Extract day from non standard Unix date - date

Question

I can see that UNIX easily handles a standard date format 'yyyy-mm-dd'. Date function can easily extract day form standard date format like this
date -d '2016-12-9 00:00:00' '+%d'
output: 09
But I can't figure out a way to get the same result using a non-standard date format. My date format is 'mm-dd-yyyy'.

Answer 1

GNU date which you are most likely using because of the -d switch, doesn't allow custom formatting of dates, unlike the FreeBSD version which supports with the -f input_fmt for converting custom date formats.
With GNU date and bit of manipulation with sed, you can achieve what you are trying to do.
Using sed converting your input date in mm-dd-yyyy to a format GNU date can understand.
customDate=$(echo "12-09-2016 00:00:00" | sed 's/\([0-9]\{2\}\)\-\([0-9]\{2\}\)\-\([0-9]\{4\}\) \(.*\)/\3-\1-\2 \4/')
Now customDate will have the date in the format 2016-12-09 00:00:00 format which you can pass to -d flag to get the required date. i.e.
date -d "$customDate" '+%d'
09
Or as a fancy "one-liner" they call it these days, without the use of the temporary variable.
date -d "$(echo "12-09-2016 00:00:00" | sed 's/\([0-9]\{2\}\)\-\([0-9]\{2\}\)\-\([0-9]\{4\}\) \(.*\)/\3-\1-\2 \4/')" '+%d'
09