 18 17:53:01.088:
Jun 18 13:17:56 routername 36806: Jun 18 15:53:01.088:
Jun 18 12:17:56 routername 36806: Jun 18 17:53:01.088:
Jun 11 17:47:56 routername 36806: Jun 18 01:53:01.088:
Jun 11 17:47:56 routername 36806: Jun 18 13:53:01.088:
Jun 07 14:17:56 routername 36806: Jun 18 00:53:01.088:

As for the code, it can be improved to make sure it is using two digits for the hour:
Code:
typeset -i mCurrHH
typeset -Z2 mCurrGrep
typeset -i mPrevHH
typeset -Z2 mPrevGrep
mCurrHH=`date +"%H"`
mPrevHH=${mCurrHH}-1
if [ ${mPrevHH} -eq -1 ]; then
  mPrevHH=23
fi
mCurrGrep=${mCurrHH}
mPrevGrep=${mPrevHH}
mFirstPart='^... .. '
egrep "${mFirstPart}${mPrevGrep}|${mFirstPart}${mCurrGrep}"

# 9  
Old 06-18-2007
Quote:
Originally Posted by earnstaf
Seems to be working. So I guess I should cron this to run at 59th minute 30 second of every hour?
My awk solution search for the current hour.
For the previous hour :
Code:
awk -v hour=$(date +%H) '
   BEGIN { 
      hour = (hour==0; 23; hour-1)
   }
   int($3) == hour && /denied/ && /gress/ {
      print $4:
   }
' file | sort -u

# 10  
Old 06-18-2007
Quote:
Originally Posted by aigles
My awk solution search for the current hour.
For the previous hour :
Code:
awk -v hour=$(date +%H) '
   BEGIN { 
      hour = (hour==0; 23; hour-1)
   }
   int($3) == hour && /denied/ && /gress/ {
      print $4:
   }
' file | sort -u

Looks good aigles... AWK is something that I really need to dig down into and learn. Thanks for your input.
# 11  
Old 06-18-2007
Quote:
Originally Posted by Shell_Life
Edit: Quick question:
What is this part doing?
Code:
mFirstPart='^... .. '

It is a regular expression as follows:
1) Begining of the line.
2) Three characters.
3) One space.
4) Two characters.
5) One space.
This is done to make sure the "egrep" is using the first date and not the second one:
Code:
Jun 18 14:17:56 routername 36806: Jun 18 17:53:01.088:
Jun 18 13:17:56 routername 36806: Jun 18 15:53:01.088:
Jun 18 12:17:56 routername 36806: Jun 18 17:53:01.088:
Jun 11 17:47:56 routername 36806: Jun 18 01:53:01.088:
Jun 11 17:47:56 routername 36806: Jun 18 13:53:01.088:
Jun 07 14:17:56 routername 36806: Jun 18 00:53:01.088:

As for the code, it can be improved to make sure it is using two digits for the hour:
Code:
typeset -i mCurrHH
typeset -Z2 mCurrGrep
typeset -i mPrevHH
typeset -Z2 mPrevGrep
mCurrHH=`date +"%H"`
mPrevHH=${mCurrHH}-1
if [ ${mPrevHH} -eq -1 ]; then
  mPrevHH=23
fi
mCurrGrep=${mCurrHH}
mPrevGrep=${mPrevHH}
mFirstPart='^... .. '
egrep "${mFirstPart}${mPrevGrep}|${mFirstPart}${mCurrGrep}"

Shell Life,
I can see why the typeset -Z2 is needed (so it doesnt find 20 when it should find 2), but the script isnt liking it:

./time.sh: line 4: typeset: -Z: invalid option
typeset: usage: typeset [-afFirtx] [-p] name[=value] ...

I tried it a few different ways:
Code:
typeset -Z2 mCurrGrep
typeset -Z 2 mCurrGrep
typeset -Z mCurrGrep 2
typeset -Z mCurrGrep=2

...same error on each. is -Z not an option in bash?
# 12  
Old 06-18-2007
Quote:
Originally Posted by earnstaf

...same error on each. is -Z not an option in bash?
make your script a 'ksh' - '#!/bin/ksh' (on the first line of the script).
# 13  
Old 06-18-2007
Quote:
Originally Posted by vgersh99
make your script a 'ksh' - '#!/bin/ksh' (on the first line of the script).
That did it .. thanks. I had read that bash offered most everything that the other shells and then some in terms of scripting... I guess that isn't the case here, eh?
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