Question

我尝试使用缩短的版本查找并替换字符串的所有实例，并且如果找到它，我想保留对捕获的引用。

我已经写了这段代码：

extern crate regex;
use regex::{Regex, Captures};

//... get buffer from stdin

let re = Regex::new(r"(capture something1) and (capture 2)").unwrap();
let out = re.replace_all(&buffer, |caps: &Captures| {
    if let ref = caps.at(2).unwrap().to_owned() {
        refs.push(ref.to_owned());
    }

    caps.at(1).unwrap().to_owned();
});

不幸的是，编译因错误而失败：

src/bin/remove_links.rs:16:18: 16:29 error: type mismatch resolving `for<'r, 'r> <[closure@src/bin/remove_links.rs:16:39: 22:6] as std::ops::FnOnce<(&'r regex::Captures<'r>,)>>::Output == std::string::String`:
  expected (),
    found struct `std::string::String` [E0271]
src/bin/remove_links.rs:16     let out = re.replace_all(&buffer, |caps: &Captures| {
                                            ^~~~~~~~~~~
src/bin/remove_links.rs:16:18: 16:29 help: run `rustc --explain E0271` to see a detailed explanation
src/bin/remove_links.rs:16:18: 16:29 note: required because of the requirements on the impl of `regex::Replacer` for `[closure@src/bin/remove_links.rs:16:39: 22:6]`

我无法理解它。我还尝试添加use regex::{Regex, Captures, Replacer}，但这根本不会改变错误。

Answer 1

正如@ BurntSushi5指出的那样，你的闭包应该返回一个String。以下是未来参考的完整示例：

extern crate regex;
use regex::{Regex, Captures};

fn main() {
    let buffer = "abcdef";
    let re = Regex::new(r"(\w)bc(\w)").unwrap();
    let out = re.replace_all(&buffer, |caps: &Captures| {
        caps.at(1).unwrap().to_owned()
    });
    println!("{:?}", out); // => "aef"
}

不能使用正则表达式包中的`replace_all`：expected（），找到String

1 个答案: