6

获取从现在到下一个午夜之间的持续时间的惯用方法是什么？

我有这样的功能：

extern crate chrono;

use chrono::prelude::*;
use time;

fn duration_until_next_midnight() -> time::Duration {
    let now = Local::now(); // Fri Dec 08 2017 23:00:00 GMT-0300 (-03)
    // ... how to continue??
}

它应该是Duration1 小时，因为下一个午夜是Sat Dec 09 2017 00:00:00 GMT-0300 (-03)

4

3 回答 3

9

在搜索文档后，我终于找到了缺失的链接：Date::and_hms.

所以，实际上，它很简单：

fn main() {
    let now = Local::now();

    let tomorrow_midnight = (now + Duration::days(1)).date().and_hms(0, 0, 0);

    let duration = tomorrow_midnight.signed_duration_since(now).to_std().unwrap();

    println!("Duration between {:?} and {:?}: {:?}", now, tomorrow_midnight, duration);
}

这个想法很简单：

  • 增加到DateTime明天，
  • 提取Date保留时区的部分，
  • DateTime通过用 指定“00:00:00”来重建新Time的and_hms。

有一个panic!in and_hms，所以必须小心指定正确的时间。

于 2017-12-08T17:34:46.797 回答
3

只需减去两个日期：午夜和现在：

extern crate chrono;
use chrono::prelude::*;
use std::time;

fn duration_until_next_midnight() -> time::Duration {
    let now = Local::now();
    // change to next day and reset hour to 00:00
    let midnight = (now + chrono::Duration::days(1))
        .with_hour(0).unwrap()
        .with_minute(0).unwrap()
        .with_second(0).unwrap()
        .with_nanosecond(0).unwrap();

    println!("Duration between {:?} and {:?}:", now, midnight);
    midnight.signed_duration_since(now).to_std().unwrap()
}

fn main() {
    println!("{:?}", duration_until_next_midnight())
}

根据 Matthieu 的要求，您可以编写如下内容：

fn duration_until_next_midnight() -> Duration {
    let now = Local::now();
    // get the NaiveDate of tomorrow
    let midnight_naivedate = (now + chrono::Duration::days(1)).naive_utc().date();
    // create a NaiveDateTime from it
    let midnight_naivedatetime = NaiveDateTime::new(midnight_naivedate, NaiveTime::from_hms(0, 0, 0));
    // get the local DateTime of midnight
    let midnight: DateTime<Local> = DateTime::from_utc(midnight_naivedatetime, *now.offset());

    println!("Duration between {:?} and {:?}:", now, midnight);
    midnight.signed_duration_since(now).to_std().unwrap()
}

但我不确定它是否更好。

于 2017-12-08T10:12:21.067 回答
1

一种方法是计算到下一个午夜所缺少的秒数，请记住time::Tm同时考虑夏令时和时区：

tm_utcoff: i32

标识用于计算此细分时间值的时区，包括夏令时的任何调整。这是 UTC 以东的秒数。例如，对于美国太平洋夏令时间，该值为 -7*60*60 = -25200。

extern crate time;
use std::time::Duration;

fn duration_until_next_midnight() -> Duration {
    let tnow = time::now();

    Duration::new(
        (86400 - tnow.to_timespec().sec % 86400 - 
        i64::from(tnow.tm_utcoff)) as u64,
        0,
    )
}

如果你想要纳秒级的精度，你必须做更多的数学......

于 2017-12-08T13:04:28.907 回答