我用sort的解决方案:

let list = {
    "you": 100, 
    "me": 75, 
    "foo": 116, 
    "bar": 15
};

let sorted = Object.entries(list).sort((a,b) => a[1] - b[1]);

for(let element of sorted) {
    console.log(element[0]+ ": " + element[1]);
}

a = { b: 1, p: 8, c: 2, g: 1 }
Object.keys(a)
  .sort((c,b) => {
    return a[b]-a[c]
  })
  .reduce((acc, cur) => {
    let o = {}
    o[cur] = a[cur]
    acc.push(o)
    return acc
   } , [])

输出= [{p: 8}, {c: 2}, {b: 1}, {g: 1}]

许多类似和有用的功能: https://github.com/shimondoodkin/groupbyfunctions/

function sortobj(obj)
{
    var keys=Object.keys(obj);
    var kva= keys.map(function(k,i)
    {
        return [k,obj[k]];
    });
    kva.sort(function(a,b){
        if(a[1]>b[1]) return -1;if(a[1]<b[1]) return 1;
        return 0
    });
    var o={}
    kva.forEach(function(a){ o[a[0]]=a[1]})
    return o;
}

function sortobjkey(obj,key)
{
    var keys=Object.keys(obj);
    var kva= keys.map(function(k,i)
    {
        return [k,obj[k]];
    });
    kva.sort(function(a,b){
        k=key;      if(a[1][k]>b[1][k]) return -1;if(a[1][k]<b[1][k]) return 1;
        return 0
    });
    var o={}
    kva.forEach(function(a){ o[a[0]]=a[1]})
    return o;
}

谢谢你，继续回答@Nosredna

现在我们知道对象需要转换为数组，然后对数组排序。这对于按字符串排序数组(或转换对象为数组)非常有用:

Object {6: Object, 7: Object, 8: Object, 9: Object, 10: Object, 11: Object, 12: Object}
   6: Object
   id: "6"
   name: "PhD"
   obe_service_type_id: "2"
   __proto__: Object
   7: Object
   id: "7"
   name: "BVC (BPTC)"
   obe_service_type_id: "2"
   __proto__: Object


    //Sort options
    var sortable = [];
    for (var vehicle in options)
    sortable.push([vehicle, options[vehicle]]);
    sortable.sort(function(a, b) {
        return a[1].name < b[1].name ? -1 : 1;
    });


    //sortable => prints  
[Array[2], Array[2], Array[2], Array[2], Array[2], Array[2], Array[2]]
    0: Array[2]
    0: "11"
    1: Object
        id: "11"
        name: "AS/A2"
        obe_service_type_id: "2"
        __proto__: Object
        length: 2
        __proto__: Array[0]
    1: Array[2]
    0: "7"
    1: Object
        id: "7"
        name: "BVC (BPTC)"
        obe_service_type_id: "2"
        __proto__: Object
        length: 2

我遵循slebetman给出的解决方案(去阅读它的所有细节)，但调整，因为你的对象是非嵌套的。

// First create the array of keys/values so that we can sort it:
var sort_array = [];
for (var key in list) {
    sort_array.push({key:key,value:list[key]});
}

// Now sort it:
sort_array.sort(function(x,y){return x.value - y.value});

// Now process that object with it:
for (var i=0;i<sort_array.length;i++) {
    var item = list[sort_array[i].key];

    // now do stuff with each item
}

为了完整起见，这个函数返回对象属性的排序数组:

function sortObject(obj) {
    var arr = [];
    for (var prop in obj) {
        if (obj.hasOwnProperty(prop)) {
            arr.push({
                'key': prop,
                'value': obj[prop]
            });
        }
    }
    arr.sort(function(a, b) { return a.value - b.value; });
    //arr.sort(function(a, b) { return a.value.toLowerCase().localeCompare(b.value.toLowerCase()); }); //use this to sort as strings
    return arr; // returns array
}

var list = {"you": 100, "me": 75, "foo": 116, "bar": 15};
var arr = sortObject(list);
console.log(arr); // [{key:"bar", value:15}, {key:"me", value:75}, {key:"you", value:100}, {key:"foo", value:116}]

JSFiddle上面的代码在这里。此解决方案基于本文。

更新的小提琴排序字符串是在这里。您可以从它中删除额外的. tolowercase()转换，以便区分大小写的字符串比较。