如果你正在使用下划线或Lodash库，你可以使用_。findKey功能:

var users = {
  'barney':  { 'age': 36, 'active': true },
  'fred':    { 'age': 40, 'active': false },
  'pebbles': { 'age': 1,  'active': true }
};

_.findKey(users, function(o) { return o.age < 40; });
// => 'barney' (iteration order is not guaranteed)

// The `_.matches` iteratee shorthand.
_.findKey(users, { 'age': 1, 'active': true });
// => 'pebbles'

// The `_.matchesProperty` iteratee shorthand.
_.findKey(users, ['active', false]);
// => 'fred'

// The `_.property` iteratee shorthand.
_.findKey(users, 'active');
// => 'barney'

这为我获得对象的键/值工作。

让obj = { “key1”:“value1”, “key2”:“value2”, “key3”:“value3”, “key4”:“value4” ｝ 种(obj) . map函数(k) { Console.log ("key with value: "+k +" = "+obj[k]) })

没有可用的标准方法。你需要迭代，你可以创建一个简单的helper:

Object.prototype.getKeyByValue = function( value ) {
    for( var prop in this ) {
        if( this.hasOwnProperty( prop ) ) {
             if( this[ prop ] === value )
                 return prop;
        }
    }
}

var test = {
   key1: 42,
   key2: 'foo'
};

test.getKeyByValue( 42 );  // returns 'key1'

提醒一句:即使上述方法有效，扩展任何主机或本机对象的.prototype通常也不是一个好主意。我这样做是因为它很适合这个问题。不管怎样，你应该在.prototype之外使用这个函数，并将对象传递给它。

如果你有一个数组值的对象。这里有一个很好的例子。让我们假设您希望根据所拥有的文件的扩展名显示一个图标。具有相同图标的所有扩展都在相同的对象值下。

注意:将这里的case包装在一个对象中要比使用大量case进行切换要好。

检查下面的代码片段(用es6编写)，看看我们如何为特定的扩展返回特定的键。

我从这个git仓库拿到了扩展名列表

// Oject that contains different icons for different extentions const icons = { "music": ["mp3", "m4a", "ogg", "acc", "flac","m3u", "wav"], "video": ["mp4","webm", "mkv", "avi", "mov", "m4v", "mpeg"], "image": ["jpg", "gif", "png", "jpeg", "tif", "psd", "raw", "ico"], "archives": ["zip", "rar", "tar", "dmg", "jar"], "3d-files": ["3ds", "dwg", "obj", "dae", "skp", "fbx"], "text": ["doc", "rtf", "txt", "odt", "tex"], "vector-graphics":["ai", "svg"], "pdf": ["pdf"], "data": ["xml", "csv", "xls"] } const get_icon_Key =( icons_object,file_extention) => { // For each key we chack if the value is contained in the list of values let key = Object.keys(icons_object).find( k=> icons[k].find( // At this leve we check if the extention exist in the array of the specific object value ie. 'music', 'video' ... icons_ext => icons_ext === file_extention) // if we find it means this is the key we are looking for ? true: false); return key } console.log(`The icon of for mp3 extention is: => ${get_icon_Key(icons,"mp3")}`) console.log(`The icon of for mp4 extention is: => ${get_icon_Key(icons,"mp4")}`)

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

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))