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shell arithmetic least significant place

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# 1  
Old 11-02-2011
shell arithmetic least significant place
I need help on arithmetic
Code:
root@server [~]# hour=`date | awk {'print $4'} | cut -d: -f 1`; echo $hour
04

Now I subtract this result by 1 or 01 I get "3" as the answer. I need "03" as the answer, ie last two significant numbers should be there.
Code:
root@server [~]# hour=`date | awk {'print $4'} | cut -d: -f 1`;lasthour=`expr $hour - 01`; echo $lasthour
3

10 - 8 = 02
15 - 10 = 05

25 - 10 = 15
30 - 12 - 18

Like above
Last edited by anilcliff; 11-02-2011 at 08:14 AM..
# 2  
Old 11-02-2011
printf %02d $lasthour instead of echo.
This User Gave Thanks to CarloM For This Post:
anilcliff
# 3  
Old 11-02-2011
thanks it worked
# 4  
Old 11-02-2011
If you are in ksh, you can force the format of your variable:-
Code:
typeset -Z2 currhour=`date +%H`         # Just get the hour, check the format for your OS with man date
((prevhour=$currhour-1))
echo "$currhour - 1 = $prevhour

You need to consider how to handle hour zero though.



I hope that this helps,
Robin
Liverpool/Blackburn
UK
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