我并不是建议你去扩展/修改String的原型，但这是我所做的:

String.prototype.includes ()

String.prototype.includes = function (includes) { console.warn("String.prototype.includes() has been modified."); return function (searchString, position) { if (searchString instanceof Array) { for (var i = 0; i < searchString.length; i++) { if (includes.call(this, searchString[i], position)) { return true; } } return false; } else { return includes.call(this, searchString, position); } } }(String.prototype.includes); console.log('"Hello, World!".includes("foo");', "Hello, World!".includes("foo") ); // false console.log('"Hello, World!".includes(",");', "Hello, World!".includes(",") ); // true console.log('"Hello, World!".includes(["foo", ","])', "Hello, World!".includes(["foo", ","]) ); // true console.log('"Hello, World!".includes(["foo", ","], 6)', "Hello, World!".includes(["foo", ","], 6) ); // false

单线解决方案

substringsArray.some(substring=>yourBigString.includes(substring))

如果子字符串存在\不存在，则返回true\false

需要ES6支持

如果您正在处理一长串子字符串，这些子字符串由空格或任何其他常见字符分隔的完整“单词”组成，那么您可以在搜索中使用一些聪明的方法。

首先把你的字符串分成X组，然后X+1，然后X+2，…X和Y应分别为子字符串中单词最少和最多的单词数。例如，如果X是1,Y是4，“Alpha Beta Gamma Delta”变成:

"阿尔法" "贝塔" "伽马" "德尔塔"

"阿尔法·贝塔" "贝塔·伽马" "伽马·德尔塔"

"Alpha Beta Gamma" "Beta Gamma Delta"

"Alpha Beta Gamma Delta"

如果X是2 Y是3，那么你可以省略第一行和最后一行。

现在，如果您将它插入到Set(或Map)中，就可以快速搜索这个列表，比字符串比较快得多。

缺点是你不能搜索像“ta Gamm”这样的子字符串。当然，你可以通过按字符而不是按单词进行分割来实现这一点，但这样你通常需要构建一个庞大的Set，并且这样做所花费的时间/内存超过了好处。

var yourstring = 'tasty food'; // the string to check against


var substrings = ['foo','bar'],
    length = substrings.length;
while(length--) {
   if (yourstring.indexOf(substrings[length])!=-1) {
       // one of the substrings is in yourstring
   }
}

let obj = [{name : 'amit'},{name : 'arti'},{name : 'sumit'}];
let input = 'it';

使用滤镜:

obj.filter((n)=> n.name.trim().toLowerCase().includes(input.trim().toLowerCase()))

我并不是建议你去扩展/修改String的原型，但这是我所做的:

String.prototype.includes ()

String.prototype.includes = function (includes) { console.warn("String.prototype.includes() has been modified."); return function (searchString, position) { if (searchString instanceof Array) { for (var i = 0; i < searchString.length; i++) { if (includes.call(this, searchString[i], position)) { return true; } } return false; } else { return includes.call(this, searchString, position); } } }(String.prototype.includes); console.log('"Hello, World!".includes("foo");', "Hello, World!".includes("foo") ); // false console.log('"Hello, World!".includes(",");', "Hello, World!".includes(",") ); // true console.log('"Hello, World!".includes(["foo", ","])', "Hello, World!".includes(["foo", ","]) ); // true console.log('"Hello, World!".includes(["foo", ","], 6)', "Hello, World!".includes(["foo", ","], 6) ); // false