使用 Chrono 时 BadFormat 将字符串解析为日期

Question

I am trying to convert a string to a date using the Chrono library. I always get a BadFormat or NotEnough error:

extern crate chrono;

use chrono::prelude::*;

fn main() {
    let dt1 = DateTime::parse_from_str(
        "2017-08-30 18:34:06.638997932 UTC", 
        "%Y-%m-%d %H:%M:%S.%9f"
    );
    println!("{:?}", dt1);
}

I'm not sure what I am doing wrong.

Answer 1

1. As mentioned in the comments, your format string doesn't allow for the " UTC" part of your string. That's why you get the BadFormat error.

2. If you add " UTC" to your format string, you still get a BadFormat error because you've typed .%9f when it should be %.9f .

3. Once you fix that, you get a NotEnough error because we aren't actually parsing a timezone.

I'd use NaiveDateTime to always parse in UTC and then add the " UTC" onto the format string to ignore it, correcting the typo:

use chrono::prelude::*; // 0.4.9

fn main() {
    let dt1 = NaiveDateTime::parse_from_str(
        "2017-08-30 18:34:06.638997932 UTC",
        "%Y-%m-%d %H:%M:%S%.9f UTC",
    );

    println!("{:?}", dt1); // Ok(2017-08-30T18:34:06.638997932)
}