Is there a way to use the day of year as an input format for the date command?

Question

Using the command

$ date +%j

gives an output of the day of the year (001-366). However I need to use the day of the current year as an input format i.e.

$ date --date='066 day'  +%F

Which for year 2021 I would expect 2021-03-07 as the output. Instead I get 2021-05-13. Does anyone know what is going on and if there is a way to get what I want using date.

Answer 1

To get the 66th day of the year with GNU date:

$ date --date='jan 1 + 65 days' +%F
2021-03-07

or

$ date --date='dec 31 last year + 66 days' +%F
2021-03-07

With date --date='066 day' +%F you get the date 66 days from today. On the 8th of March, this happens to be the 13th of May.

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The above uses GNU date. On OpenBSD (not that you asked, but anyway), you could do something like the following:

$ date -u -r "$(( $(date -ju -f %m%d 0101 +%s) + 65*86400 ))" +%F
2021-03-07

which would be somewhat similar to the first GNU date command above in that it adds some time from January 1st.

Answer 2

I propose this option: if for example you want to know the date of the 80th day:

# take the number of day today
$ n=$(date +%j)
# set some number of days you want to know the date
$ t=80
# subtract the 'n', and add the 't'
$ date -d "-$n day +$t day" %F
2021-03-21

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Adapted to your example:

$ n=$(date +%j)
$ t=66
$ date -d "-$n day +$t day" %F
2021-03-07