How to *return* *only* the nth occurrence of a regex match using sed?

Question

Info

I have a string that has time info in it. I need to get only the number of months from this string. I'm trying to accomplish this using sed, but I haven't had too much luck yet. Using bash + command line tools is a requirement for me.

Current attempt

echo "1969 years 12 months 25 days 19 hours 38 minutes 24 seconds since last release" | sed -r "s/^([0-9]+).*/\1/"

returns 1969

Bonus

More generally, my problem is the entire "1969 years 12 months" portion of the string. I'm using date to get a time differential but I have to account for the 1970 start date. I find it odd that my current output expresses 1970 as 1969 year and 12 months.

Edit: I'm attempting to use nunk's solution from this post:

How to find the difference in days between two dates?

It works fine for everything, except for the months. The quantity of months is off by (12-printed number of months).

Answer 1

Does this suit:

echo "1969 years 12 months 25 days 19 hours 38 minutes 24 seconds since last release" | sed -r 's/.* ([0-9]+) months.*/\1/'

gives

12

What's happening here is this:

• The "match" part of the expression will succeed if there are, in order, all of:
  • .* any stuff
  • a space
  • [0-9]+ one or more digit ("captured" using ( and ))
  • another space
  • months the word "months"
  • .* any stuff
• If it succeeds, it's guaranteed to match the whole of the line, because of the .* at both ends.
• Also, if it succeeds, the match (the whole line) is replaced by \1 which is a special code meaning "the contents of the first set of capture brackets. You'll see that the value captured by the brackets above is the number preceding the word "months".

Answer 2

I would do this as next:

str="1969 years 12 months 25 days 19 hours 38 minutes 24 seconds since last release"

read y m d H M S < <(echo $(grep -oP '\d+' <<<"$str"))
echo "month: $m"
echo "year: $y"
#etc for $d $H $M $S

prints:

month: 12
year: 1969

• the grep filter out all numbers (each to alone line)
• the echo make from the numbers an space separated string
• what will read the read and it will assign each number to the corresponsing variable

Answer 3

It sounds like you're trying to fix the output from running a buggy tool to diff dates. Just use GNU awk instead:

$ echo "2014-09-03T14:44:48+00:00" |
gawk '{gsub(/[-T:+]/," "); print (systime() - mktime($0)) / (24*60*60)}'
5.86991

Do whatever rounding you like on the number of days output, e.g.:

$ echo "2014-09-03T14:44:48+00:00" |
gawk '{gsub(/[-T:+]/," "); printf "%.0f\n", (systime() - mktime($0)) / (24*60*60)}'
6

Note that the implementation of rounding using "%.0f" is system dependent - it might round 0.5 up or it might round it towards the nearest even number. If that's a concern check your system and write your own rounding function if necessary.

Answer 4

You could use Perl to separate out the key, value pairs:

$ str="1969 years 12 months 25 days 19 hours 38 minutes 24 seconds since last release"
$ echo $str | perl -lane 'print "$1 $2"  while /(\d+)\s(\w+)/g'
1969 years
12 months
25 days
19 hours
38 minutes
24 seconds

Then use grep to grab the one you want:

$ echo $str | perl -lane 'print "$1 $2"  while /(\d+)\s(\w+)/g' | grep 'months'
12 months