javascript - 如何从此日志输出中找到最里面的失败模式

Question

鉴于此日志记录输出。我想匹配所有失败命令的path/to/*.command。在本例中是第三个和第四个命令。

    Starting.. path/to/first.command
      Some Text..
    Done

    Starting.. other/path/to/second.command
      Some Other Text..
    Done

    Starting.. other/path/to/third.command
      Some Text..
    Fail

    Starting.. other/path/to/forth.command
      Some Other Text..
    Fail

这就是我想出的开始..(.+\.command)[\s\S]+?失败

但这还不够好。不情愿的量词与最内部的匹配third.command 不匹配。但它匹配封闭的first.command(就正则表达式而言，这是正确的，但不是期望的)

此处演示:https://regex101.com/r/fl3eaz/1

Answer 1

[\s\S]+将贪婪地匹配任何字符序列，包括换行符，但您只想搜索到 Fail 所在的位置。或Done遇到。因为Some Text行始终恰好是一行长，通过(在命令之后)匹配 single [\s\S] 来利用这一点(换行符)，后跟一行字符，再后跟另一个 [\s\S]+ (换行符)，后跟 Fail .

const input = `
    Starting.. path/to/first.command
      Some Text..
    Done

    Starting.. other/path/to/second.command
      Some Other Text..
    Done

    Starting.. other/path/to/third.command
      Some Text..
    Fail

    Starting.. other/path/to/forth.command
      Some Other Text..
    Fail
    `;
const re = /Starting\.\. (.+\.command)[\s\S].+[\s\S] +Fail/g;
let match;
while (match = re.exec(input)) {
  console.log(match[1]);
}

如果您使用(较新、支持较少的)lookbehind，则会更简单:

const input = `
    Starting.. path/to/first.command
      Some Text..
    Done

    Starting.. other/path/to/second.command
      Some Other Text..
    Done

    Starting.. other/path/to/third.command
      Some Text..
    Fail

    Starting.. other/path/to/forth.command
      Some Other Text..
    Fail
    `;
const re = /(?<=Starting\.\. +).+\.command(?=[\s\S].+[\s\S] +Fail)/g;
console.log(input.match(re));