help with date command

Shell Programming and Scripting
1

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.

2

This should do what you're looking for:

JD=`date "+%y %m"|awk '{$2--}$2==0{$1--;$2=12}{printf("%.2d %.2d\n", $1, $2)}'`

Regards

3

Thanks Franklin, it gets me: 08 02

Is there a way to get the output: 802 (not 08 02)?

Thanks in advance.

4

Change the format in the printf statement:

printf("%d%.2d\n", $1, $2)

Regards

5

Great, works, many thanks!

6

Here is another way without all the print, cut, and tr stuff

JD=$(date --date="last month" +%Y%m | cut " 4 5 6")

Works in Linux Not Solaris.

7

This is GNUism

8

JD=`date +'%y %m'|cut -d' ' -f1,2`

try this command

9

Try this one

JD=`date +'%y %m'|cut -d' ' -f1,2`

10

Hi,

JD=`date +'%y %m'|cut -d' ' -f1,2`