如何在JS(Javascript)中重载对象的构造函数?

如何在JS(Javascript)中重载对象的构造函数?,javascript,constructor,overriding,Javascript,Constructor,Overriding,我能做点什么吗 function User(form) { this._username = form.username.value; this._password = form.password.value; this._surname = form.surname.value; this._lastname = form.lastname.value; this._birthdate = form.b_day.value+"-"+form.b_mont

我能做点什么吗

function User(form) {
    this._username = form.username.value;
    this._password = form.password.value;
    this._surname = form.surname.value;
    this._lastname = form.lastname.value;
    this._birthdate = form.b_day.value+"-"+form.b_month.value+"-"+form.b_year.value;
    this._avatar = form.avatar;
    this._messages = new Array();
    this._messagesCount=0;
}

function User(userName,password,surname,lastName,birthdate) {
    this._username = userName;
    this._password = password;
    this._surname = surname;
    this._lastname = lastName;
    this._birthdate = birthdate;
    this._avatar = form.avatar;
    this._messages = new Array();
    this._messagesCount=0;
}

不可以,JavaScript不支持任何类型的重载

您可以做的是将已经填充了值的对象传递到构造函数中,然后从对象中获取值,但这会复制代码


或者,您可以创建一个默认构造函数并添加方法,如
initFromUser
setFromForm
,然后获取相应的参数并设置对象值,
new User()。initFormForm(form)
在我看来非常干净。

您不能这样做,因为JavaScript不是强类型语言,所以它不会看到表单和用户名之间的区别。您可以创建多个函数,如
createUserFromForm(form)
createUserFromUserInfo(用户名、密码等)
或者,您可以尝试使用未指定参数的单数构造函数,然后使用arguments集合检查输入并决定要执行的操作。

我喜欢Ilya Volodins的答案,我想我会添加以下内容作为示例:

function foo() {
    var evt = window.event || arguments[1] || arguments.callee.caller.arguments[0];
    var target = evt.target || evt.srcElement;

    var options = {};

    if (arguments[0]) options = arguments[0];

    var default_args = {
        'myNumber'      :   42,
        'myString'      :   'Hello',
        'myBoolean'     :   true
    }
    for (var index in default_args) {
        if (typeof options[index] == "undefined") options[index] = default_args[index];
    }

    //Do your thing

}

//then you call it like this
foo();

//or

foo({'myString' : 'World'});

//or

foo({'myNumber' : 666, 'myString' : 'World', 'myBoolean' : false});

可能有更好的方法可以做到这一点,但这只是一个例子。

通过计算参数的数量来重载构造函数或任何其他Javascript函数:

function FooString()
{   if(arguments.length>0)
    {   this.str=arguments[0];
        return;
    }
    this.str="";
}

var s1=new FooString;
var s2=new FooString("hello world");

您还可以通过检测缺少多少参数来设置默认参数。

您可以使用JSON字符串和typeof命令的组合轻松模拟重载方法和构造函数。请参见下面的示例-您可以根据输入的数据类型来确定val属性的形状:

function test(vals)
    {
        this.initialise = function (vals) {

            if (typeof (vals) == 'undefined')
            {
                this.value = 10;
            }
            else if (Object.prototype.toString.call(vals) === '[object Array]')
            {
                this.value = vals[0];
            }
            else if (typeof (vals) === 'object') {
                if (vals.hasOwnProperty('x')) {
                    this.value = vals.x;
                }
                else if (vals.hasOwnProperty('y')) {
                    this.value = vals.y;
                }
            }
            else {
                this.value = vals; // e.g. it might be a string or number
            }

        }

        this.otherMethods = function () {
            // other methods in the class
        }

        this.initialise(vals);
    }

    var obj1 = test(); // obj1.val = 10;
    var obj2 = test([30, 40, 50]); // obj1.val = 30;
    var obj3 = test({ x: 60, y: 70 }); // obj1.val = 60;
    var obj4 = test({ y: 80 }); // obj1.val = 80;
    var obj5 = test('value'); // obj1.val = 'value';
    var obj6 = test(90); // obj1.val = 90;

您可以使用ES6功能创建构造函数,如下所示

class Person {
  constructor(name, surname) {
    if (typeof name === "object") {
      this.name = name.name;
      this.surname = name.surname;
    } else {
      this.name = name;
      this.surname = surname;
    }
  }
}

const person1 = new Person("Rufat", "Gulabli");
const person2 = new Person({ name: "Rufat", surname: "Gulabli" });
const person3 = new Person();
console.log(person1);
console.log(person2);
console.log(person3);
打印输出:

  • 人{姓名:'Rufat',姓氏:'Gulabli'}
  • 人{姓名:'Rufat',姓氏:'Gulabli'}
  • 人{姓名:未定义,姓氏:未定义}