UNIX for Dummies Questions & Answers

I guess that could be further reduced to:

Code:

prev_month=$(date +%m | awk '{print ($1==1) ? 12 : $1-1}')

or

Code:

cur_month=$(date +%m)
prev_month=$(( ${cur_month#0}==1 ? 12 : ${cur_month#0} - 1 ))

--
Note: 10#<num> is bash/ksh93

Except, of course, that:

@Don, the only thing that was required is the number of the previous month (following the question in post #3), so year and day do not matter in this case..

The question in post #3 was:
Where the 1st two posts were talking about changing the year in a date in the format DD/MM/YYYY. I interpreted the as well to mean that the month as well as the year were to be decremented. I didn't see anything saying that the day was no longer desired.

Looking at it more closely, I agree that the specification was not clear as to whether the request was for a date 13 months ago, a date 1 month ago, or a date in the month before the current month in the year before the current year (which could be 13 months ago or, in January, 1 month ago). Did the as well mean there was a request to get the date corresponding to today in the previous year as well as the date corresponding to today in the previous month; or to get the date corresponding to the current date in the previous month of the previous year.

From post #1, I thought the original request was to get the behavior of the command:

Code:

date +%d/%m/%Y -d "-1 years"

that works on systems where the date utility uses the -d option to specify a time other than now, on a system where the date utility does not have that option.

I may have read much too much between the lines.