原型对象/方法代码 - 不在控制台上显示答案

【问题标题】：prototype object/method code - not displaying on cosole原型对象/方法代码 - 不在控制台上显示
【发布时间】：2016-05-23 22:09:59
【问题描述】：

所以这段代码是用于我的 javascript 类的，所以它对于网站本身来说并不是必不可少的，但我仍然被困住了。所以我正在尝试编写一个代码，当我调用某个名称和方法（名称、类型 1、类型 2、身高或体重）时，我会得到我在代码中提供的答案。我越来越不确定，但是在查看我的讲师的演示时，它可以工作。

我希望我没有错过一些非常简单的东西。非常感谢任何帮助！

 var Pokemon = function Pokemon(name , type1 , type2 , height , weight) {
  this.name = name;
  this.type1 = type1;
  this.type2 = type2;
  this.height = height;
  this.weight = weight;
}

Pokemon.prototype.whichPoke = function whichPoke() {
  console.log("Hey, I am " + this.name);
}
Pokemon.prototype.whichType1 = function whichType1() {
  console.log("Hey, I am " + this.type1);
}
Pokemon.prototype.whichType2 = function whichType2() {
  console.log ("Hey, I am " + this.type2);
}
Pokemon.prototype.whichHeight = function whichHeight() {
  console.log ("Hey, I am " + this.height + " inches tall");
}
Pokemon.prototype.whichWeight = function whichWeight() {
  console.log ("Hey, I am " + this.weight + " pounds")
}

var Horsea = new Pokemon({
  name: "Horsea",
  type1: "Water",
  type2: "none",
  height: "16",
  weight: "17.6"
})
var Seadra = new Pokemon({
  name: "Seadra",
  type1: "Water",
  type2: "none",
  height: "47",
  weight: "55.1"
})
var Kingdra = new Pokemon({
  name: "Kingdra",
  type1: "Water",
  type2: "Dragon",
  Height: "71",
  Weight: "335.1"
})

【问题讨论】：

标签： javascript function methods console prototype

【解决方案1】：

 var Pokemon = function Pokemon(name , type1 , type2 , height , weight) {
  this.name = name;
  this.type1 = type1;
  this.type2 = type2;
  this.height = height;
  this.weight = weight;
}

Pokemon.prototype.whichPoke = function whichPoke() {
  console.log("Hey, I am " + this.name);
}
Pokemon.prototype.whichType1 = function whichType1() {
  console.log("Hey, I am " + this.type1);
}
Pokemon.prototype.whichType2 = function whichType2() {
  console.log ("Hey, I am " + this.type2);
}
Pokemon.prototype.whichHeight = function whichHeight() {
  console.log ("Hey, I am " + this.height + " inches tall");
}
Pokemon.prototype.whichWeight = function whichWeight() {
  console.log ("Hey, I am " + this.weight + " pounds")
}

var Horsea = new Pokemon("Horsea", "Water", "none", "16", "17.6");
var Seadra = new Pokemon("Seadra", "Water", "none", "47", "55.1");
var Kingdra = new Pokemon("Kingdra", "Water", "Dragon", "71", "335.1");

【讨论】：

您传递给创建 Pokemon 对象的参数 {...} 是一个对象而不是 5 个数据。它被识别为一个参数 - “名称”，而不是 5 个参数。所以剩下的参数 type1, type2, height, weight 都是未定义的。
tinypic.com/r/sl4rj9/9 好了，这是我的问题。我没有得到任何具体信息，仅此而已。哈哈
我很尴尬。我的代码确实有效，只是我没有把我的 () 放在我的函数调用之后。
永远不要为任何类型的错误感到难过:)