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date=`/usr/ucb/expr $date1 - 1`

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# 1  
Old 08-09-2005
Data date=`/usr/ucb/expr $date1 - 1`
Hi I need to subtract one day from date1=`/bin/date +%d`
So I used
date=`/usr/ucb/expr $date1 - 1`

The only thing is if date1 is a single digit like 08, date will be 8 instead of 08.
How can I avoid losing 0?

Thanks for all your help!!!
# 2  
Old 08-09-2005
one way using Bourne:
Code:
echo "`date +%d` - 1`" | bc | sed 's/^.$/0&/'

Q: what will happen on the first of a month?
Last edited by vgersh99; 08-10-2005 at 11:29 AM.. Reason: correction
# 3  
Old 08-10-2005
Just a little correction to vgersh99 code
Code:
echo "`date +'%d - 1' | bc | sed 's/^.$/0&/'`"

or with 'expr' :
Code:
date1=`/bin/date +%d`
yesterday=`/usr/bin/expr $date1 - 1 | sed 's/^.$/0&/'`

Another way with KSH
Code:
typeset -Z2 yesterday
(( yesterday = $(date +%d) - 1 ))

# 4  
Old 08-10-2005
date +%d -1
Thanks a lot.
vgersh99,
I don't think there is any difference with first of the month.
I thought first of the month would be 1.

Quote:
Originally Posted by vgersh99
one way using Bourne:
Code:
echo "`date +%d) - 1` | bc | sed 's/^.$/0&/'

Q: what will happen on the first of a month?
# 5  
Old 08-10-2005
Quote:
Originally Posted by whatisthis
Thanks a lot.
vgersh99,
I don't think there is any difference with first of the month.
I thought first of the month would be 1.
$ echo '01 - 1' | bc | sed 's/^.$/0&/'
00
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