Question

在我的应用程序中，用户必须输入控制台名称，用户名和密码。稍后程序应检查登录到动态数组的参与，如果没有，则添加它，如果是，则显示错误警告。当你跑步时，一个输入框，静静地输入上面的名字＆＃34;，＆＃34;用户名＆＃34;和＃34;密码＆＃34;。但是，程序不会对数组中的参与进行登录检查并显示错误消息，并且它不会自带数组字段。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;

public class User  {
    public static void main(String[] args) throws IOException {
        BufferedReader bReader =new BufferedReader(new InputStreamReader(System.in));
        String name = bReader.readLine();
        String login = bReader.readLine();
        String password = bReader.readLine();
        UserADD(name,login,password);

    }
    public static ArrayList<String> UserADD(String name,String login,String password) {
        ArrayList <String> users = new ArrayList<String>();

        for (int i = 0; i<users.size(); i++) {
            if (users.contains(login)) {
                System.out.println("Это имя пользователя уже занято");
            }
            else users.add(name);

            users.add(login);
            users.add(password);

            System.out.println(users);
        }

        return users;
    }
}

Answer 1

您的SELECT st.title, count(q.id) AS question_count, sum( CASE WHEN a.answer IS NOT NULL THEN 1 ELSE 0 END ) AS answer_count, g.user_id AS game_user_id, a.maxid, a.game_id AS a_game_id, modifiedAnswer.modified AS finished, FROM games g LEFT JOIN (SELECT MAX(answers.id) AS maxid, game_id FROM answers GROUP BY answers.game_id) AS a ON (a.game_id = g.id) LEFT JOIN answers AS modifiedAnswer ON modifiedAnswer.id = a.maxid LEFT JOIN questions q ON (a.question_id = q.id) LEFT JOIN sessions s ON (s.id = q.session_id) LEFT JOIN sessiontypes st ON (st.id = s.sessiontype_id) WHERE g.user_id = 21 GROUP BY g.id HAVING ( question_count = answer_count ) ORDER BY finished DESC;是本地方法。这意味着每次调用方法时，都会创建另一个对象，并且它最初是空的。

可能的解决方案是在ArrayList<String> users方法中创建ArrayList并将其作为参数传递给main。

在您学习userAdd()时，请尝试遵循最佳做法。不要使用大写字母启动函数名称。它们是为类名保留的。

Answer 2

ArrayList users在UserADD方法下是本地的。将ArrayList users声明为全局将解决问题。

public class User  {

//Global declaration of ArrayList users
public static ArrayList <String> users = new ArrayList<String>();

public static void main(String[] args) throws IOException {
    BufferedReader bReader =new BufferedReader(new InputStreamReader(System.in));
    String name = bReader.readLine();
    String login = bReader.readLine();
    String password = bReader.readLine();
    UserADD(name,login,password);

}
public static ArrayList<String> UserADD(String name,String login,String password) {
    //don't initialize users ArrayList here
    for (int i = 0; i<users.size(); i++) {
        if (users.contains(login)) {
            System.out.println("Это имя пользователя уже занято");
        }
        else users.add(name);   users.add (login);users.add(password);

        System.out.println(users);
}


    return users;
}
}

Answer 3

您必须让类对象中的用户比函数调用更长寿。有点像：

public class UserRegistration  {

    private List<String> users = new ArrayList<>();

public static void main(String[] args) throws IOException {
    new UserRegistration().run();
}

private void run() {
    BufferedReader bReader =new BufferedReader(new InputStreamReader(System.in));
    String name = bReader.readLine();
    String login = bReader.readLine();
    String password = bReader.readLine();
    addUser(name,login,password);
}

public boolean addUser(String name,String login,String password) {
    for (int i = 0; i<users.size(); i++) {
        if (users.contains("\t" + login + "\t")) {
            System.out.println("Это имя пользователя уже занято");
            return false;
        }
    }
    users.add(name + "\t" + login + "\t" +  password);
    return true;
}
}

Answer 4

最佳解决方案是

public class User {

    Set<UserLogin> users = new HashSet<UserLogin>();

    public static void main(String[] arg) throws IOException {
        BufferedReader bReader =new BufferedReader(new InputStreamReader(System.in));
        String name = bReader.readLine();
        String login = bReader.readLine();
        String password = bReader.readLine();
        new User().UserADD(new User(). new UserLogin(name, login, password));                
    }

    private void UserADD(final UserLogin user) {

        if (!users.contains(user)) {
            users.add(user);
        } else {
            System.out.println("Это имя пользователя уже занято");
        }
    }

    public class UserLogin {

        private String name;
        private String login;
        private String pass;

        public UserLogin(String name, String login, String pass) {
            this.name = name;
            this.login = login;
            this.pass = pass;
        }

        public String getName() {
            return name;
        }

        public String getLogin() {
            return login;
        }

        public String getPass() {
            return pass;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;

            UserLogin user = (UserLogin) o;

            return !(login != null ? !login.equals(user.login) : user.login != null);

        }

        @Override
        public int hashCode() {
            return login != null ? login.hashCode() : 0;
        }
    }
}

Answer 5

非常感谢你，但问题仍然存在。当您尝试第二个用户首先输入用户名时，程序会平静地接受它，忽略任何限制并且屏幕上的警告不会出现。我尝试了所有的选择。

ArrayList似乎不包含任何东西

5 个答案: