If there are dates as 2010-06-01 and another as 2010-05-15
Using shell script or date command how to get the number of days between the two dates
Thanks..
A:
Got it
d1=`date +%s -d $1`
d2=`date +%s -d $2`
((diff_sec=d2-d1))
echo - | awk -v SECS=$diff_sec '{printf "Number of days : %d",SECS/(60*60*24)}'
thanks..
Hulk
2010-08-02 04:46:49
If you're going to use `(())` you might as well also use `$()` instead of backticks. Also, there's no need to pipe something into `awk`, just use `awk ... 'BEGIN {printf ...}'`. And you don't really need to use `awk` if it's just integer math: `echo "Number of days: $(( ( d2 - d1 ) / ( 60 * 60 * 24 ) ))"`
Dennis Williamson
2010-08-02 11:16:16
Thanks.................
Hulk
2010-08-03 05:14:59
+1
A:
Using only date and shell arithmetics:
echo $(($(($(date -d "2010-06-01" "+%s") - $(date -d "2010-05-15" "+%s"))) / 86400))
smichak
2010-08-02 05:03:07
It's not necessary to use `$(())` twice. Just use a set of grouping parentheses: `echo $(( ( $(date -d "2010-06-01" "+%s") - $(date -d "2010-05-15" "+%s") ) / 86400))` (optional spaces added for emphasis and readability)
Dennis Williamson
2010-08-02 11:05:43
Note that `%s` is specific to GNU date, so it might not be available on non-Linux systems. The [comp.unix.shell FAQ](http://cfajohnson.com/shell/cus-faq.html#6) has a long discussion on date calculations (basically, it's hard unless you have GNU date).
Gilles
2010-08-02 19:21:55
In fact there's a bigger problem: it sometimes fails in the spring… [My answer explains why.](http://stackoverflow.com/questions/3385003/shell-script-to-get-difference-in-two-dates/3391111#3391111)
Gilles
2010-08-02 19:32:14
+2
A:
There's a solution that almost works: use the %s date format of GNU date, which prints the number of seconds since 1970-01-01 00:00. These can be subtracted to find the time difference between two dates.
echo $(( ($(date -d 2010-06-01 +%s) - $(date -d 2010-05-15 +%s)) / 86400))
But the following displays 0 in some locations:
echo $((($(date -d 2010-03-29 +%s) - $(date -d 2010-03-28 +%s)) / 86400))
Because of daylight savings time, there are only 23 hours between those times. You need to add at least one hour (and at most 23) to be safe.
echo $((($(date -d 2010-03-29 +%s) - $(date -d 2010-03-28 +%s)) + 43200 / 86400))
Or you can tell date to work in a timezone without DST.
echo $((($(date -u -d 2010-03-29 +%s) - $(date -u -d 2010-03-28 +%s)) / 86400))
(POSIX says to call the reference timezone is UTC, but it also says not to count leap seconds, so the number of seconds in a day is always exactly 86400 in a GMT+xx timezone.)
Gilles
2010-08-02 19:31:21