Question

if (standOrTerraceTickets != 1 && standOrTerraceTickets != 2) {
    System.out.print("Invalid input. Please enter 1 for stand tickets or 2 for terrace tickets. ");
}
if (standOrTerraceTickets == 1 || standOrTerraceTickets == 2) {
    System.out.print("Now, how many adult tickets do you require? ");
}
int adultTickets = aScanner.nextInt();
System.out.print("How many children's tickets do you require? ");

所以在上面的这篇文章中，当用户最初输入3（而不是1或2）时，会打印错误消息invalid input。当他们输入1时，它会跳过此打印

Now, how many adult tickets do you require?

并直接转到儿童票上。

为什么？

Answer 1

你需要改善你的条件。

这里需要的是nested if-else。截至目前，对于任何输入，您的最后一个语句System.out.print("How many children's tickets do you require? ");将始终运行。

int adultTickets = 0;
if (standOrTerraceTickets != 1 && standOrTerraceTickets != 2) {
    System.out.print("Invalid input. Please enter 1 for stand tickets or 2 for terrace tickets. ");
}
else if (standOrTerraceTickets == 1 || standOrTerraceTickets == 2) {
    System.out.print("Now, how many adult tickets do you require? ");
    adultTickets = aScanner.nextInt();
    System.out.print("How many children's tickets do you require? ");
}

Answer 2

请使用以下开关案例构造。

switch(standOrTerraceTickets) 
{
    case 1:
    case 2:
        System.out.print("Now, how many adult tickets do you require? ");
        break;
    default:
        System.out.print("Invalid input. Please enter 1 for stand tickets or 2 for terrace tickets. ");
        break;
}

Answer 3

如果输入为3，则显示错误消息 - 输入不正确时，应显示错误消息。如果输入为1或2（两个有效输入），则不会显示错误消息并输入成人票号。然后显示儿童票的提示。

while (standOrTerraceTickets != 1 && standOrTerraceTickets != 2)
{
   System.out.print("Invalid input. Please enter 1 for stand tickets or 2 for terrace tickets. ");
   standOrTerraceTickets = aScanner.nextInt();
}
System.out.print("Now, how many adult tickets do you require? ");
int adultTickets = aScanner.nextInt();
System.out.print("How many children's tickets do you require? ");

Answer 4

只是因为输入是3而if语句仅在standOrTerraceTickets == 1 || standOrTerraceTickets == 2

时执行

检查无效的用户输入java

4 个答案: