250

有没有办法获取对象的函数名?

function alertClassOrObject (o) {
   window.alert(o.objectName); //"myObj" OR "myClass" as a String
}

function myClass () {
   this.foo = function () {
       alertClassOrObject(this);
   }
}

var myObj = new myClass();
myObj.foo();

for (var k in this) {...}- 没有关于classNameor的信息ObjectName。有可能得到其中之一吗?

4

8 回答 8

431

获取对象的构造函数,然后检查其名称属性。

myObj.constructor.name

返回“我的班级”。

于 2012-04-25T11:15:17.303 回答
32

例子:

function Foo () { console.log('Foo function'); }
var f = new Foo();
console.log('f', f.constructor.name); // -> "Foo"

var Bar = function () { console.log('Anonymous function (as Bar)'); };
var b = new Bar();
console.log('b', b.constructor.name); // -> "Bar"

var Abc = function Xyz() { console.log('Xyz function (as Abc)'); };
var a = new Abc();
console.log('a', a.constructor.name); // -> "Xyz"

class Clazz { constructor() { console.log('Clazz class'); } }
var c = new Clazz();
console.log('c', c.constructor.name); // -> "Clazz"

var otherClass = class Cla2 { constructor() { console.log('Cla2 class (as otherClass)'); } }
var c2 = new otherClass();
console.log('c2', c2.constructor.name); // -> "Cla2"

于 2013-11-21T13:27:05.460 回答
6

由于已经回答了这个问题,我只想指出在 JavaScript 中获取对象构造函数的方法的不同。构造函数和实际的对象/类名称之间存在差异。如果以下内容增加了您决定的复杂性,那么您可能正在寻找instanceof. 或者也许你应该问自己“我为什么要这样做?这真的是我想要解决的问题吗?”

笔记:

在较旧的obj.constructor.name浏览器上不可用。匹配(\w+)应该满足 ES6 样式类。

代码:

var what = function(obj) {
  return obj.toString().match(/ (\w+)/)[1];
};

var p;

// Normal obj with constructor.
function Entity() {}
p = new Entity();
console.log("constructor:", what(p.constructor), "name:", p.constructor.name , "class:", what(p));

// Obj with prototype overriden.
function Player() { console.warn('Player constructor called.'); }
Player.prototype = new Entity();
p = new Player();
console.log("constructor:", what(p.constructor), "name:", p.constructor.name, "class:", what(p));

// Obj with constructor property overriden.
function OtherPlayer() { console.warn('OtherPlayer constructor called.'); }
OtherPlayer.constructor = new Player();
p = new OtherPlayer();
console.log("constructor:", what(p.constructor), "name:", p.constructor.name, "class:", what(p));

// Anonymous function obj.
p = new Function("");
console.log("constructor:", what(p.constructor), "name:", p.constructor.name, "class:", what(p));

// No constructor here.
p = {};
console.log("constructor:", what(p.constructor), "name:", p.constructor.name, "class:", what(p));

// ES6 class.
class NPC { 
  constructor() {
  }
}
p = new NPC();
console.log("constructor:", what(p.constructor), "name:", p.constructor.name , "class:", what(p));

// ES6 class extended
class Boss extends NPC {
  constructor() {
    super();
  }
}
p = new Boss();
console.log("constructor:", what(p.constructor), "name:", p.constructor.name , "class:", what(p));

结果:

在此处输入图像描述

代码:https ://jsbin.com/wikiji/edit?js,console

于 2016-10-25T21:43:46.780 回答
5

如果您使用标准 IIFE(例如使用 TypeScript)

var Zamboch;
(function (_Zamboch) {
    (function (Web) {
        (function (Common) {
            var App = (function () {
                function App() {
                }
                App.prototype.hello = function () {
                    console.log('Hello App');
                };
                return App;
            })();
            Common.App = App;
        })(Web.Common || (Web.Common = {}));
        var Common = Web.Common;
    })(_Zamboch.Web || (_Zamboch.Web = {}));
    var Web = _Zamboch.Web;
})(Zamboch || (Zamboch = {}));

你可以预先注释原型

setupReflection(Zamboch, 'Zamboch', 'Zamboch');

然后使用 _fullname 和 _classname 字段。

var app=new Zamboch.Web.Common.App();
console.log(app._fullname);

注释功能在这里:

function setupReflection(ns, fullname, name) {
    // I have only classes and namespaces starting with capital letter
    if (name[0] >= 'A' && name[0] <= 'Z') {
        var type = typeof ns;
        if (type == 'object') {
            ns._refmark = ns._refmark || 0;
            ns._fullname = fullname;
            var keys = Object.keys(ns);
            if (keys.length != ns._refmark) {
                // set marker to avoid recusion, just in case 
                ns._refmark = keys.length;
                for (var nested in ns) {
                    var nestedvalue = ns[nested];
                    setupReflection(nestedvalue, fullname + '.' + nested, nested);
                }
            }
        } else if (type == 'function' && ns.prototype) {
            ns._fullname = fullname;
            ns._classname = name;
            ns.prototype._fullname = fullname;
            ns.prototype._classname = name;
        }
    }
}

提琴手

于 2014-04-04T12:08:17.023 回答
3

试试这个:

var classname = ("" + obj.constructor).split("function ")[1].split("(")[0];
于 2015-03-26T12:06:35.887 回答
3

我面临着类似的困难,这里提出的解决方案都不是我正在研究的最佳解决方案。我所拥有的是一系列以模式显示内容的函数,我试图在单个对象定义下重构它,从而制作类的函数和方法。当我发现其中一种方法在模态本身内部创建了一些导航按钮时,问题就出现了,这些导航按钮使用 onClick 到其中一个函数——现在是该类的一个对象。我已经考虑(并且仍在考虑)处理这些导航按钮的其他方法,但是通过扫描父窗口中定义的变量,我能够找到类本身的变量名称。我所做的是搜索与我班级的“实例”匹配的任何内容,以防万一,

var myClass = function(varName)
{
    this.instanceName = ((varName != null) && (typeof(varName) == 'string') && (varName != '')) ? varName : null;

    /**
     * caching autosweep of window to try to find this instance's variable name
     **/
    this.getInstanceName = function() {
        if(this.instanceName == null)
        {
            for(z in window) {
                if((window[z] instanceof myClass) && (window[z].uniqueProperty === this.uniqueProperty)) {
                    this.instanceName = z;
                    break;
                }
            }
        }
        return this.instanceName;
    }
}
于 2015-12-10T17:38:49.817 回答
1

我们所需要的:

  1. 在函数中包装一个常量(其中函数的名称等于我们要获取的对象的名称)
  2. 在对象内部使用箭头函数

console.clear();
function App(){ // name of my constant is App
  return {
  a: {
    b: {
      c: ()=>{ // very important here, use arrow function 
        console.log(this.constructor.name)
      }
    }
  }
}
}
const obj = new App(); // usage

obj.a.b.c(); // App

// usage with react props etc, 
// For instance, we want to pass this callback to some component

const myComponent = {};
myComponent.customProps = obj.a.b.c;
myComponent.customProps(); // App

于 2020-03-31T08:34:13.107 回答
1

在运行时获取类名的最有效方法

let className = this.constructor.name

于 2021-03-10T04:23:58.507 回答