最短一行

let key = Object.keys(obj).find(k=>obj[k]===value);

返回值为:

let keys = Object.keys(obj).filter(k=>obj[k]===value);

如果value为数组或对象:

let keys = Object.keys(obj).filter(k=>JSON.stringify(obj[k])===JSON.stringify(value));

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))

没有看到以下内容:

Const obj = { id: 1、 名称:“窝” }; 函数getKeyByValue(obj, value) { 返回Object.entries (obj)。Find (([， name]) => value === name); ｝ const [key] = getKeyByValue(obj， 'Den'); console.log(关键)

我使用这个函数:

Object.prototype.getKey = function(value){
  for(var key in this){
    if(this[key] == value){
      return key;
    }
  }
  return null;
};

用法:

// ISO 639: 2-letter codes
var languageCodes = {
  DA: 'Danish',
  DE: 'German',
  DZ: 'Bhutani',
  EL: 'Greek',
  EN: 'English',
  EO: 'Esperanto',
  ES: 'Spanish'
};

var key = languageCodes.getKey('Greek');
console.log(key); // EL

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>

给定输入={"a":"x"， "b":"y"， "c":"x"}…

使用第一个值(如输出={“x”:“一”,“y”:“b”}):

输入= { “一个”:“x”, “b”:“y”, “c”:“x” ｝ output = Object.keys(input)。reduceRight(函数(accum, key, i) { Accum [input[key]] = key; 返回accum; }, {}) console.log(输出)

使用最后一个值(例如输出={“x”:“c”、“y”:“b”}):

输入= { “一个”:“x”, “b”:“y”, “c”:“x” ｝ output = Object.keys(input)。Reduce(函数(accum, key, i) { Accum [input[key]] = key; 返回accum; }, {}) console.log(输出)

为每个值数组的键(例如输出={“x”:“c”,“a”,“y”:[b]}):

输入= { “一个”:“x”, “b”:“y”, “c”:“x” ｝ output = Object.keys(input)。reduceRight(函数(accum, key, i) { accum[输入[主要]]= (accum[输入[键 ]] || []). concat(关键); 返回accum; }, {}) console.log(输出)