javascript - 如何在 JavaScript 中调用原型(prototype)之外的函数？

Question

如何在 TopDownGame 范围之外调用 setTile 函数？我尝试了 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

如果您已添加到原型(prototype)，则必须创建对象的实例来调用方法:

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);

您试图像调用静态方法一样调用它。如果这就是您真正想要做的，请将方法定义为函数的属性，而不是原型(prototype)的属性。

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

TopDownGame.Lesson31.staticMethod();

但是，如果您确实想保留 setTile 作为原型(prototype)方法，但仍调用它，则可以使用 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