2

我更愿意利用 Hyperhyper::header::Headers#get方法的类型安全性,而不是get_raw使用&str.

实现这一目标的最佳方法是什么?

4

1 回答 1

3

挖掘hyper::header::Headers源代码,我发现有一个用于生成代码的简洁宏:header!. 但是,您将需要一些咒语才能使其有用:

#[macro_use]
extern crate hyper;

use hyper::{Body, Method, Request, Response};
use std::fmt::{self, Display};
use std::str::FromStr;
use std::num::ParseIntError;

// For a header that looks like this:
//    x-arbitrary-header-with-an-integer: 8

#[derive(Clone, Debug, Eq, PartialEq)]
pub struct ArbitraryNumber(i8);

impl Display for ArbitraryNumber {
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        write!(f, "Arbitrary Protocol v{}", self.0)
    }
}

impl FromStr for ArbitraryNumber {
    type Err = ParseIntError;

    fn from_str(s: &str) -> Result<Self, Self::Err> {
        s.parse::<i8>().map(|int| ArbitraryNumber(int))
    }
}

//impl Header for ArbitraryNumberHeader
header! { (ArbitraryNumberHeader, "x-arbitrary-header-with-an-integer") => [ArbitraryNumber] }

一旦你在范围内有了一个Response命名res,你可以像这样访问这个标头:

let arbitrary_header: AribitraryNumber = res.headers().get::<ArbitraryNumberHeader>().unwrap();
于 2018-02-16T20:07:42.420 回答