javascript - 正则表达式:多次匹配相同的表达式并返回表达式前的完整字符串

Question

我有一个看起来像这样的字符串:

 /a/b/entry/0/c/d/entry/0

我想将 /entry/0 与 /entry/[0-9]+ 匹配。对于每次出现 单独并返回一个数组

 ["/a/b/entry/0/", "/a/b/entry/0/c/d/entry/0"]

对于 /entry/[0-9]+ 在字符串中出现的任意次数。 另请注意，我想将表达式保留在输出字符串中。

是否可以使用单个 Regex 表达式和 JavaScript 中的 string.match 或 string.split 函数？

Answer 1

您也可以使用split 和reduce。

var input = "/a/b/entry/0/c/d/entry/0";
var output = input.split( /(?<=entry\/[0-9]+)/ )
                  .reduce( (a, c) => { 
                       a.push( (a[a.length - 1] || "") + c ); //push new item after appending c to last item in accumulator
                       return a; return accumulator
                  } ,[])//initialize accumulator

演示

var input = "/a/b/entry/0/c/d/entry/0";
var output = input.split(/(?<=entry\/[0-9]+)/)
  .reduce((a, c) => {
    a.push((a[a.length - 1] || "") + c); //push new item after appending c to last item in accumulator
    return a;
    return accumulator
  }, []) //initialize accumulator

console.log(output);

编辑

使用@Wiktor的建议match然后reduce

var input = "/a/b/entry/0/c/d/entry/0";
var output = (input.match(/.*?\/entry\/\d+(?:\/|$)/g) || []) //check for null
                  .reduce( (a, c) => { 
                       a.push( (a[a.length - 1] || "") + c ); //push new item after appending c to last item in accumulator
                       return a; return accumulator
                  } ,[])//initialize accumulator