Question

程序将要求用户输入他想要搜索的项目的代码。如果该项目的代码存在，它将打印到屏幕上，所有工作正常，直到这里。问题是当用户输入不存在的代码时，程序将无法运行。它不会打印“找不到项目”

这是代码

public void searchItem(){
    boolean invalidInput;
    int q = -1;

    do {        
        try {   
            boolean found = false;
            invalidInput = false;

            System.out.println("Enter the item's code you want to search for : ");
            q = s.nextInt();

            out: for (int i = 0; i<items.length; i++){

                if(q == items[i].getCode()){
                    System.out.println(items[i].toString());
                    found = true;
                    System.exit(2);
                }
                counter++;
            }
            if(!found)
                System.out.print("Item not found");


        } catch (InputMismatchException e) {
            System.out.println("Please enter a valid code [Numbers Only]");
            s.next();
            invalidInput = true;  // This is what will get the program to loop back
        }
    } while (invalidInput);  
}

Answer 1

如果我使用你的代码的这个（浓缩形式）它可以工作，并打印＆＃34;项目未找到＆＃34;正如我们所期望的那样......问题就在于我感觉到的其他地方......

如果您输入缺失（但有效）的商品编号，请提供进一步的信息！

public static void main(String[] args) {
    boolean invalidInput;
    int q = -1;
    int[] items = { 1, 2, 3, 4 };

    do {
        boolean found = false;
        invalidInput = false;

        System.out.println("Enter the item's code you want to search for : ");
        q = 5;

        for (int i = 0; i < items.length; i++) {
            if (q == items[i]) {
                System.out.println(items[i]);
                found = true;
                System.exit(2);
            }
        }
        if (!found)
            System.out.print("Item not found");
    } while (invalidInput);
}

if（！found）语句无法正常工作

1 个答案: