Scanner类throwFor(Unknown Source)及跳过下一个扫描器分析

在使用Scanner类时遇到一个问题：

Exception in thread "main" java.util.NoSuchElementException
at java.util.Scanner.throwFor(Unknown Source)
at java.util.Scanner.next(Unknown Source)

在执行scanner.next()时遇到异常。Stack Overflow上给出了问题原因与解决办法。

原因：当一个类中两个及两个以上的Scanner实例时，其中一个Scanner类的实例执行scanner.close()方法会关闭其他潜在的InputStream流，导致其他Scanner的扫描器无法读取输入流。

解决办法：对于控制台程序，在程序运行中只注册一个Scanner类的实例从System.in中读取信息。

问题二：使用Scanner#nextInt()时跳过下一个扫描器。

产生原因：在使用Scanner#nextInt()时，nextInt()在遇到 '\n'之前结束，但“\n"会被下一个扫描器所接收，如Scanner#nextLine()，从而直接跳过Scanner#nextLine()。

解决办法：统一使用Scanner#nextLine()代替所有扫描函数。然后进行强制类型转换。

String nextIntString = keyboard.nextLine(); //get the number as a single line
int nextInt = Integer.parseInt(nextIntString); //convert the string to an int

补充：在使用Scanner#hasNextInt(),hasNextDouble()...函数时，如果返回值为false则应该在else语句中增加Scanner#nextLine()以抵消 '\n'。

public void showMenu() {
System.out.println("****************************");
System.out.println("Option Menu");
System.out.println("1、登录");
System.out.println("2、注册");
System.out.println("3、退出");
System.out.println("请选择:");
System.out.println("****************************");
if (scanner.hasNextInt()) {
int index = Integer.parseInt(scanner.nextLine());
choice(index);
} else {
scanner.nextLine();
System.out.println("请输入数字...");
showMenu();
}
}

问题一解释：

You close the second

Scanner

which closes the underlying

InputStream

, therefore the first

Scanner

can no longer read from the same

InputStream

and a

NoSuchElementException

results.

The solution: For console apps, use a single

Scanner

to read from

System.in

.

Aside: As stated already, be aware that

Scanner#nextInt

does not consume newline characters. Ensure that these are consumed before attempting to call

nextLine

again by using

Scanner#newLine()

.

问题二解释：

The

nextInt()

method leaves the

\n

(end line) symbol and is picked up immediately by

nextLine()

, skipping over the next input. What you want to do is use

nextLine()

for everything, and parse it later:

String nextIntString = keyboard.nextLine(); //get the number as a single line
int nextInt = Integer.parseInt(nextIntString); //convert the string to an int

This is by far the easiest way to avoid problems--don't mix your "next" methods. Use only

nextLine()

and then parse

int

s or separate words afterwards.

Also, make sure you use only one

Scanner

if your are only using one terminal for input. That could be another reason for the exception.

Last note: compare a

String

with the

.equals()

function, not the

==

operator.

if (playAgain == "yes"); // Causes problems
if (playAgain.equals("yes")); // Works every time