如何用java编写一个非常简单的登录系统

Question

我需要创建一个系统来检查文件中的用户名和密码，如果正确，它会说明是否在标签中。 到目前为止，我已经能够简单地使一个用户名和密码等于变量，但需要以某种方式将其链接到一个文件。 我是一个菜鸟程序员，所以非常感谢你的帮助。 这是我的身份验证按钮下的内容。

String pass;
String user;
user = txtUser.getText();
pass = txtPass.getText();

if(pass.equals("blue") && user.equals("bob") ){
    lblDisplay.setText("Credentials Accepted.");
}
else{
    lblDisplay.setText("Please try again.");
}     

Answer 1

您将需要使用java.util.Scanner来解决此问题。

这是一个很好的控制台登录程序：

import java.util.Scanner; // I use scanner because it's command line.

public class Login {
public void run() {
    Scanner scan = new Scanner (new File("the\\dir\\myFile.extension"));
    Scanner keyboard = new Scanner (System.in);
    String user = scan.nextLine();
    String pass = scan.nextLine(); // looks at selected file in scan

    String inpUser = keyboard.nextLine();
    String inpPass = keyboard.nextLine(); // gets input from user

    if (inpUser.equals(user) && inpPass.equals(pass)) {
        System.out.print("your login message");
    } else {
        System.out.print("your error message");
    }
}
} 

当然，你会使用Scanner scanner = new Scanner (File toScan); 但不适用于用户输入。

快乐编码！

最后要注意的是，如果您可以制作 Swing 组件，那么您至少是一个体面的程序员。

Answer 2

这是我在这个网站上的第一个代码试试这个

import java.util.Scanner;
public class BATM {

public static void main(String[] args) {
    Scanner input = new Scanner(System.in);

    String username;
    String password;

    System.out.println("Log in:");
    System.out.println("username: ");
    username = input.next();

    System.out.println("password: ");
    password = input.next();

    //users check = new users(username, password);

    if(username.equals(username) && password.equals(password)) 
        System.out.println("You are logged in");



}

}

Answer 3

代码

import java.util.Scanner;

public class LoginMain {

public static void main(String[] args) {

    String Username;
    String Password;

    Password = "123";
    Username = "wisdom";

    Scanner input1 = new Scanner(System.in);
    System.out.println("Enter Username : ");
    String username = input1.next();

    Scanner input2 = new Scanner(System.in);
    System.out.println("Enter Password : ");
    String password = input2.next();

    if (username.equals(Username) && password.equals(Password)) {

        System.out.println("Access Granted! Welcome!");
    }

    else if (username.equals(Username)) {
        System.out.println("Invalid Password!");
    } else if (password.equals(Password)) {
        System.out.println("Invalid Username!");
    } else {
        System.out.println("Invalid Username & Password!");
    }

}

}

Answer 4

检查此代码：

导入 java.util.Scanner;

公共课主要{

public static void main(String[] args) throws IllegalAccessException {

    String username ;
    String password;
    String yes_0r_no;
    String scann;
    String passscan;

    Scanner scan = new Scanner(System.in);
    Scanner scanner = new Scanner(System.in);

    Scanner name = new Scanner(System.in);
    System.out.println("Username:");
    username = name.next().toLowerCase();

    Scanner pass = new Scanner(System.in);
    System.out.println("Password:");
    password = pass.next().toLowerCase();


    System.out.println("You are logged in");

    Scanner ask = new Scanner(System.in);
    System.out.println("Do you want to check this or not(yes or no) :");

     yes_0r_no = ask.next().toLowerCase();



    while (true){
        if (yes_0r_no.equals("yes")){
            System.out.println("Username:");
          scann = scan.next().toLowerCase();
          if (scann == username) {
              continue;
          }
        System.out.println("Password");
        passscan = scanner.next().toLowerCase();
        if (passscan.equals(password)) {
            System.out.println("You are logged in");
            break;

        }if (!password.equals(passscan)) {
            throw new IllegalAccessException();
            }
        }

        if (yes_0r_no.equals("no"))
            break ;



    }


}

}

Answer 5

import java.util.Scanner;

public class LoginMain {

public static void main(String[] args) {

    String Username;
    String Password;

    Password = "123";
    Username = "wisdom";

    Scanner input1 = new Scanner(System.in);
    System.out.println("Enter Username : ");
    String username = input1.next();

    Scanner input2 = new Scanner(System.in);
    System.out.println("Enter Password : ");
    String password = input2.next();

    if (username.equals(Username) && password.equals(Password)) {

        System.out.println("Access Granted! Welcome!");
    }

    else if (username.equals(Username)) {
        System.out.println("Invalid Password!");
    } else if (password.equals(Password)) {
        System.out.println("Invalid Username!");
    } else {
        System.out.println("Invalid Username & Password!");
    }

Answer 6

Map<String, String> d = new HashMap<>();

void input(String u, String p, String e) {
    read();
    if (e.equals("login")) login(u, p);
    else if (e.equals("register")) register(u, p);
    write();
}

void read() {
    d = new HashMap<>();
    String s = "";
    try {
        s = new String(Files.readAllBytes(Paths.get("data.txt")));
    }catch(IOException e) {
        e.printStackTrace();
    }
    String [] pairs = s.split("\n");
    for (int i = 0; i < pairs.length; i++) {
        d.put(pairs[i].split(",")[0], pairs[i].split(",")[1]);
    }
}

void write() {
    try (FileWriter m = new FileWriter("data.txt")) {
        for (Map.Entry<String, String> entry : d.entrySet()) {
            m.write(entry.getKey() + "," + entry.getValue() + "\n");    
        }
        m.close();
    }catch (IOException e) {
        e.printStackTrace();
    }
}

boolean login(String u, String p) {
    return (d.get(u).equals(p)) ? true : false;
}

boolean register(String u, String p) {
    if (d.containsKey(u)) return false;
    d.put(u, p);
    return true;
}

Answer 7

提示：1. String.Contains 2.文件类

Answer 8

您可以这样做的一种方法是拥有一个带有用户名的文件并直接在其下传递。 然后使用 Scanner 类，并在创建它时将该文件作为 Scanner 的参数。 然后使用方法 hasNext(); 和 nextLine 验证用户名和密码；

String user;
String pass;

Scanner scan = new Scanner(new File("File.txt"));

while(scan.hasNext){ //While the file still has more lines remaining
     if(scan.nextLine() == user){
          if(scan.nextLine == pass){
                   lblDisplay.setText("Credentials Accepted.");
          }
          else{
               lblDisplay.setText("Please try again.");
          }
     }
}

Answer 9

import java.<span class="q39pbqr9" id="q39pbqr9_9">net</span>.*;
import java.io.*;

<span class="q39pbqr9" id="q39pbqr9_1">public class</span> A
{
    static String user = "user";
    static String pass = "pass";
    static String param_user = "username";
    static String param_pass = "password";
    static String content = "";
    static String action = "action_url";
    static String urlName = "url_name";
    public static void main(String[] args)
    {
        try
        {
            user = URLEncoder.encode(user, "UTF-8");
            pass = URLEncoder.encode(pass, "UTF-8");
            content = "action=" + action +"&amp;" + param_user +"=" + user + "&amp;" + param_pass + "=" + pass;
            URL url = new URL(urlName);
            HttpURLConnection urlConnection = (HttpURLConnection)(url.openConnection());
            urlConnection.setDoInput(true);
            urlConnection.setDoOutput(true);
            urlConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
            urlConnection.setRequestMethod("POST");
            DataOutputStream dataOutputStream = new DataOutputStream(urlConnection.getOutputStream());
            dataOutputStream.writeBytes(content);
            dataOutputStream.flush();
            dataOutputStream.close();

            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(urlConnection.getInputStream()));
            String responeLine;
            StringBuilder response = new StringBuilder();
            while ((responeLine = bufferedReader.readLine()) != null)
            {
                response.append(responeLine);
            }
            System.out.println(response);
        }catch(Exception ex){ex.printStackTrace();}
    }