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

function sortAssocObject(list) {
    var sortable = [];
    for (var key in list) {
        sortable.push([key, list[key]]);
    }
    // [["you",100],["me",75],["foo",116],["bar",15]]

    sortable.sort(function(a, b) {
        return (a[1] < b[1] ? -1 : (a[1] > b[1] ? 1 : 0));
    });
    // [["bar",15],["me",75],["you",100],["foo",116]]

    var orderedList = {};
    for (var idx in sortable) {
        orderedList[sortable[idx][0]] = sortable[idx][1];
    }

    return orderedList;
}

sortAssocObject(list);

// {bar: 15, me: 75, you: 100, foo: 116}

感谢@orad在TypeScript中提供了答案。现在，我们可以在JavaScript中使用下面的代码片断。

function sort(obj,valSelector) { const sortedEntries = Object.entries(obj) .sort((a, b) => valSelector(a[1]) > valSelector(b[1]) ? 1 : valSelector(a[1]) < valSelector(b[1]) ? -1 : 0); return new Map(sortedEntries); } const Countries = { "AD": { "name": "Andorra", }, "AE": { "name": "United Arab Emirates", }, "IN": { "name": "India", }} // Sort the object inside object. var sortedMap = sort(Countries, val => val.name); // Convert to object. var sortedObj = {}; sortedMap.forEach((v,k) => { sortedObj[k] = v }); console.log(sortedObj); //Output: {"AD": {"name": "Andorra"},"IN": {"name": "India"},"AE": {"name": "United Arab Emirates"}}

一个过时问题的后续答案。我写了两个函数，一个是按键排序，另一个是按值排序，并在两个函数中以排序形式返回对象。它也应该在字符串上工作，因为这就是我张贴这个的原因(如果值不是数字的话，上面的一些按值排序有困难)。

const a = { absolutely: "works", entirely: 'zen', best: 'player', average: 'joe' } const prop_sort = obj => { return Object.keys(obj) .sort() .reduce((a, v) => { a[v] = obj[v]; return a; }, {}); } const value_sort = obj => { const ret = {} Object.values(obj) .sort() .forEach(val => { const key = Object.keys(obj).find(key => obj[key] == val) ret[key] = val }) return ret } console.log(prop_sort(a)) console.log(value_sort(a))

你的对象可以有任意数量的属性，如果你把对象放在数组中，你可以选择根据你想要的任何对象属性进行排序，数字或字符串。考虑这个数组:

var arrayOfObjects = [   
    {
        name: 'Diana',
        born: 1373925600000, // Mon, Jul 15 2013
        num: 4,
        sex: 'female'
    },
    {

        name: 'Beyonce',
        born: 1366832953000, // Wed, Apr 24 2013
        num: 2,
        sex: 'female'
    },
    {            
        name: 'Albert',
        born: 1370288700000, // Mon, Jun 3 2013
        num: 3,
        sex: 'male'
    },    
    {
        name: 'Doris',
        born: 1354412087000, // Sat, Dec 1 2012
        num: 1,
        sex: 'female'
    }
];

按出生日期排序，最年长的先

// use slice() to copy the array and not just make a reference
var byDate = arrayOfObjects.slice(0);
byDate.sort(function(a,b) {
    return a.born - b.born;
});
console.log('by date:');
console.log(byDate);

按名称排序

var byName = arrayOfObjects.slice(0);
byName.sort(function(a,b) {
    var x = a.name.toLowerCase();
    var y = b.name.toLowerCase();
    return x < y ? -1 : x > y ? 1 : 0;
});

console.log('by name:');
console.log(byName);

http://jsfiddle.net/xsM5s/16/

另一种解决方法:-

var res = [{"s1":5},{"s2":3},{"s3":8}].sort(function(obj1,obj2){ 
 var prop1;
 var prop2;
 for(prop in obj1) {
  prop1=prop;
 }
 for(prop in obj2) {
  prop2=prop;
 }
 //the above two for loops will iterate only once because we use it to find the key
 return obj1[prop1]-obj2[prop2];
});

//res将有结果数组

试试这个。即使你的对象没有你试图排序的属性也会被处理。

只需通过发送属性和对象来调用它。

var sortObjectByProperty = function(property,object){

    console.time("Sorting");
    var  sortedList      = [];
         emptyProperty   = [];
         tempObject      = [];
         nullProperty    = [];
    $.each(object,function(index,entry){
        if(entry.hasOwnProperty(property)){
            var propertyValue = entry[property];
            if(propertyValue!="" && propertyValue!=null){
              sortedList.push({key:propertyValue.toLowerCase().trim(),value:entry});  
            }else{
                emptyProperty.push(entry);
           }
        }else{
            nullProperty.push(entry);
        }
    });

      sortedList.sort(function(a,b){
           return a.key < b.key ? -1 : 1;
         //return a.key < b.key?-1:1;   // Asc 
         //return a.key < b.key?1:-1;  // Desc
      });


    $.each(sortedList,function(key,entry){
        tempObject[tempObject.length] = entry.value;
     });

    if(emptyProperty.length>0){
        tempObject.concat(emptyProperty);
    }
    if(nullProperty.length>0){
        tempObject.concat(nullProperty);
    }
    console.timeEnd("Sorting");
    return tempObject;
}