问题

我需要解析邮件的解析器，并且我需要从中提取发件人的详细信息。我使用了Rust的mail_parser库，其中通过from方法提取了解析后的发件人信息。from方法返回了一个名为HeaderValue的枚举类型。但我不明白的是，如果我需要获取Address(Addr<'x>)，Address是一个包含以下选项的结构体：

pub struct Addr<'x> {
    pub name: Option<Cow<'x, str>>,
    pub address: Option<Cow<'x, str>>,
}

我遇到的问题是无法通过match语句从Address(Addr<'x>)中提取数据并获取选项中的name和address字段。我应该如何编写match语句来实现这个目的？

   pub struct Addr<'x> {
    pub name: Option<Cow<'x, str>>,
    pub address: Option<Cow<'x, str>>,
}

答案1

得分: 0

它不难，只是标准语法。你具体遇到了什么错误？

use std::borrow::Cow;

pub struct Addr<'x> {
pub name: Option<Cow<'x, str>>,
pub address: Option<Cow<'x, str>>,
}

pub enum HeaderValue<'x> {
Address(Addr<'x>),
AddressList(Vec<Addr<'x>>),
Empty,
}

fn main() {
let value = HeaderValue::Address(Addr{name: Some("somename".into()), address: None});
match value {
HeaderValue::Address(Addr{name, address}) => println!("Parsed name={:?} address={:?}", name, address),
_ => println!("Couldn't match"),
}
}

打印
> Parsed name=Some("somename") address=None
或者对匹配不同组合有更多控制的话：
```rust
match value {
        HeaderValue::Address(Addr{name:Some(name), address:Some(address)}) => println!("Parsed both name={:?} address={:?}", name, address),
        HeaderValue::Address(Addr{address:Some(address), ..}) => println!("Parsed only address={:?}", address),
        HeaderValue::Address(Addr{name:Some(name), ..}) => println!("Parsed only name={:?}", name),
        HeaderValue::Address(Addr{..}) => println!("Parsed empty"),
        _ => println!("Couldn't match"),
}

打印
> Parsed only name="somename"

对评论的回答：

我认为你不能在单个 match 语句中实现这个，但你可以使用多个 if let 语句，像这样：

let value = HeaderValue::Address(Addr{name: Some("somename".into()), address: None});
if let HeaderValue::Address(Addr{address:Some(address), ..}) = &value {
    println!("Parsed address={:?}", address);
}
if let HeaderValue::Address(Addr{name:Some(name), ..}) = &value {
    println!("Parsed name={:?}", name);
}

打印:
> Parsed name="somename"

<details>
<summary>英文:</summary>
It&#39;s not any hard, just the standard syntax. What errors did you face exactly?
 

use std::borrow::Cow;

pub struct Addr<'x> {
pub name: Option<Cow<'x, str>>,
pub address: Option<Cow<'x, str>>,
}

pub enum HeaderValue<'x> {
Address(Addr<'x>),
AddressList(Vec<Addr<'x>>),
Empty,
}

fn main() {
let value = HeaderValue::Address(Addr{name: Some("somename".into()), address: None});
match value {
HeaderValue::Address(Addr{name, address}) => println!("Parsed name={:?} address={:?}", name, address),
_ => println!("Couldn't match"),
}
}

Prints
> Parsed name=Some("somename") address=None
Or for more control over matching different combinations:

match value {
HeaderValue::Address(Addr{name:Some(name), address:Some(address)}) => println!("Parsed both name={:?} address={:?}", name, address),
HeaderValue::Address(Addr{address:Some(address), ..}) => println!("Parsed only address={:?}", address),
HeaderValue::Address(Addr{name:Some(name), ..}) => println!("Parsed only name={:?}", name),
HeaderValue::Address(Addr{..}) => println!("Parsed empty"),
_ => println!("Couldn't match"),
}

Prints
> Parsed only name="somename"
## Answer to the comment:
I don't think you can do that with a single `match` statement, but you can do multiple `if let` statements like so

let value = HeaderValue::Address(Addr{name: Some("somename".into()), address: None});
if let HeaderValue::Address(Addr{address:Some(address), ..}) = &value {
println!("Parsed address={:?}", address);
}
if let HeaderValue::Address(Addr{name:Some(name), ..}) = &value {
println!("Parsed name={:?}", name);
}

Prints:
> Parsed name="somename"
</details>