如何检查字符串是否包含JavaScript中的子字符串？

Question

通常我会期待一种String.contains()方法,但似乎没有一种方法.

检查这个的合理方法是什么？

Answer 1

以下列出了当前的可能性:

1.(ES6)String.prototype.includes - 去回答(没有IE支持)

var string = "foo",
    substring = "oo";

console.log(string.includes(substring));

2. ES5和更老 includes

var string = "foo",
    substring = "oo";

console.log(string.indexOf(substring) !== -1);

String.prototype.indexOf返回另一个字符串中字符串的位置.如果没有找到,它将返回String.prototype.includes.

3.includes - 去回答

var string = "foo",
    substring = "oo";

console.log(string.includes(substring));

4. lodash包括 - 去回答

var string = "foo",
    substring = "oo";

console.log(string.indexOf(substring) !== -1);

5. RegExp - 去回答

var string = "foo",
    substring = "oo";

console.log(string.includes(substring));

6.匹配 - 去回答

var string = "foo",
    substring = "oo";

console.log(string.indexOf(substring) !== -1);

性能测试显示String.prototype.indexOf可能是最佳选择,如果它涉及速度问题.

Answer 2

String.prototype.includesES6中有一个:

"potato".includes("to");
> true

请注意,您可能需要加载String.prototype.includes或类似才能在旧版浏览器上使用它.

if (!String.prototype.includes) {
  String.prototype.includes = function(search, start) {
    'use strict';
    if (typeof start !== 'number') {
      start = 0;
    }

    if (start + search.length > this.length) {
      return false;
    } else {
      return this.indexOf(search, start) !== -1;
    }
  };
}

Answer 3

另一种选择是KMP（Knuth–Morris–Pratt）。

与单纯算法O（n？m）的最坏情况相比，KMP算法在最坏情况的O（n + m）时间中搜索长度为n的字符串中的长度为m的子字符串，因此使用KMP可能会如果您担心最坏的情况下的时间复杂度，则是合理的。

这是Nayuki项目的JavaScript实现，取自https://www.nayuki.io/res/knuth-morris-pratt-string-matching/kmp-string-matcher.js：

// Searches for the given pattern string in the given text string using the Knuth-Morris-Pratt string matching algorithm.
// If the pattern is found, this returns the index of the start of the earliest match in 'text'. Otherwise -1 is returned.

// Searches for the given pattern string in the given text string using the Knuth-Morris-Pratt string matching algorithm.
// If the pattern is found, this returns the index of the start of the earliest match in 'text'. Otherwise -1 is returned.