检查无效的用户输入java

因此，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，将跳过此打印

现在，你需要多少张成人票

然后直接进入儿童票

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为什么呢

仅仅因为输入是3，并且只有当standerracetickets==1 | | standerracetickets==2时才会执行if语句，所以您需要改善您的条件

这里需要的是嵌套的。到目前为止，您的上一张账单System.out.Print您需要多少张儿童票；对于任何输入，将始终运行

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? ");

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请使用以下开关盒构造

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? ");
}

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如果输入为3，则显示错误消息-当输入不正确时，应显示错误消息。如果输入为1或2，则两个有效输入均不显示错误消息，并输入成人票号。然后将显示子票据提示

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;
}

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你必须在一个循环中调用它，直到用户输入正确的输入。你需要在打印无效输入后扫描响应…是的，你需要一个循环，我认为后一部分应该进入第二个if块。第二个if块可以是if的else块，它可以提高性能。因为您是这里的新用户，让我告诉您，如果您得到问题的解决方案，您应该在这里接受答案。这有助于其他人了解什么对你有用。要接受答案，您只需单击要选择的答案左侧的右勾号。

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? ");