Parse ISO8601 duration format.
https://en.wikipedia.org/wiki/ISO_8601#Durations
iso8601-duration = "0.2.0"
use iso8601_duration::Duration;
assert_eq!(
"P3Y6M4DT12H30M5S".parse(),
Ok(Duration::new(3., 6., 4., 12., 30., 5.))
);
assert_eq!("P23DT23H".parse::<Duration>().unwrap().num_hours(), Some(575.));
assert_eq!("P0.5Y".parse::<Duration>().unwrap().num_years(), Some(0.5));
assert_eq!("P0.5Y0.5M".parse::<Duration>().unwrap().num_months(), Some(6.5));
assert_eq!("P12W".parse::<Duration>().unwrap().num_days(), Some(84.));
assert!("PT".parse::<Duration>().is_err());
assert!("P12WT12H30M5S".parse::<Duration>().is_err());
assert!("P0.5S0.5M".parse::<Duration>().is_err());
assert!("P0.5A".parse::<Duration>().is_err());
Duration
can be converted to either std::time::Duration
or
chrono::Duration
by calling to_std
or to_chrono
.
Both to_std
and to_chrono
will return None
if the duration
includes year
and month
. Because ISO8601 duration format allows
the usage of year
and month
, and these durations are non-standard.
Since months can have 28, 29 30, 31 days, and years can have either
365 or 366 days.
To perform a lossless conversion, a starting date must be specified:
// requires `chrono` feature
use iso8601_duration::Duration;
use chrono::DateTime;
let one_month: Duration = "P1M".parse().unwrap();
let date = DateTime::parse_from_rfc3339("2000-02-01T00:00:00Z").unwrap();
assert_eq!(
one_month.to_chrono_at_datetime(date).num_days(),
29 // 2000 is a leap year
);
License: MIT