Java 忽略/清除用户输入中的额外整数

我被困在程序的这一部分，用户输入多个整数，只选择第一个整数。在我当前的代码中，输出如下

There are 10 sticks on the board.
Jerry: How many sticks do you take (1-3)? *1 3 4 2*
There are 9 sticks on the board.
Sherry: How many sticks do you take (1-3)? There are 6 sticks on the board.
Jerry: How many sticks do you take (1-3)? Please enter a number between 1 and 3.
Jerry: How many sticks do you take (1-3)? There are 4 sticks on the board.
Sherry: How many sticks do you take (1-3)? 

Jerry: How many sticks do you take (1-3)? *1 3 4 2*
There are 8 sticks on the board.
Pat: How many sticks do you take (1-3)? *3*
There are 5 sticks on the board.

而在给定的示例中，输出是这样的

There are 10 sticks on the board.
Jerry: How many sticks do you take (1-3)? *1 3 4 2*
There are 9 sticks on the board.
Sherry: How many sticks do you take (1-3)? There are 6 sticks on the board.
Jerry: How many sticks do you take (1-3)? Please enter a number between 1 and 3.
Jerry: How many sticks do you take (1-3)? There are 4 sticks on the board.
Sherry: How many sticks do you take (1-3)? 

Jerry: How many sticks do you take (1-3)? *1 3 4 2*
There are 8 sticks on the board.
Pat: How many sticks do you take (1-3)? *3*
There are 5 sticks on the board.

似乎额外的整数已排队等待下一个扫描仪输入命令。在给定的示例中，多余的整数被清除。这是如何做到的？是否有清除所有当前用户输入的命令，或者是否使用.hasNextInt和任何超过第一个输入的整数都被简单地丢弃？此外，hasNextInt在输入中的位置如何？在得到true或false后，如果没有添加新输入，下一个hasNextInt命令是否会在相同的输入位置查找

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用**括起来的文本是用户输入。

如果不能确定用户是否正确输入，可以通过接受字符串而不是整数来完成。像这样：

Scanner s = new Scanner(System.in);
System.out.print("How many?: ");
int many = Integer.parseInt(s.nextLine().trim().split(" ")[0]);
System.out.println(many);

这甚至可以处理输入，如2 5 6 5以及您的示例，方法是获取整个输入字符串，修剪任何前导和尾随的s，将字符串拆分为围绕s的字符串数组，然后返回第一个，即您希望在索引0处返回的字符串

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正如所写的那样，它没有做的是处理foobar42之类的输入，或者处理拆分字符串的第一个索引无法解析为整数的任何其他输入。尝试这样做将导致NumberFormatException。

请在问题中包含您的代码。添加您的代码。我猜您使用whilesc.hasNextInt{}而不是if。