[英]Java accepting only numbers from user with Scanner
I am trying to understand how to only accept numbers from the user, and I have attempted to do so using try catch blocks, but I would still get errors with it. 我试图了解如何仅接受来自用户的数字,并且尝试使用try catch块来这样做,但是我仍然会收到错误。
Scanner scan = new Scanner(System.in);
boolean bidding;
int startbid;
int bid;
bidding = true;
System.out.println("Alright folks, who wants this unit?" +
"\nHow much. How much. How much money where?" );
startbid = scan.nextInt();
try{
while(bidding){
System.out.println("$" + startbid + "! Whose going to bid higher?");
startbid =+ scan.nextInt();
}
}catch(NumberFormatException nfe){
System.out.println("Please enter a bid");
}
I am trying to understand why it is not working. 我试图了解为什么它不起作用。
I tested it out by inputting a into the console, and I would receive an error instead of the hopeful "Please enter a bid" resolution. 我通过在控制台中输入a进行了测试,我将收到错误消息,而不是希望的“请输入出价”解决方案。
Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Scanner.java:909)
at java.util.Scanner.next(Scanner.java:1530)
at java.util.Scanner.nextInt(Scanner.java:2160)
at java.util.Scanner.nextInt(Scanner.java:2119)
at Auction.test.main(test.java:25)
尝试捕获引发的异常类型,而不是NumberFormatException
( InputMismatchException
)。
The message is pretty clear: Scanner.nextInt()
throws an InputMismatchException
, but your code catches NumberFormatException
. 消息很清楚: Scanner.nextInt()
引发InputMismatchException
,但是您的代码捕获了NumberFormatException
。 Catch the appropriate exception type. 捕获适当的异常类型。
While using Scanner.nextInt()
, it causes some problems. 使用Scanner.nextInt()
,它会引起一些问题。 When you use Scanner.nextInt()
, it does not consume the new line (or other delimiter) itself so the next token returned will typically be an empty string. 当您使用Scanner.nextInt()
,它本身不会占用换行符(或其他定界符),因此返回的下一个标记通常是一个空字符串。 Thus, you need to follow it with a Scanner.nextLine()
. 因此,您需要使用Scanner.nextLine()
跟随它。 You can discard the result. 您可以放弃结果。
It's for this reason that I suggest always using nextLine
(or BufferedReader.readLine()
) and doing the parsing after using Integer.parseInt()
. 出于这个原因,我建议始终使用nextLine
(或BufferedReader.readLine()
)并在使用Integer.parseInt()
之后进行解析。 Your code should be as follows. 您的代码应如下所示。
Scanner scan = new Scanner(System.in);
boolean bidding;
int startbid;
int bid;
bidding = true;
System.out.print("Alright folks, who wants this unit?" +
"\nHow much. How much. How much money where?" );
try
{
startbid = Integer.parseInt(scan.nextLine());
while(bidding)
{
System.out.println("$" + startbid + "! Whose going to bid higher?");
startbid =+ Integer.parseInt(scan.nextLine());
}
}
catch(NumberFormatException nfe)
{
System.out.println("Please enter a bid");
}
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