你可以使用更高阶的函数来返回你的提取器的“编译”版本，这样更快。

使用正则表达式，并在闭包中编译一次正则表达式，Javascript的匹配将返回所有匹配项。

这让我们只需要删除我们用作标记的东西(例如:{{)，我们可以使用字符串长度来实现slice。

function extract([beg, end]) {
    const matcher = new RegExp(`${beg}(.*?)${end}`,'gm');
    const normalise = (str) => str.slice(beg.length,end.length*-1);
    return function(str) {
        return str.match(matcher).map(normalise);
    }
}

编译一次，使用多次…

const stringExtractor = extract(['{','}']);
const stuffIneed = stringExtractor('this {is} some {text} that can be {extracted} with a {reusable} function');
// Outputs: [ 'is', 'text', 'extracted', 'reusable' ]

或者一次性使用……

const stuffIneed = extract(['{','}'])('this {is} some {text} that can be {extracted} with a {reusable} function');
// Outputs: [ 'is', 'text', 'extracted', 'reusable' ]

也可以看看Javascript的replace函数，但使用了替换参数的函数(如果你在做一个迷你模板引擎(字符串插值)，你会这样做……lodash。Get也可以帮助你得到你想要替换的值?.．.

我的回答太长了，但它可能会帮助到一些人!

获取两个子字符串之间的字符串(包含多于1个字符)

function substrInBetween(whole_str, str1, str2){
   if (whole_str.indexOf(str1) === -1 || whole_str.indexOf(str2) === -1) {
       return undefined; // or ""
  }
  var strlength1 = str1.length;
  return whole_str.substring(
                whole_str.indexOf(str1) + strlength1, 
                whole_str.indexOf(str2)
               );

   }

注意，我使用indexOf()而不是lastIndexOf()，因此它将检查这些字符串的第一次出现

这是我刚做的东西。

注意，如果在启动后没有找到end，函数将返回启动后的所有内容。它还期望只有一次出现的开始和结束，如果有多次，它将只考虑第一次。

许可:公共领域

/**
 * Extracts a string from `source` that is placed between `start` and `end`. The function
 * considers only one instance of start and before, or the first instance and does not support
 * multiple occurences otherwise. If end string is not found, it will return everything after
 * `start` to the end of the string.
 */
export function stringBetween(source, start, end) {
  if (source.indexOf(start) === -1) {
    return null;
  }

  const sourceSplitByStartString = source.split(start);

  // Note: If start string is the very first occurence in source string, the result will be an
  // array where the first item is an empty string and the next item is of interest.

  if (
    sourceSplitByStartString.length === 1
    || sourceSplitByStartString[1] === ''
  ) {
    // It means that start is either the entire string or is at the very end of the string, so there
    // is not anything between
    return '';
  }

  const afterStart = sourceSplitByStartString[1];

  // If the after separator is not found, return everything after the start separator to the end
  // of the string
  if (afterStart.indexOf(end) === -1) {
    return afterStart;
  }

  const afterStartSplitByEnd = afterStart.split(end);

  if (afterStartSplitByEnd[0] === '') {
    return '';
  }

  return afterStartSplitByEnd[0];
}

测试:

import { stringBetween } from './string';

describe('string utlities', () => {
  describe('stringBetween', () => {
    it('Extracts a substring between 2 other substrings', () => {
      const sample1 = stringBetween('Black cat climbed the tree fast.', 'cat ', ' the tree');
      expect(sample1).toBe('climbed');

      const sample2 = stringBetween('Black cat climbed the tree fast.', 'Black ', ' fast.');
      expect(sample2).toBe('cat climbed the tree');
    });

    it('extracts everything after start if end is not found', () => {
      const sample2 = stringBetween('Black cat climbed the tree fast.', 'Black ', 'not-there');
      expect(sample2).toBe('cat climbed the tree fast.');
    });

    it('returns empty string if start string occurs at the end', () => {
      const sample = stringBetween('Black cat climbed the tree fast.', 'fast.', 'climbed');
      expect(sample).toBe('');
    });

    it('returns empty string if start string is the entire string', () => {
      const sample = stringBetween('Black cat', 'Black cat', 'climbed');
      expect(sample).toBe('');
    });

    it('returns empty string if there is not anything between start and end', () => {
      const sample = stringBetween('Black cat climbed the tree fast.', 'climbed ', 'the tree');
      expect(sample).toBe('');
    });

    it('returns null if start string does not exist in the source string', () => {
      const sample = stringBetween('Black cat climbed the tree fast.', 'not-there ', 'the tree');
      expect(sample).toBe(null);
    });
  });
});

我使用@tsds的方式，但只使用分裂函数。

var str = 'one:two;three';    
str.split(':')[1].split(';')[0] // returns 'two'

警告:如果字符串中没有“:”，访问数组的“1”索引将抛出错误!str.split(“:”)[1]

因此，如果存在不确定性，@tsds的方式更安全

str.split(':').pop().split(';')[0]

var s = 'MyLongString:StringIWant;';
/:([^;]+);/.exec(s)[1]; // StringIWant

var str = '[basic_salary]+100/[basic_salary]';
var arr = str.split('');
var myArr = [];
for(var i=0;i<arr.length;i++){
    if(arr[i] == '['){
        var a = '';
        for(var j=i+1;j<arr.length;j++){
            if(arr[j] == ']'){
                var i = j-1;
                break;
            }else{
                a += arr[j];
            }
        }
        myArr.push(a);
    }
    var operatorsArr = ['+','-','*','/','%'];
    if(operatorsArr.includes(arr[i])){
        myArr.push(arr[i]);
    }
    var numbArr = ['0','1','2','3','4','5','6','7','8','9'];
    if(numbArr.includes(arr[i])){
        var a = '';
        for(var j=i;j<arr.length;j++){
            if(numbArr.includes(arr[j])){
                a += arr[j];
            }else{
                var i = j-1;
                break;
            }
        }
        myArr.push(a);
    }
}
myArr = ["basic_salary", "+", "100", "/", "basic_salary"]