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

以下是目前为止(在我看来)最好的解决方案。这是一个现代的(ES6)解决方案,它:

是高效的(一行!) 避免for循环 与其他答案中使用的some()函数不同,这个函数不仅返回一个布尔值(true/false) 相反,它要么返回子字符串(如果它在数组中找到),要么返回undefined 更进一步,允许您选择是否需要部分子字符串匹配(示例如下)

享受吧!



const arrayOfStrings = ['abc', 'def', 'xyz'];
const str = 'abc';
const found = arrayOfStrings.find(v => (str === v));

在这里,found将被设置为'abc'。这将适用于精确的字符串匹配。

如果你用:

const found = arrayOfStrings.find(v => str.includes(v));

同样,found在本例中被设置为'abc'。这不允许部分匹配,所以如果str被设置为'ab', found将是未定义的。


And, if you want partial matches to work, simply flip it so you're doing:
const found = arrayOfStrings.find(v => v.includes(str));

代替。如果str被设为'ab' found就会被设为'abc'

容易peasy !



其他回答

对于用谷歌搜索的人来说,

确切的答案应该是。

const substrings = ['connect', 'ready'];
const str = 'disconnect';
if (substrings.some(v => str === v)) {
   // Will only return when the `str` is included in the `substrings`
}

没有内置的东西可以帮你做这个,你必须为它写一个函数,尽管它可以只是一个对某些数组方法的回调。

有两种方法:

数组一些方法 正则表达式

数组的一些

数组some方法(在ES5中添加)使这非常简单:

if (substrings.some(function(v) { return str.indexOf(v) >= 0; })) {
    // There's at least one
}

更好的箭头函数和newish includes方法(都是ES2015+):

if (substrings.some(v => str.includes(v))) {
    // There's at least one
}

生活例子:

const substrings = ["one", "two", "three"]; let str; // Setup console.log(`Substrings: ${substrings}`); // Try it where we expect a match str = "this has one"; if (substrings.some(v => str.includes(v))) { console.log(`Match using "${str}"`); } else { console.log(`No match using "${str}"`); } // Try it where we DON'T expect a match str = "this doesn't have any"; if (substrings.some(v => str.includes(v))) { console.log(`Match using "${str}"`); } else { console.log(`No match using "${str}"`); }

正则表达式

如果你知道字符串不包含任何正则表达式中的特殊字符,那么你可以欺骗一下,像这样:

if (new RegExp(substrings.join("|")).test(string)) {
    // At least one match
}

...这将创建一个正则表达式,它是您正在寻找的子字符串的一系列更改(例如,one|two),并测试是否有匹配它们中的任何一个,但如果任何子字符串包含正则表达式中的任何特殊字符(*,[等),您必须首先转义它们,您最好只做无聊的循环。有关如何逃离它们的信息,请参阅这个问题的答案。

生活例子:

const substrings = ["one", "two", "three"]; let str; // Setup console.log(`Substrings: ${substrings}`); // Try it where we expect a match str = "this has one"; if (new RegExp(substrings.join("|")).test(str)) { console.log(`Match using "${str}"`); } else { console.log(`No match using "${str}"`); } // Try it where we DON'T expect a match str = "this doesn't have any"; if (new RegExp(substrings.join("|")).test(str)) { console.log(`Match using "${str}"`); } else { console.log(`No match using "${str}"`); }

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
function containsAny(str, substrings) {
    for (var i = 0; i != substrings.length; i++) {
       var substring = substrings[i];
       if (str.indexOf(substring) != - 1) {
         return substring;
       }
    }
    return null; 
}

var result = containsAny("defg", ["ab", "cd", "ef"]);
console.log("String was found in substring " + result);

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