这并不难，只是标准语法。你到底犯了哪些错误？

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
解析的名称=Some（“somename”）地址=无
或者对于匹配不同组合的更多控制：

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
仅解析name=“somename”

评论回复：

我不认为您可以用一个match语句做到这一点，但您可以像这样做多个if let语句

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);
}

图纸：
解析名称=“somename”