更新2(2020年9月28日):这种语言在不断发展，所以如果你可以使用Partial(在v2.1中引入)，那么这是我现在最喜欢的实现这一目标的方式。

class Box {
   x: number;
   y: number;
   height: number;
   width: number;

   public constructor(b: Partial<Box> = {}) {
      Object.assign(this, b);
   }
}

// Example use
const a = new Box();
const b = new Box({x: 10, height: 99});
const c = new Box({foo: 10});          // Will fail to compile

更新(2017年6月8日):guyarad和snolflake在他们的评论中对我的回答提出了有效的观点。我建议读者看看Benson, Joe和snolflake的答案，他们的答案比我的更好

原答案(2014年1月27日)

另一个如何实现构造函数重载的例子:

class DateHour {

  private date: Date;
  private relativeHour: number;

  constructor(year: number, month: number, day: number, relativeHour: number);
  constructor(date: Date, relativeHour: number);
  constructor(dateOrYear: any, monthOrRelativeHour: number, day?: number, relativeHour?: number) {
    if (typeof dateOrYear === "number") {
      this.date = new Date(dateOrYear, monthOrRelativeHour, day);
      this.relativeHour = relativeHour;
    } else {
      var date = <Date> dateOrYear;
      this.date = new Date(date.getFullYear(), date.getMonth(), date.getDate());
      this.relativeHour = monthOrRelativeHour;
    }
  }
}

来源:http://mimosite.com/blog/post/2013/04/08/Overloading-in-TypeScript

我知道这是一个老问题，但1.4中的新功能是联合类型;对所有函数重载(包括构造函数)使用这些。例子:

class foo {
    private _name: any;
    constructor(name: string | number) {
        this._name = name;
    }
}
var f1 = new foo("bar");
var f2 = new foo(1);

As commented in @Benson answer, I used this example in my code and I found it very useful. However I found with the Object is possibly 'undefined'.ts(2532) error when I tried to make calculations with my class variable types, as the question mark leads them to be of type AssignedType | undefined. Even if undefined case is handled in later execution or with the compiler type enforce <AssignedType> I could not get rid of the error, so could not make the args optional.I solved creating a separated type for the arguments with the question mark params and the class variables without the question marks. Verbose, but worked.

下面是原始代码，给出了类方法()中的错误，如下所示:

/** @class */

class Box {
  public x?: number;
  public y?: number;
  public height?: number;
  public width?: number;

  // The Box class can work double-duty as the interface here since they are identical
  // If you choose to add methods or modify this class, you will need to
  // define and reference a new interface for the incoming parameters object 
  // e.g.:  `constructor(params: BoxObjI = {} as BoxObjI)` 
  constructor(params: Box = {} as Box) {
    // Define the properties of the incoming `params` object here. 
    // Setting a default value with the `= 0` syntax is optional for each parameter
    const {
      x = 0,
      y = 0,
      height = 1,
      width = 1,
    } = params;

    //  If needed, make the parameters publicly accessible
    //  on the class ex.: 'this.var = var'.
    /**  Use jsdoc comments here for inline ide auto-documentation */
    this.x = x;
    this.y = y;
    this.height = height;
    this.width = width;
  }

  method(): void {
    const total = this.x + 1; // ERROR. Object is possibly 'undefined'.ts(2532)
  }
}

const box1 = new Box();
const box2 = new Box({});
const box3 = new Box({ x: 0 });
const box4 = new Box({ x: 0, height: 10 });
const box5 = new Box({ x: 0, y: 87, width: 4, height: 0 });

所以变量不能在类方法中使用。 如果像这样纠正，例如:

method(): void {
    const total = <number> this.x + 1;
}

现在出现这个错误:

Argument of type '{ x: number; y: number; width: number; height: number; }' is not 
assignable to parameter of type 'Box'.
Property 'method' is missing in type '{ x: number; y: number; width: number; height: 
number; }' but required in type 'Box'.ts(2345)

好像整个arguments bundle不再是可选的了。

因此，如果创建了带有可选参数的类型，并且从可选参数中删除了类变量，我就实现了我想要的，参数是可选的，并且能够在类方法中使用它们。下面是解决方案代码:

type BoxParams = {
  x?: number;
  y?: number;
  height?: number;
  width?: number;
}

/** @class */
class Box {
  public x: number;
  public y: number;
  public height: number;
  public width: number;

  // The Box class can work double-duty as the interface here since they are identical
  // If you choose to add methods or modify this class, you will need to
  // define and reference a new interface for the incoming parameters object 
  // e.g.:  `constructor(params: BoxObjI = {} as BoxObjI)` 
  constructor(params: BoxParams = {} as BoxParams) {
    // Define the properties of the incoming `params` object here. 
    // Setting a default value with the `= 0` syntax is optional for each parameter
    const {
      x = 0,
      y = 0,
      height = 1,
      width = 1,
    } = params;

    //  If needed, make the parameters publicly accessible
    //  on the class ex.: 'this.var = var'.
    /**  Use jsdoc comments here for inline ide auto-documentation */
    this.x = x;
    this.y = y;
    this.height = height;
    this.width = width;
  }

  method(): void {
    const total = this.x + 1;
  }
}

const box1 = new Box();
const box2 = new Box({});
const box3 = new Box({ x: 0 });
const box4 = new Box({ x: 0, height: 10 });
const box5 = new Box({ x: 0, y: 87, width: 4, height: 0 });

感谢任何花时间阅读并试图理解我想要表达的观点的人的评论。

提前谢谢你。

一般来说，对于N个重载，最好使用:

constructor(obj?: {fromType1: IType1} | {fromType2: IType2}) {    
    if(obj){
      if(obj.fromType1){
        //must be of form IType1
      } else if(obj.fromType2){
        //must have used a IType2
      } else {
        throw "Invalid argument 1"
      }
    } else {
      //obj not given
    }
}   

至少现在我们可以确定走哪条路，并采取相应的行动

我们可以使用守卫来模拟构造函数重载

interface IUser {
  name: string;
  lastName: string;
}

interface IUserRaw {
  UserName: string;
  UserLastName: string;
}

function isUserRaw(user): user is IUserRaw {
  return !!(user.UserName && user.UserLastName);
}

class User {
  name: string;
  lastName: string;

  constructor(data: IUser | IUserRaw) {
    if (isUserRaw(data)) {
      this.name = data.UserName;
      this.lastName = data.UserLastName;
    } else {
      this.name = data.name;
      this.lastName = data.lastName;
    }
  }
}

const user  = new User({ name: "Jhon", lastName: "Doe" })
const user2 = new User({ UserName: "Jhon", UserLastName: "Doe" })

在可选的类型参数足够好的情况下，考虑以下代码，它可以在不重复属性或定义接口的情况下实现相同的功能:

export class Track {
   public title: string;
   public artist: string;
   public lyrics: string;

   constructor(track?: Track) {
     Object.assign(this, track);
   }
}

请记住，这将分配所有传递到track的属性，即使它们没有在track上定义。