捕获同一行复杂表达式中的多个实例

我认为在这种情况下你不需要正则表达式，如果你的日志消息的格式一致，一些好的旧字符串操作就可以解决问题并且更易于维护。

我不知道性能怎么样，但这肯定更容易理解，

我知道这并没有真正回答您关于正则表达式的问题，如果您需要使用正则表达式，我深表歉意。

fn main() {
    let line = "[2025-01-01T08:17:29.791951550Z INFO] Values: [(311154184, Some(389971313710868251)), (311154187, Some(389967898428572732)), (311154182, Some(389971313710868251)), (311154174, Some(389971313710868251)), (311154178, Some(389971313710868251)), (311154197, Some(389811146843151022)), (311154171, Some(389971313710868251)), (311154167, Some(389971313710868251)), (311154185, Some(389967898428572732)), (311154168, Some(389971313710868251)), (311154191, Some(389967898428572732)), (311154196, Some(389811259875653181)), (311154199, Some(0)), (311154192, Some(389967898428572732)), (311154172, Some(389971313710868251)), (311154181, Some(389971313710868251)), (311154177, Some(389971313710868251)), (311154176, Some(389971313710868251)), (311154179, Some(389971313710868251)), (311154183, Some(389971313710868251)), (311154186, Some(389967898428572732)), (311154175, Some(389971313710868251)), (311154173, Some(389971313710868251)), (311154180, Some(389971313710868251)), (311154190, Some(389967898428572732)), (311154170, Some(389971313710868251)), (311154165, None), (311154189, Some(389967898428572732)), (311154169, Some(389971313710868251)), (311154166, None), (311154195, Some(389967898428572732)), (311154198, Some(287087126717485928)), (311154193, Some(389967898428572732)), (311154188, Some(389967898428572732)), (311154194, Some(389967898428572732))]";

    let line = line
        .chars()
        .skip_while(|c| *c != ']')
        .skip_while(|c| *c != '[')
        .collect::<String>();

    let mut result: Vec<(u64, Option<u64>)> = vec![];

    for snippet in line.split("),") {
        let s = snippet.replace(['[', ']', ' ', '(', ')'], "");
        let mut s = s.split(',');

        let a = s.next().unwrap().parse::<u64>().unwrap();
        let b = s
            .next()
            .unwrap()
            .strip_prefix("Some")
            .map(|s| s.parse::<u64>().unwrap());

        result.push((a, b));
    }

    println!("{:?}", result);
}

这是使用正则表达式的解决方案。这将生成一个(u64, Option<u64>)类型为的向量。

use regex::Regex;

fn main() {
    let string = "<YOUR_STRING_HERE>";

    let re = Regex::new(r"\((\d+), (Some\(\d+\)|None)\)").unwrap();
    let result: Vec<(u64, Option<u64>)> = re
        .captures_iter(&string)
        .map(|c| {
            let (_, [first, second]): (&str, [&str; 2]) = c.extract();
            (
                first.parse::<u64>().unwrap(),
                if second == "None" {
                    None
                } else {
                    second[5..(second.len() - 1)].parse::<u64>().ok()
                },
            )
        })
        .collect();
    println!("{:?}", result);
}