I have a date in YYYY.MM.DD HH:SS format (e.g. 2014.02.14 13:30). I'd like to convert it in seconds since epoch using the date command.
The command
date -d"2014.02.14 13:30" +%s
won't work, because of the dots separation.
Any Ideas?
5 Answers
Why don't you make the date format acceptable? Just replace dots with dashes:
$ date --date="`echo '2014.02.14 13:30' | sed 's/\./-/g'`" +%s 1392370200 Here I first change the format:
$ echo '2014.02.14 13:30' | sed 's/\./-/g' 2014-02-14 13:30 and then use the result as a parameter for date.
Note that the result depends on your timezone.
2 Comments
Tomás Atómico
I should've mentioned that it acted piped in an awk command: awk 'BEGIN{getline}{command=("date --date=\""$1" "$2"\""); while (....
afenster
Then make the same replacement in
awk using its gsub function.You can use:
s='2014.02.14 13:30' date -d "${s//./}" Fri Feb 14 13:30:00 EST 2014 To get EPOCH value:
date -d "${s//./}" '+%s' 1392402600 
2 Comments
Hackaholic
why no of seconds differ?? in each ans??
Hackaholic
ohh its due to EST AND IST got it :)
using awk :
s=`echo "2014.02.14 13:30" | awk '{gsub(/\./,"-",$0);print $0}'` echo -d "$s" date -d "$s" +%s output:
Fri Feb 14 13:30:00 IST 2014 1392364800 Comments
Perl: does not require you to munge the string
d="2014.02.14 13:30" epoch_time=$(perl -MTime::Piece -E 'say Time::Piece->strptime(shift, "%Y.%m.%d %H:%M")->epoch' "$d") echo $epoch_time 1392384600 Timezone: Canada/Eastern
1 Comment
Tomás Atómico
thanks, perl seems powerful, but one language at a time.
I Finally solved it using
awk 'BEGIN{FS=","}{ gsub(/./," ",$1);gsub(/:/," ",$2); var=sprintf("%s %s 00",$1,$2); print mktime(var), $3,$4,$5,$6,$7 }' myfile | less
so myfile:
2014.09.24,15:15,1.27921,1.27933,1.279,1.27924,234
became
1411582500 1.27921 1.27933 1.279 1.27924 234
:)
-f