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date command

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Name: netrom
Date: March 3, 2008 at 06:48:09 Pacific
OS: AIX
CPU/Ram: n/a
Product: n/a
Comment:

Hi guys, I know it may sound silly. I have this in my script:

JD=`date +%Y +%m|cut -c "4 5 6"`
echo $JD

the JD when I run it today outputs 803. Which is 8=part of year 2008, 03 is current month.

Is there a (simple) way of having outputted: 802 (that is always the year AND PREVIOUS month??).

e.g. when comes 1.1.2009, I'd like the JD output 812.

when it's 1.4.2008, I'd like to have variable JD = 803.

Please help!

Many thanks.



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Response Number 1
Name: pklinger
Date: March 18, 2008 at 09:03:05 Pacific
Reply:

Referencing http://computing.net/unix/wwwboard/...

#!/bin/ksh
typeset -R3 JD
((HRS=$(date +%d)*24))
JD=$(TZ=aaa$HRS date +%y%m)
print $JD


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