Question

我在输出两个单独的问题时遇到问题。代码是：

System.out.println("Please enter full name: ");

String name = keyboard.nextLine();

while (name.length() >= 21) 
{
    System.out.println("Name is invalid, please re enter:");
    name = keyboard.nextLine();
}


System.out.println("Please enter reference number: ");
String reference = keyboard.nextLine();

while (reference.length() > 6) {
    System.out.println("Refrence incorrect, please re enter");
    reference = keyboard.nextLine();
}
while (!reference.matches("(?i)[A-Z]{2}[0-9]{3}[A-Z]")) {
    System.out.println("Reference is incorrect, please re-enter:");
    reference = keyboard.nextLine();

然而输出的内容类似于：

Please enter full name:

Please enter reference number:

没有空间让我输入名称或参考。即使我这样做，也要求再次参考。有人能在我的代码中发现任何问题吗？（如果你不能告诉我优雅的编码哈哈，我是初学者）

之前的代码是：

    Scanner keyboard = new Scanner(System.in);

    System.out.println("Please select from the following options:");
    System.out.println("1. Enter new Policy");
    System.out.println("2. Display summary of policies");
    System.out.println("3. Display summary of policies for selected month");
    System.out.println("4. Find and display Policy");
    System.out.println("0. Exit");
    int option = keyboard.nextInt();

    if (option == 1) {
        System.out.println("Please enter full name: ");

...

Answer 1

问题在于你打电话

int option = keyboard.nextInt();

它不会读取最后一个换行符，您可以通过调用

来解决此问题

int option = Integer.parseInt(keyboard.nextLine());

nextLine()也会使用换行符，但会返回String，因此您需要将其解析为Integer。

修改

如果输入（在nextInter中）不是整数，则会得到NumberFormatException。要处理此问题，您必须使用try-catch子句：

int option; try { option = Integer.parseInt(keyboard.nextLine()); } catch (NumberFormatException e) { e.printStackTrace(); }

输出在同一条线上？

1 个答案: