如何在JavaScript中调用原型之外的函数？

Question

如何调用setTile范围之外的函数TopDownGame？我试过TopDownGame.Lesson31.setTile(x,y),但它不起作用.

var TopDownGame = TopDownGame || {};

TopDownGame.Lesson31 = function() {};

TopDownGame.Lesson31.prototype = {
    setTile: function(x, y) {
        console.log(tile);
    }
};

Answer 1

如果已添加到原型,则必须创建对象的实例以调用方法:

var TopDownGame = TopDownGame || {};

TopDownGame.Lesson31 = function() {};
TopDownGame.Lesson31.prototype = {

setTile: function(x, y) {
        console.log("setTile invoked");
    },
};


var instance = new TopDownGame.Lesson31();
instance.setTile(3, 4);

你试图调用它,就像它是一个静态方法.如果这是您真正想要做的,请将方法定义为函数的属性,而不是原型的属性.

TopDownGame.Lesson31 = function() {};
TopDownGame.Lesson31.staticMethod = function() {
    console.log('Static method invoked');
}

TopDownGame.Lesson31.staticMethod();

但是如果你真的想保留setTile原型方法,但仍然调用它,你可以使用该apply方法.

var TopDownGame = TopDownGame || {};

TopDownGame.Lesson31 = function() {};
TopDownGame.Lesson31.prototype = {
    setTile: function(x, y) {
        console.log(`setTile invoked, this='${this}', x=${x}, y=${y}`);
    },
};

new TopDownGame.Lesson31().setTile(3, 4);
TopDownGame.prototype.setTile.apply('actually a string', [5, 6]);

这将导致:

setTile invoked, this='[object Object]', x=3, y=4
setTile invoked, this='actually a string', x=5, y=7