我知道我迟到了，但是你觉得我今天做的这个EMCMAScript 2017解决方案怎么样?它处理多个匹配，因为如果两个键有相同的值会发生什么?这就是我创建这个小片段的原因。

当有一个匹配时，它只返回一个字符串，但当有几个匹配时，它返回一个数组。

let object = { nine_eleven_was_a_inside_job: false, javascript_isnt_useful: false } // Complex, dirty but useful. Handle mutiple matchs which is the main difficulty. Object.prototype.getKeyByValue = function (val) { let array = []; let array2 = []; // Get all the key in the object. for(const [key] of Object.entries(this)) { if (this[key] == val) { // Putting them in the 1st array. array.push(key) } } // List all the value of the 1st array. for(key of array) { // "If one of the key in the array is equal to the value passed in the function (val), it means that 'val' correspond to it." if(this[key] == val) { // Push all the matchs. array2.push(key); } } // Check the lenght of the array. if (array2.length < 2) { // If it's under 2, only return the single value but not in the array. return array2[0]; } else { // If it's above or equal to 2, return the entire array. return array2; } } /* Basic way to do it wich doesn't handle multiple matchs. let getKeyByValue = function (object, val) { for(const [key, content] of Object.entries(object)) { if (object[key] === val) { return key } } } */ console.log(object.getKeyByValue(false))

或者，更简单的是—按您想要的顺序创建一个具有键和值的新对象，然后查找该对象。我们在使用上面的原型代码时发生了冲突。你不必在键周围使用String函数，那是可选的。

 newLookUpObj = {};
 $.each(oldLookUpObj,function(key,value){
        newLookUpObj[value] = String(key);
    });

lodash way https://lodash.com/docs#findKey

Var用户= { 'barney':{'年龄':36，'活跃':true}， 'fred': {'age': 40， 'active': false}， '鹅卵石':{'年龄':1，'活动':true} }; _。findKey(users， {'age': 1， 'active': true}); //→'鹅卵石'

以下是我的解决方案:

例如，我假设我们有一个包含三个值对的对象:

function findKey(object, value) {

    for (let key in object)
        if (object[key] === value) return key;

    return "key is not found";
}

const object = { id_1: "apple", id_2: "pear", id_3: "peach" };

console.log(findKey(object, "pear"));
//expected output: id_2

我们可以简单地编写一个findKey(数组，value)，它接受两个参数，一个是对象，一个是你正在寻找的键的值。因此，该方法是可重用的，您不需要每次都手动迭代对象，只需为该函数传递两个参数。

这里有一个Lodash解决方案，适用于flat key =>值对象，而不是嵌套对象。已接受的答案建议使用_。findKey适用于具有嵌套对象的对象，但在这种常见情况下不起作用。

这种方法将对象颠倒，将键交换为值，然后通过查找新(颠倒的)对象上的值来查找键。如果没有找到键，则返回false，我更喜欢undefined，但你可以很容易地在_的第三个参数中交换它。getKey()中的get方法。

// Get an object's key by value var getKey = function( obj, value ) { var inverse = _.invert( obj ); return _.get( inverse, value, false ); }; // US states used as an example var states = { "AL": "Alabama", "AK": "Alaska", "AS": "American Samoa", "AZ": "Arizona", "AR": "Arkansas", "CA": "California", "CO": "Colorado", "CT": "Connecticut", "DE": "Delaware", "DC": "District Of Columbia", "FM": "Federated States Of Micronesia", "FL": "Florida", "GA": "Georgia", "GU": "Guam", "HI": "Hawaii", "ID": "Idaho", "IL": "Illinois", "IN": "Indiana", "IA": "Iowa", "KS": "Kansas", "KY": "Kentucky", "LA": "Louisiana", "ME": "Maine", "MH": "Marshall Islands", "MD": "Maryland", "MA": "Massachusetts", "MI": "Michigan", "MN": "Minnesota", "MS": "Mississippi", "MO": "Missouri", "MT": "Montana", "NE": "Nebraska", "NV": "Nevada", "NH": "New Hampshire", "NJ": "New Jersey", "NM": "New Mexico", "NY": "New York", "NC": "North Carolina", "ND": "North Dakota", "MP": "Northern Mariana Islands", "OH": "Ohio", "OK": "Oklahoma", "OR": "Oregon", "PW": "Palau", "PA": "Pennsylvania", "PR": "Puerto Rico", "RI": "Rhode Island", "SC": "South Carolina", "SD": "South Dakota", "TN": "Tennessee", "TX": "Texas", "UT": "Utah", "VT": "Vermont", "VI": "Virgin Islands", "VA": "Virginia", "WA": "Washington", "WV": "West Virginia", "WI": "Wisconsin", "WY": "Wyoming" }; console.log( 'The key for "Massachusetts" is "' + getKey( states, 'Massachusetts' ) + '"' ); <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js"></script>

我创建了bimap库(https://github.com/alethes/bimap)，它实现了一个强大、灵活和高效的JavaScript双向地图接口。它没有依赖关系，在服务器端(在Node.js中，你可以用npm install bimap安装它)和浏览器中(通过链接到lib/bimap.js)都可以使用。

基本操作非常简单:

var bimap = new BiMap;
bimap.push("k", "v");
bimap.key("k") // => "v"
bimap.val("v") // => "k"

bimap.push("UK", ["London", "Manchester"]);
bimap.key("UK"); // => ["London", "Manchester"]
bimap.val("London"); // => "UK"
bimap.val("Manchester"); // => "UK"

在两个方向上，键值映射的检索同样快。底层没有昂贵的对象/数组遍历，因此无论数据大小如何，平均访问时间都保持不变。