How can I get the 1st and last date of the previous month in a Bash script?

Question

I have scheduled a Bash script to run on the 1st of the month but I need to create 2 variables in it with the 1st and last date of the previous month, whatever those may be.

Answer 1

Unlike some answers, this will work for the 31st and any other day of the month. I use it to output unix timestamps but the output format is easily adjusted.

first=$(date --date="$(date +'%Y-%m-01') - 1 month" +%s)
last=$(date --date="$(date +'%Y-%m-01') - 1 second" +%s)

Example (today's date is Feb 14, 2019):

echo $first $last

1546300800 1548979199

To output in other formats, change final +%s to a different format such as +%Y-%m-%d or omit for default format in your locale.

In case you need, you can also back up an arbitrary number of months like this:

    # variable must be >= 1
    monthsago=23
    date --date="$(date +'%Y-%m-01') - ${monthsago} month"
    date --date="$(date +'%Y-%m-01') - $(( ${monthsago} - 1 )) month - 1 second"

Example output (today's date is Feb 15, 2019):

Wed Mar 1 00:00:00 UTC 2017
Fri Mar 31 23:59:59 UTC 2017