Javascript中對象和方法的錯誤

Question

我剛剛開始使用Javascript上的OOP。我對編程世界很陌生。你能幫我下面的代碼嗎？我的文本編輯器在'else'塊上顯示語法錯誤。

function Dog(name, breed, weight) { 
    this.name = name; 
    this.breed = breed; 
    this.weight = weight; 
    this.bark = function() { 
     if (this.weight > 25) alert(this.name + " says Woof") 
    } else { 
     alert(this.name + " says Poof"); 
    } 
} 

var fido = new Dog("Fido", "Mixed", 38); 

fido.bark(); 

Answer 1

試試這個

function Dog(name, breed, weight){ 
this.name = name; 
this.breed = breed; 
this.weight = weight; 
this.bark = function(){ 
     if (this.weight > 25){ 
      alert(this.name + " says Woof"); 
     } 
     else { 
      alert(this.name + " says Poof"); 
     } 
    }; 
} 

var fido = new Dog("Fido", "Mixed", 38); 

fido.bark(); 

Answer 2

if (this.weight > 25) 
    alert(this.name + " says Woof") 
} 
else { 
    alert(this.name + " says Poof"); 
} 

你如果{

Answer 3

你缺少不打開一個{後if (this.weight > 25)以及一個.和bark();

之間 fido

function Dog(name, breed, weight){ 
    this.name = name; 
    this.breed = breed; 
    this.weight = weight; 
    this.bark = function(){ 
     if (this.weight > 25){ 
      alert(this.name + " says Woof") 
     } else { 
      alert(this.name + " says Poof"); 
     } 
    } 
} 

var fido = new Dog("Fido", "Mixed", 38); 
fido.bark(); 

1. 您需要適當的縮進到更容易看到這樣的東西。
2. 我認爲你的運行時抱怨其他問題，因爲它在函數之外而不是「附加」到if，因爲缺少括號。
3. 也許JavaScript不是學習編程的語言。你只是玩弄學習，或者你嘗試完成某些事情？

Answer 4

if (this.weight > 25) 
    alert(this.name + " says Woof") 
} 

就是它出了問題，你就錯過一個{。它應該是：

if (this.weight > 25) 
{ 
    alert(this.name + " says Woof") 
} 

Answer 5

function Dog(name, breed, weight){ 
this.name = name; 
this.breed = breed; 
this.weight = weight; 
this.bark = function(){ 
if (this.weight > 25){ 
alert(this.name + " says Woof") 
} 
else { 
alert(this.name + " says Poof"); 
} 
} 
} 

var fido = new Dog("Fido", "Mixed", 38); 

fido bark();