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

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)

其他回答

objArray.sort( (a, b) => a.id.localeCompare(b.id, 'en', {'sensitivity': 'base'}));

这将按字母顺序对它们进行排序,并且不区分大小写。它也非常干净,易于阅读:D

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;
});

这是一个简单的函数,你可以用它来排序数组对象的属性;不管属性是字符串类型还是整数类型,它都可以工作。

var cars = [ {make:"AMC", model:"Pacer", year:1978}, {make:"Koenigsegg", model:"CCGT", year:2011}, {make:"Pagani", model:"Zonda", year:2006}, ]; function sortObjectsByProp(objectsArr, prop, ascending = true) { let objectsHaveProp = objectsArr.every(object => object.hasOwnProperty(prop)); if(objectsHaveProp) { let newObjectsArr = objectsArr.slice(); newObjectsArr.sort((a, b) => { if(isNaN(Number(a[prop]))) { let textA = a[prop].toUpperCase(), textB = b[prop].toUpperCase(); if(ascending) { return textA < textB ? -1 : textA > textB ? 1 : 0; } else { return textB < textA ? -1 : textB > textA ? 1 : 0; } } else { return ascending ? a[prop] - b[prop] : b[prop] - a[prop]; } }); return newObjectsArr; } return objectsArr; } let sortedByMake = sortObjectsByProp(cars, "make"); // returns ascending order by its make; let sortedByYear = sortObjectsByProp(cars, "year", false); // returns descending order by its year,since we put false as a third argument; console.log(sortedByMake); console.log(sortedByYear);

// 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
*/

ES6代码更短

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