线程"main"java.util.NoSuchElementException中的异常

Question

每当我运行它,该chooseCave()功能与in.nextInt().当我选择洞穴时,消息会以2秒的间隔弹出,然后一旦超过该部分,它就会给我错误:

Exception in thread "main" java.util.NoSuchElementException: No line found
    at java.util.Scanner.nextLine(Unknown Source)
    at Dragon.main(Dragon.java:81)

我曾尝试hasNextLine()和hasNextInt(),当我使用while hasNextLine()的main方法,我得到一吨多的错误.当我while hasNextInt()在chooseCave()方法中使用时,它不接受我的输入.

当我if hasNextInt()在chooseCave()方法中使用时,它不接受我对playAgain字符串的输入,并直接进入另一个游戏,但随后hasNextInt()布尔返回false并且无限地发送"哪个洞穴......".

我已经完成了错误报告以及类似问题的Java-docs和Stack Overflow.请帮忙.

import java.util.Scanner;
public class Dragon {

public static void displayIntro() {
    System.out.println("You are in a land full of dragons. In front of you, ");
    System.out.println("You see two caves. In one cave, the dragon is friendly");
    System.out.println("and will share his treasure with you. The other dragon");
    System.out.println("is greedy and hungry, and will eat you on sight");
    System.out.println(' ');
}

public static int chooseCave() {
    Scanner in = new Scanner(System.in);
    int cave = 0;
    while (cave != 1 && cave != 2) {
        System.out.println("Which cave will you go into? (1 or 2)");

        cave = in.nextInt();

    }
    in.close();
    return cave;
} 

public static void checkCave(int chosenCave) {
    System.out.println("You approach the cave...");
    try
       {
       // Sleep at least n milliseconds.
       // 1 millisecond = 1/1000 of a second.
       Thread.sleep( 2000 );
       }
    catch ( InterruptedException e )
       {
       System.out.println( "awakened prematurely" );
       }
    System.out.println("It is dark and spooky...");
    try
       {
       // Sleep at least n milliseconds.
       // 1 millisecond = 1/1000 of a second.
       Thread.sleep( 2000 );
       }
    catch ( InterruptedException e )
       {
       System.out.println( "awakened prematurely" );
       }
    System.out.println("A large dragon jumps out in front of you! He opens his jaws and...");
    try
       {
       // Sleep at least n milliseconds.
       // 1 millisecond = 1/1000 of a second.
       Thread.sleep( 2000 );
       }
    catch ( InterruptedException e )
       {
       System.out.println( "awakened prematurely" );
       }

    double friendlyCave = Math.ceil(Math.random() * 2);

    if (chosenCave == friendlyCave) {
        System.out.println("Gives you his treasure!");
    }
    else {
        System.out.println("Gobbles you down in one bite!");
    }



}
public static void main(String[] args) {
    Scanner inner = new Scanner(System.in);
    String playAgain = "yes";
    boolean play = true;
    while (play) {
        displayIntro();
        int caveNumber = chooseCave();
        checkCave(caveNumber);
        System.out.println("Do you want to play again? (yes or no)");
        playAgain = inner.nextLine();
        if (playAgain == "yes") {
            play = true;
        }
        else {
            play = false;
        }
    }
    inner.close();

}

}

Answer 1

你关闭第二个Scanner关闭底层的东西InputStream,因此第一个Scanner不能再读取相同InputStream的NoSuchElementException结果.

解决方案:对于控制台应用程序,使用单个Scanner来读取System.in.

旁白:如前所述,请注意Scanner#nextInt不要使用换行符.在尝试使用之前nextLine再次呼叫之前,请确保消耗这些消耗Scanner#newLine().

请参阅:不要在单个InputStream上创建多个缓冲包装器

Answer 2

该nextInt()方法离开\n(结束行)符号并立即被拾取nextLine(),跳过下一个输入.你想要做的是nextLine()用于所有事情,并在以后解析它:

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

这是迄今为止避免问题的最简单方法 - 不要混淆你的"下一步"方法.仅使用nextLine(),然后解析int或单独的单词.

此外,Scanner如果您只使用一个终端进行输入,请确保只使用一个.这可能是例外的另一个原因.

最后注意:将a String与.equals()函数进行比较,而不是==运算符.

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

Answer 3

只是不要靠近

in.close()从您的代码中删除。