根据数组的一个属性按字母顺序对数组中的对象排序

假设你有一个这样的JavaScript类

var DepartmentFactory = function(data) {
    this.id = data.Id;
    this.name = data.DepartmentName;
    this.active = data.Active;
}

假设您随后创建了该类的许多实例，并将它们存储在一个数组中

var objArray = [];
objArray.push(DepartmentFactory({Id: 1, DepartmentName: 'Marketing', Active: true}));
objArray.push(DepartmentFactory({Id: 2, DepartmentName: 'Sales', Active: true}));
objArray.push(DepartmentFactory({Id: 3, DepartmentName: 'Development', Active: true}));
objArray.push(DepartmentFactory({Id: 4, DepartmentName: 'Accounting', Active: true}));

现在我将拥有一个由DepartmentFactory创建的对象数组。如何使用array.sort()方法按每个对象的DepartmentName属性对这个对象数组进行排序?

array.sort()方法在对字符串数组排序时工作得很好

var myarray=["Bob", "Bully", "Amy"];
myarray.sort(); //Array now becomes ["Amy", "Bob", "Bully"]

但是我如何让它与对象列表一起工作呢?

当前回答

// Sorts an array of objects "in place". (Meaning that the original array will be modified and nothing gets returned.)
function sortOn (arr, prop) {
    arr.sort (
        function (a, b) {
            if (a[prop] < b[prop]){
                return -1;
            } else if (a[prop] > b[prop]){
                return 1;
            } else {
                return 0;   
            }
        }
    );
}

//Usage example:

var cars = [
        {make:"AMC",        model:"Pacer",  year:1978},
        {make:"Koenigsegg", model:"CCGT",   year:2011},
        {make:"Pagani",     model:"Zonda",  year:2006},
        ];

// ------- make -------
sortOn(cars, "make");
console.log(cars);

/* OUTPUT:
AMC         : Pacer : 1978
Koenigsegg  : CCGT  : 2011
Pagani      : Zonda : 2006
*/



// ------- model -------
sortOn(cars, "model");
console.log(cars);

/* OUTPUT:
Koenigsegg  : CCGT  : 2011
AMC         : Pacer : 1978
Pagani      : Zonda : 2006
*/



// ------- year -------
sortOn(cars, "year");
console.log(cars);

/* OUTPUT:
AMC         : Pacer : 1978
Pagani      : Zonda : 2006
Koenigsegg  : CCGT  : 2011
*/

2013-01-09 08:23:21

其他回答

objArray.sort((a, b) => a.DepartmentName.localeCompare(b.DepartmentName))

2020-04-04 18:56:01

一个简单的答案:

objArray.sort(function(obj1, obj2) {
   return obj1.DepartmentName > obj2.DepartmentName;
});

ES6道:

objArray.sort((obj1, obj2) => {return obj1.DepartmentName > obj2.DepartmentName};

如果你需要让它小写或大写等，只要这样做，并将结果存储在一个变量中，而不是比较该变量。例子:

objArray.sort((obj1, obj2) => {
   var firstObj = obj1.toLowerCase();
   var secondObj = obj2.toLowerCase();
   return firstObj.DepartmentName > secondObj.DepartmentName;
});

2016-07-29 01:15:53

要支持unicode:

objArray.sort(function(a, b) {
   return a.DepartmentName.localeCompare(b.DepartmentName);
});

2012-01-17 20:02:07

像这样做

objArrayy.sort(function(a, b){
 var nameA=a.name.toLowerCase(), nameB=b.name.toLowerCase()
 if (nameA < nameB) //sort string ascending
  return -1
 if (nameA > nameB)
  return 1
 return 0 //default return value (no sorting)
});
console.log(objArray)

2013-01-04 06:55:57

ES6代码更短

objArray.sort((a, b) => a.DepartmentName.toLowerCase().localeCompare(b.DepartmentName.toLowerCase()))

2020-08-21 14:42:57

根据数组的一个属性按字母顺序对数组中的对象排序

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