Java &引用；没有这样的元素“；第二次运行方法时出错

抱歉，如果这是一个愚蠢的问题，我是一个编程高手，下面的代码有点问题

public static void main(String[] args) throws IOException {
    addName();
    addName();      //Line "10" in error
}

public static void addName() throws IOException {
    //Initializing Variables and Scanner
    Scanner keyboard = new Scanner(System.in);
    FileWriter fwriter = new FileWriter("Data.txt", true);
    PrintWriter outputFile = new PrintWriter(fwriter);
    int numFriends;
    String firstName;
    String lastName;
    String email;

    //Get number of entries to add
    System.out.print("How many people do you wish to add? ");
    numFriends = Integer.parseInt(keyboard.nextLine());        //Line "25" in error

    //Prompt for info and add it to the file
    for(int i = 1; i <= numFriends; i++){
        System.out.printf("Enter the first name of person %d: ",i);
        firstName = keyboard.nextLine();
        System.out.printf("Enter the last name of person %d: ",i);
        lastName = keyboard.nextLine();
        System.out.printf("Enter the email of person %d: ",i);
        email = keyboard.nextLine();

        outputFile.println(firstName + " " + lastName + " " + email);
    }

    //Wrap things up and give a confirmation
    outputFile.close();
    keyboard.close();
    System.out.println("Data written to the file.");;
}

---

我原本以为这是出自一个没有在某处消费过的新词角色，但我似乎无法理解。为什么我会犯这个错误？我怎样才能修好它？谢谢大家!

您不应该在方法结束时关闭

扫描仪

，理想情况下，您应该在整个执行过程中重复使用相同的

扫描仪

通过关闭

扫描仪

，您也将关闭其输入源，即本例中的

系统。因此，用户不能再输入任何内容

以下是更正后的代码段：

public static void main(String[] args) throws IOException {
    Scanner keyboard = new Scanner(System.in);
    addName(keyboard);
    addName(keyboard);
}

public static void addName(Scanner keyboard) throws IOException {
    FileWriter fwriter = new FileWriter("Data.txt", true);
    PrintWriter outputFile = new PrintWriter(fwriter);
    int numFriends;
    String firstName;
    String lastName;
    String email;

    //Get number of entries to add
    System.out.print("How many people do you wish to add? ");
    numFriends = Integer.parseInt(keyboard.nextLine());        //Line "25" in error

    //Prompt for info and add it to the file
    for(int i = 1; i <= numFriends; i++){
        System.out.printf("Enter the first name of person %d: ",i);
        firstName = keyboard.nextLine();
        System.out.printf("Enter the last name of person %d: ",i);
        lastName = keyboard.nextLine();
        System.out.printf("Enter the email of person %d: ",i);
        email = keyboard.nextLine();

        outputFile.println(firstName + " " + lastName + " " + email);
    }

    //Wrap things up and give a confirmation
    outputFile.close();
    System.out.println("Data written to the file.");
}

publicstaticvoidmain（字符串[]args）引发IOException{
扫描仪键盘=新扫描仪（System.in）；
地址名（键盘）；
地址名（键盘）；
}
公共静态void addName（扫描仪键盘）引发IOException{
FileWriter fwriter=新的FileWriter（“Data.txt”，true）；
PrintWriter outputFile=新的PrintWriter（fwriter）；
国际女朋友；
字符串名；
字符串lastName；
字符串电子邮件；
//获取要添加的条目数
System.out.print（“您希望添加多少人？”）；
numFriends=Integer.parseInt（keyboard.nextLine（））；//行“25”出错
//提示输入信息并将其添加到文件中
例如（int i=1；我哇，我现在觉得自己很笨。非常感谢！关闭阅读器也会关闭输入源并不一定是直观的，但随着您经验的增加，您会意识到这通常是Java中的默认行为（我实际上不知道标准库中的任何反例）。如果这个答案有帮助，请随意投票/接受。原则是，你负责关闭你打开的任何东西。因此，你打开了文件
Data.txt
；你负责关闭它。你没有打开
系统。在
，你只在它周围包装了一个
扫描仪，但它已经由Java为你打开了（或者，实际上是操作系统）所以你不应该关闭它。
public static void main(String[] args) throws IOException {
    Scanner keyboard = new Scanner(System.in);
    addName(keyboard);
    addName(keyboard);
}

public static void addName(Scanner keyboard) throws IOException {
    FileWriter fwriter = new FileWriter("Data.txt", true);
    PrintWriter outputFile = new PrintWriter(fwriter);
    int numFriends;
    String firstName;
    String lastName;
    String email;

    //Get number of entries to add
    System.out.print("How many people do you wish to add? ");
    numFriends = Integer.parseInt(keyboard.nextLine());        //Line "25" in error

    //Prompt for info and add it to the file
    for(int i = 1; i <= numFriends; i++){
        System.out.printf("Enter the first name of person %d: ",i);
        firstName = keyboard.nextLine();
        System.out.printf("Enter the last name of person %d: ",i);
        lastName = keyboard.nextLine();
        System.out.printf("Enter the email of person %d: ",i);
        email = keyboard.nextLine();

        outputFile.println(firstName + " " + lastName + " " + email);
    }

    //Wrap things up and give a confirmation
    outputFile.close();
    System.out.println("Data written to the file.");
}