如何将计时`DateTime< UTC>`实例转换为`DateTime< Local>`？ [英] How do I convert a chrono `DateTime<UTC>` instance to `DateTime<Local>`?

问题描述

我的目标是将 utc 转换为 loc ：

use chrono::{Local, UTC, TimeZone};

let utc = chrono::UTC::now();
let loc = chrono::Local::now();

println!("{:?}", utc);
println!("{:?}", loc);

println!("{:?}", utc.with_timezone(&Local));
println!("{:?}", Local.from_utc_datetime(&utc.naive_local()));

...产生以下输出：

... which produced the following output:

2015-02-26T16:22:27.873593Z
2015-02-26T17:22:27.873663+01:00
2015-02-26T15:22:27.873593+00:00
2015-02-26T15:22:27.873593+00:00

第二行中显示的 loc 时间是我转换 utc 。

The loc time shown in the second row is what I want to see when converting utc.

如何正确地将 DateTime< UTC> 实例转换为 DateTime< ; Local> ？

How do I properly convert a DateTime<UTC> instance to DateTime<Local>?

我正在使用 chrono 0.2.2 。在 DateTime中。 from_utc 方法甚至告诉我我应该使用 TimeZone 特性。但是，我缺少了一些东西。

I am using chrono 0.2.2. In the DateTime.from_utc method it's even telling me I should use the TimeZone trait. However, I am missing something.

推荐答案

糟糕，感谢您的举报。这是一个错误，并已记录为问题＃26 。这应该在Chrono 0.2.3。中修复。

Oops, thank you for reporting. This is a bug and registered as the issue #26. This should be fixed in Chrono 0.2.3.

除了该错误外， utc.with_timezone（& Local）确实是转换为当地时间的正确方法。有一个重要的身份，就是 utc.with_timezone（& Local）.with_timezone（& UTC）应该等于 utc （特殊情况下，当地时区已更改）。

Besides from the bug, utc.with_timezone(&Local) is indeed a correct way to convert to the local time. There is an important identity that utc.with_timezone(&Local).with_timezone(&UTC) should be equal to utc (except for the exceptional case, where the local time zone has been changed).

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