代碼一直給人一種InputMismatchException時當我輸入2個字

Question

if (userOption == 2) { 
    System.out.println("You have chosen produce! Please enter (1) for organic or (0) for non-organic."); 
    type = sc.nextInt(); 
    System.out.println("Please enter a name for this produce."); 
    name = sc.next(); 
    sc.nextLine(); 
    System.out.println("Please enter the amount of calories."); 
    calories = sc.nextInt(); 
    System.out.println("Please enter the amount of carbohydrates."); 
    carbohydrates = sc.nextInt(); 

    list.add(new Produce(name, calories, carbohydrates, type)); 
} 

你好，當我把我的話之間的空間爲輸入「姓名」，它給了我一個錯誤InputMismatchError的熱量時，即時通訊中的熱量沒有推杆，在用戶輸入「姓名」之前，卡路里甚至不應該得到輸入。謝謝:)

Answer 1

看來你試圖輸入一個String到Integer。捕捉到了這個，你可以更改您的代碼是這樣的：

if (userOption == 2) { 
      System.out.println("You have chosen produce! Please enter (1) for organic or (0) for non-organic."); 
      try { 
       type = sc.nextInt(); 
      } catch (InputMismatchException e) { 
       System.out.println("Wrong format entered."); 
       // ask question again or move on. 
      } 
      System.out.println("Please enter a name for this produce."); 
      name = sc.next(); 
      sc.nextLine(); 
      System.out.println("Please enter the amount of calories."); 
      calories = sc.nextInt(); 
      System.out.println("Please enter the amount of carbohydrates."); 
      carbohydrates = sc.nextInt(); 

     list.add(new Produce(name, calories, carbohydrates, type)); 
    } 

的try/catch語句將捕獲該異常，並告訴他們有什麼做錯了用戶。

Answer 2

您的輸入是「牛肉炸玉米餅」，其中包含一個白色空間。所述Java-Doc狀態：

掃描器斷開其輸入到使用定界符圖案， 它默認與空白匹配。

因此，您的sc.next();返回「牛肉」，在流上留下「Taco」。你的下一個sc.nextInt();然後返回「塔科」，這是沒有整數並導致成InputMismatchException其中states：

InputMismatchException - 如果下一個標記不匹配Integer正則表達式，或者超出範圍

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要修復它試試這個：

System.out.println("You have chosen produce! Please enter (1) for organic or (0) for non-organic."); 
int type = sc.nextInt(); 
// Clear the input 
sc.nextLine(); 
System.out.println("Please enter a name for this produce."); 
// Read in the whole next line (so nothing is left that can cause an exception) 
String name = sc.nextLine();