我有一个非常简单的JavaScript对象,我将其用作关联数组。是否有一个简单的函数允许我获取值的键,或者我必须迭代对象并手动找到它?


当前回答

http://jsfiddle.net/rTazZ/2/

var a = new Array(); 
    a.push({"1": "apple", "2": "banana"}); 
    a.push({"3": "coconut", "4": "mango"});

    GetIndexByValue(a, "coconut");

    function GetIndexByValue(arrayName, value) {  
    var keyName = "";
    var index = -1;
    for (var i = 0; i < arrayName.length; i++) { 
       var obj = arrayName[i]; 
            for (var key in obj) {          
                if (obj[key] == value) { 
                    keyName = key; 
                    index = i;
                } 
            } 
        }
        //console.log(index); 
        return index;
    } 

其他回答

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。

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))

function extractKeyValue(obj, value) {
    return Object.keys(obj)[Object.values(obj).indexOf(value)];
}

用于闭包编译器提取编译后未知的键名

更性感的版本,但使用未来对象。项功能

function objectKeyByValue (obj, val) {
  return Object.entries(obj).find(i => i[1] === val);
}

使用Underscore.js库:

var hash = {
  foo: 1,
  bar: 2
};

(_.invert(hash))[1]; // => 'foo'

下划线js解决方案

let samplLst = [{id:1,title:Lorem},{id:2,title:Ipsum}]
let sampleKey = _.findLastIndex(samplLst,{_id:2});
//result would be 1
console.log(samplLst[sampleKey])
//output - {id:2,title:Ipsum}