非常直截了当。在javascript中,我需要检查字符串是否包含数组中持有的任何子字符串。


当前回答

如果数组不大,可以使用indexOf()循环并逐个检查每个子字符串。或者,您可以构造一个带有子字符串作为替代的正则表达式,这可能更有效,也可能不更有效。

其他回答

var str = "texttexttext";
var arr = ["asd", "ghj", "xtte"];
for (var i = 0, len = arr.length; i < len; ++i) {
    if (str.indexOf(arr[i]) != -1) {
        // str contains arr[i]
    }
}

编辑: 如果测试的顺序不重要,你可以使用这个(只有一个循环变量):

var str = "texttexttext";
var arr = ["asd", "ghj", "xtte"];
for (var i = arr.length - 1; i >= 0; --i) {
    if (str.indexOf(arr[i]) != -1) {
        // str contains arr[i]
    }
}

借鉴T.J. Crowder的解决方案,我创建了一个原型来处理这个问题:

Array.prototype.check = function (s) {
  return this.some((v) => {
    return s.indexOf(v) >= 0;
  });
};

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

这太迟了,但我刚刚遇到了一个问题。在我自己的项目中,我使用以下方法来检查字符串是否在数组中:

["a","b"].includes('a')     // true
["a","b"].includes('b')     // true
["a","b"].includes('c')     // false

通过这种方式,你可以获取一个预定义数组并检查它是否包含字符串:

var parameters = ['a','b']
parameters.includes('a')    // true
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
   }
}