Java.util.scanner错误处理 [英] Java.util.scanner error handling

问题描述

我正在帮助一个Java问题的朋友.但是，我们遇到了障碍.我们正在使用Java.Util.Scanner.nextInt()从用户那里获取一个数字，不断询问用户是否还提供了其他任何东西.唯一的问题是，我们无法弄清楚如何进行错误处理.

I'm helping a friend with a java problem. However, we've hit a snag. We're using Java.Util.Scanner.nextInt() to get a number from the user, asking continiously if the user gives anything else. Only problem is, we can't figure out how to do the error handeling.

我们尝试过的事情:

do {
  int reloop = 0;
  try {
    number = nextInt(); 
  } catch (Exception e) {
    System.out.println ("Please enter a number!");
    reloop ++; 
  }
} while(reloop != 0);

唯一的问题是，如果您输入的不是数字，则会无限循环.

Only problem is, this loops indefinatly if you enter in something not a number.

有帮助吗?

推荐答案

您可以使用nextLine().

You can use hasNextInt() to verify that the Scanner will succeed if you do a nextInt(). You can also call and discard nextLine() if you want to skip the "garbage".

所以，像这样:

Scanner sc = new Scanner(System.in);
while (!sc.hasNextInt()) {
   System.out.println("int, please!");
   sc.nextLine();
}
int num = sc.nextInt();
System.out.println("Thank you! (" + num + ")");

另请参见:

• See also:

  • How do I keep a scanner from throwing exceptions when the wrong type is entered? (java)

  除了不必要的冗长的错误处理(因为让nextInt()抛出InputMismatchException而不是检查hasNextInt())外，代码问题还在于，当执行时，例外，您不要将Scanner移过有问题的输入！这就是为什么您会陷入无限循环！

  The problem with your code, in addition to the unnecessarily verbose error handling because you let nextInt() throw an InputMismatchException instead of checking for hasNextInt(), is that when it does throw an exception, you don't advance the Scanner past the problematic input! That's why you get an infinite loop!

  您可以调用并丢弃nextLine()来解决此问题，但更妙的是，如果改用上面介绍的无异常hasNextInt()预检查技术.

  You can call and discard the nextLine() to fix this, but even better is if you use the exception-free hasNextInt() pre-check technique presented above instead.

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