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。

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

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}));

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

在这个问题上做了一些尝试，并尽量减少循环之后，我最终得到了这个解决方案:

codeen上的演示

const items = [
      {
        name: 'One'
      },
      {
        name: 'Maria is here'
      },
      {
        name: 'Another'
      },
      {
        name: 'Z with a z'
      },
      {
        name: '1 number'
      },
      {
        name: 'Two not a number'
      },
      {
        name: 'Third'
      },
      {
        name: 'Giant'
      }
    ];

    const sorted = items.sort((a, b) => {
      return a[name] > b[name];
    });

    let sortedAlphabetically = {};

    for(var item in sorted) {
      const firstLetter = sorted[item].name[0];
      if(sortedAlphabetically[firstLetter]) {
        sortedAlphabetically[firstLetter].push(sorted[item]);
      } else {
        sortedAlphabetically[firstLetter] = [sorted[item]]; 
      }
    }

    console.log('sorted', sortedAlphabetically);