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

DEMO

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

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

console.log(objArray.sort(function(a, b) { return a.name > b.name}));

其他回答

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;
});

你必须传递一个接受两个参数的函数,比较它们,并返回一个数字,所以假设你想要根据ID对它们排序,你会写…

objArray.sort(function(a,b) {
    return a.id-b.id;
});
// objArray is now sorted by Id
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

你必须这样做:

objArray.sort(function(a, b) {
    var textA = a.DepartmentName.toUpperCase();
    var textB = b.DepartmentName.toUpperCase();
    return (textA < textB) ? -1 : (textA > textB) ? 1 : 0;
});

注意:改变大小写(改为大写或小写)可以确保不区分大小写。

因为这里给出的所有解决方案都没有null/undefined安全操作,所以我用这种方式处理(你可以根据自己的需要处理null):

ES5

objArray.sort(
  function(a, b) {
    var departmentNameA = a.DepartmentName ? a.DepartmentName : '';
    var departmentNameB = b.DepartmentName ? b.DepartmentName : '';

    departmentNameA.localeCompare(departmentNameB);
  }
);

ES6+

objArray.sort(
 (a: DepartmentFactory, b: DepartmentFactory): number => {
   const departmentNameA = a.DepartmentName ? a.DepartmentName : '';
   const departmentNameB = b.DepartmentName ? b.DepartmentName : '';

   departmentNameA.localeCompare(departmentNameB);
 }
);

我还删除了其他人使用的toLowerCase,因为localeCompare是不区分大小写的。另外,在使用Typescript或ES6+时,我更喜欢在参数上更明确一点,以便对未来的开发人员更明确。