Extract month and day from linux "date" command

Question

I would like to modify the following statement to extract the Month, as well as day:

testdate=$(date -d "2 days" +%d| sed 's/0([1-9])/ \1/')

right now it only extracts the day...

I'm currently googling how sed works. looks like a regular expression of sorts that's being used but... I thought I'd check with the forum for certain.

Thanks.

EDIT 1

now I have:

  testdate=$(date -d "2 days" "+%b %_d %Y")

Which is great... except when the leading zeros in the day are stripped out. then i end up with two spaces between the Month and day for example, for April 29, 2014, I get:

  Apr 29 2014

which is great. But for May 01, 2014 I get:

  May  1, 2014

where there's a double space between "May" and "1". I guess i can do it in two steps... by doing the date command, followed by sed to find and replace double spaces. Just wondering if there's a way to do it all in one command. Thanks.

Answer 1

date accepts a FORMAT arg which should be used here. sed is not necessary:

date +%d.%m.

Outputs:

03.04.

in European time format. If you want the month's name printed use

date +'%d. %B'

Output:

03. April

Answer 2

To read the month and day into shell variables (assuming bash/ksh/zsh)

read month day < <(date -d "2 days" "+%m %d")

If you're planning to do arithmetic on these values, be of numbers that would be treated as octal but are invalid octal numbers 08 and 09. If you want to strip off the leading zero, use "+%_m %_d"

---

Using read, the shell takes care of the excess whitespace for you:

read mon day year < <(date -d "2 days" "+%b %_d %Y")
testdate="$mon $day $year"

If you don't want to use the temp vars:

testdate="Apr  5 2014"       # $(date -d "2 days" "+%b %_d %Y")
testdate=${testdate//  / }   # globally replace 2 spaces with 1 space
echo "$testdate"