convert_to_array = function (sentence) {
     return sentence.trim().split(" ");
};

let ages = convert_to_array ("I'm a programmer in javascript writing script");

function confirmEnding(string) {
let target = "ipt";
    return  (string.substr(-target.length) === target) ? true : false;
}

function mySearchResult() {
return ages.filter(confirmEnding);
}

mySearchResult();

您可以像这样检查并使用过滤器返回匹配单词的数组

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
   }
}

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

convert_to_array = function (sentence) {
     return sentence.trim().split(" ");
};

let ages = convert_to_array ("I'm a programmer in javascript writing script");

function confirmEnding(string) {
let target = "ipt";
    return  (string.substr(-target.length) === target) ? true : false;
}

function mySearchResult() {
return ages.filter(confirmEnding);
}

mySearchResult();

您可以像这样检查并使用过滤器返回匹配单词的数组

基于t。j。克劳德的答案

使用转义的RegExp测试至少一个子字符串的“至少一次”出现。

函数buildSearch(substrings) { 返回新的RegExp( 子字符串 . map(函数(s) {s.replace返回 (/[.*+?^${}()|[\]\\]/ g , '\\$&');}) .join('{1，}|') + '{1，}' ）; ｝ var pattern = buildSearch(['hello'，'world']); console.log(模式。测试('你好')); console.log(模式。Test ('what a wonderful world')); console.log(模式。Test ('my name is…'));

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