Question

我想弄清楚为什么我的toString（）函数只会在添加另一个披萨时才能正常工作。这是第一个创建的披萨的输出，这是预期的并且运作良好

$ kitchen = new Kitchen()                                              
$ pizza = new Pizza()                                                   
$ pizza.toString()                                                      

-> "no crust pizza with no toppings and no sauce: $0.00"              
$ pizza.ingredients.add(kitchen.ingredients.get(1))                     
-> true                                                                 
$ pizza.toString()                                                     

-> "Thick crust pizza with no toppings and no sauce: $3.50"             
$ pizza.ingredients.add(kitchen.ingredients.get(2))                     
-> true                                                                
$ pizza.toString()                                                 

-> "Thick crust pizza with no toppings and Tomato sauce: $4.50"         
$ pizza.ingredients.add(kitchen.ingredients.get(5))                   
-> true                                                            
$ pizza.toString()                                                     

-> "Thick crust pizza with Olives and Tomato sauce: $6.00"           
$ pizza.ingredients.add(kitchen.ingredients.get(7))                    
-> true                                                              
$ pizza.toString()                                                     

-> "Thick crust pizza with Olives, Beef and Tomato sauce: $8.75"      
$ pizza.ingredients.add(kitchen.ingredients.get(4))                     
-> true                                                                
$ pizza.toString()                                                      

-> "Thick crust pizza with Olives, Beef, Capsicum and Tomato sauce:

但另一个披萨的输出如下

EXPECTED OUTPUT                                                         Current output

Creating new pizza                                                      Creating new pizza
Ingredient(s): Thin                                                     Ingredient(s): Thin
Thin crust pizza with no toppings and no sauce: $3.00                <
Ingredient(s): Tomato                                                   Ingredient(s): Tomato
Thin crust pizza with no toppings and Tomato sauce: $4.00            <
Ingredient(s): Jalapenos                                                Ingredient(s): Jalapenos
Thin crust pizza with Jalapenos and Tomato sauce: $5.00              <
Ingredient(s): Beef                                                     Ingredient(s): Beef
Thin crust pizza with Jalapenos, Beef and Tomato sauce: $7.75        <

=============================================== ==========================编辑

在分析我的代码之后，似乎没有修改Pizza类中的链接列表成分，因此问题将出现在我的测试方法中。

public void testing(Ingredient ingredient1){  
    for (Ingredient ingredient : ingredients){
        if (ingredient1.ingtoString().equals(ingredient)){
            ingredients.add(ingredient1);
        }
    }
}

这个方法由我的Kitchen类调用，就像这样。

  public void nameMatch(String match){
        Pizza pizza = new Pizza();
        for (Ingredient ingredient : ingredients){
            if (match.equals(ingredient.ingtoString())){
                pizza.testing(ingredient);
            }
        }
        pizza.toString();
    }

最后，这就是用于创建披萨并调用这些方法的内容

public void create() {
    System.out.println("Creating new pizza");
    String ingredient1 = "";
    while (!ingredient1.equals(".")){
        System.out.print("Ingredient(s): ");
        ingredient1 = (In.nextLine());
        kitchen.nameMatch(ingredient1);
    }
    System.out.println("ORDER SUMMARY");
    kitchen.nameMatch(ingredient1);
}

快速记录所有这些方法都在3个单独的类中

Answer 1

每当您致电nameMatch时，您都会创建一个新披萨，将该成分添加到披萨中，然后将披萨扔掉。我想你想要制作一个披萨及其所有成分。

另外，nameMatch实际上并没有打印任何东西。 pizza.toString()会将披萨转换为字符串，但如果您想打印该字符串，则需要拨打System.out.println - 例如System.out.println(pizza.toString());

toString方法似乎只运行一次，即使它被称为多个

1 个答案: