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

// 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
*/

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。

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。

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

// use `new DepartmentFactory` as given below. `new` is imporatant

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}));

function sortOn(property){
    return function(a, b){
        if(a[property] < b[property]){
            return -1;
        }else if(a[property] > b[property]){
            return 1;
        }else{
            return 0;   
        }
    }
}

//objArray.sort(sortOn("id")); // because `this.id = data.Id;`
objArray.sort(sortOn("name")); // because `this.name = data.DepartmentName;`
console.log(objArray);

演示:http://jsfiddle.net/diode/hdgeH/