如何从Android Studio的XAMPP数据库获取用户ID

我正在创建一个具有登录和注册页面的Android应用程序。有人登录后，他们可以进行应用程序提供的不同测试。我如何将他们的结果与他们的用户ID一起存储在XAMPP数据库中？我提供了login.php，LoginActivity.java，User.java和EditProfile.java代码。我提供了一个例子，其中有人可能想要编辑他们的个人资料，当他们想要保存更改时，应该用他们的用户ID保存。我正在努力。任何帮助将不胜感激，谢谢如何从Android Studio的XAMPP数据库获取用户ID

的login.php

<?php 
    $error = NULL; 

    include_once('connection.php'); 
     if(isset($_POST['txtUsername']) && isset($_POST['txtPassword'])){ 
     $username = $_POST['txtUsername']; 
     $password = $_POST['txtPassword']; 

     $query = "SELECT * FROM user WHERE username = '$username' AND password = '$password'"; 

     $result = mysqli_query($conn, $query); 

     if($username == $error || $password == $error) { 
      echo "Login Failed <br>"; 
     } 
     else if($result->num_rows > 0){ 
      if(isset($_POST['mobile']) && $_POST['mobile'] == "android"){ 
       echo "success"; 
       exit; 
      } 
      echo "Login Successful"; 
     } 
     else{ 
      echo "Login Failed <br>"; 
     } 
    } 
?> 


<html> 
<head> 
    <title>Login</title> 
</head> 
<body> 
<h1>Login </h1> 
<form action="<?PHP $_PHP_SELF ?>" method="post"> 
    Username <input type="text" name="txtUsername" value="" /> <br/> 
    Password <input type="password" name="txtPassword" value=""/><br/> 
    <input type="submit" name="btnSubmit" value="Login"/> </form> 
</body> 
</html>

LoginActivity.java

package com.delta.object.newandroidproject; 

import android.content.Intent; 
import android.support.v7.app.AppCompatActivity; 
import android.os.Bundle; 
import android.support.v7.widget.Toolbar; 
import android.view.View; 
import android.view.WindowManager; 
import android.widget.Button; 
import android.widget.EditText; 
import android.widget.Toast; 

import com.kosalgeek.asynctask.AsyncResponse; 
import com.kosalgeek.asynctask.PostResponseAsyncTask; 

import java.util.HashMap; 

public class LoginActivity extends AppCompatActivity implements  AsyncResponse, View.OnClickListener { 

    EditText etUsername, etPassword; 
    Button loginBtn; 

    @Override 
    protected void onCreate(Bundle savedInstanceState) { 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.activity_login); 
     Toolbar mToolbar = (Toolbar) findViewById(R.id.login_toolbar); 
     setSupportActionBar(mToolbar); 
     this.getWindow().setSoftInputMode(WindowManager.LayoutParams.SOFT_INPUT_STATE_ALWAYS_HIDDEN); 

     etUsername = (EditText) findViewById(R.id.etUsername); 
     etPassword = (EditText) findViewById(R.id.etPassword); 
     loginBtn = (Button) findViewById(R.id.email_login_button); 
     loginBtn.setOnClickListener(this); 
    } 

    @Override 
    public void onClick(View v) { 
     HashMap postData = new HashMap(); 
     postData.put("mobile", "android"); 
     postData.put("txtUsername", etUsername.getText().toString()); 
     postData.put("txtPassword", etPassword.getText().toString()); 

     PostResponseAsyncTask task = new PostResponseAsyncTask(this, postData); 
     task.execute("http://10.0.3.2/androidproject/login.php"); 

    } 

    @Override 
    public void processFinish(String result) { 
     if (result.equals("success")) { 
      Intent i = new Intent(this, TestActivity.class); 
      startActivity(i); 
     } 
     else{ 
      Toast.makeText(this, "Login Failed", Toast.LENGTH_LONG).show(); 
     } 

    } 

    public void registerClick(View view) { 
     Intent i = new Intent(LoginActivity.this, SignUpActivity.class); 
     startActivity(i); 
    } 
}

user.php的

package com.delta.object.newandroidproject; 

import com.google.gson.annotations.SerializedName; 

public class User { 

    @SerializedName("user_id") 
    public int user_id; 

    @SerializedName("name") 
    public String name; 

    @SerializedName("gender") 
    public String gender; 

    @SerializedName("username") 
    public String username; 

    @SerializedName("password") 
    public String password; 
}

EditProfile.java

package com.delta.object.newandroidproject; 

import android.content.Intent; 

import java.util.ArrayList; 
import java.util.HashMap; 


public class EditProfile extends AppCompatActivity { 

    EditText etName, etPassword, etUsername; 
    Button submitBtn; 
    RadioGroup rg; 
    private User user; 

    @Override 
    protected void onCreate(Bundle savedInstanceState) { 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.edit_profile); 
     Toolbar mToolbar = (Toolbar) findViewById(R.id.edit_profile_toolbar); 
     setSupportActionBar(mToolbar); 
    } 

    public void editProfileSave(View view) { 
     Intent i = new Intent(EditProfile.this, TestActivity.class); 
     startActivity(i); 
    } 

    public void cancelClick(View view) { 
     Intent i = new Intent(EditProfile.this, TestActivity.class); 
     startActivity(i); 
    } 
}

来源

2017-03-07 GleneaMan

[Little Bobby]（http://bobby-tables.com/）说*** [你的脚本存在SQL注入攻击风险。]（http://stackoverflow.com/questions/60174/如何可以防止SQL注入在PHP中）***了解[准备]（http：//en.wikipedia .org/wiki/Prepared_statement）[MySQLi]的声明（http://php.net/manual/en/mysqli.quickstart.prepared-statements.php）。即使[转义字符串]（http://stackoverflow.com/questions/5741187/sql-injection-that-gets-around-mysql-real-escape-string）是不安全的！ [不要相信它？]（http://stackoverflow.com/q/38297105/1011527） –

**不要存储纯文本密码！**请使用PHP的[内置函数]（http：// jayblanchard。 net/proper_password_hashing_with_PHP.html）来处理密码安全性。如果您使用的PHP版本低于5.5，则可以使用'password_hash（）'[兼容包]（https://github.com/ircmaxell/password_compat）。确保你*** [不要越狱密码]（http://stackoverflow.com/q/36628418/1011527）***或在哈希之前使用其他任何清理机制。这样做*更改密码并导致不必要的附加编码。 –

我想和大家分享一些我在这里的经历我如何解决我的问题，在我的项目

首先，为了让我能够从数据库中获取的用户ID，我会从我的登录解析用户ID值.php使用JSON。

这里是login.php中的样子，读码

login.php 
<?php 
require_once 'include/DB_Functions.php'; 
$db = new DB_Functions(); 

// json response array 
$response = array("error" => FALSE); 

if (isset($_POST['email']) && isset($_POST['password'])) { 

    // receiving the post params 
    $email = $_POST['email']; 
    $password = $_POST['password']; 

    // get the user by email and password 
    $user = $db->getUserByEmailAndPassword($email, $password); 

    if ($user != false) { 
     // user is found 
     //so reading all the database record 

     $response["error"] = FALSE; 
     $response["uid"] = $user["unique_id"]; //here is the user ID,will fetch along 
     $response["user"]["name"] = $user["name"]; 
     $response["user"]["email"] = $user["email"]; 
     $response["user"]["created_at"] = $user["created_at"]; 
     $response["user"]["updated_at"] = $user["updated_at"]; 
     echo json_encode($response); 
    } else { 
     // user is not found with the credentials 
     $response["error"] = TRUE; 
     $response["error_msg"] = "Login credentials are wrong. Please try again!"; 

     echo json_encode($response); //encode this to JSON,so later on can be use in Android code 
    } 
} else { 
    // required post params is missing 
    $response["error"] = TRUE; 
    $response["error_msg"] = "Required parameters email or password is missing!"; 
    echo json_encode($response); 
} 
?>

的评语所以，现在你可以在你的processFinsh（）

@Override 
     public void processFinish(JSONArray jarray) { 
      // TODO Auto-generated method stub 
      try{ 
      JSONObject jObj = new JSONObject(response); 
       boolean error = jObj.getBoolean("error"); 

       // Check for error node in json 
       if (!error) { 
        // user successfully logged in 


        // Get your user ID here and other details 
        String uid = jObj.getString("uid"); 

        JSONObject user = jObj.getJSONObject("user"); 
        String name = user.getString("name"); 
        String email = user.getString("email"); 
        String created_at = user 
          .getString("created_at"); 

        //Here you can set your userId to the global value,so you can use it in your EditProfile.java 
        User userId = (User)getApplicationContext(); 
        userId.setUserId(uid); 

        // Launch main activity 
        Intent intent = new Intent(LoginActivity.this, 
          MainActivity.class); 
        startActivity(intent); 
        finish(); 

       } 
       catch (JSONException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 

     }

让你的用户名在你User.java创建一个getter和setter来存储你的用户ID

 private String userId; 

     public String getUserId(){ 
      return userId; 
     } 

     public void setUserId(String user_id){ 
      this.userId = user_id; 
     }

我不知道你想用你的用户ID，BU只要你想吨至使用它，你可以这样

//Get your userId 
    User userId = (User)getApplicationContext(); 
    String userId= userId.getUserId(); 
    Log.d("userId",userId) // log to the android monitor to see whether is the value you want or not

呼吁为了让你有一些基本的概念，JSON编码工作，你可以看看在下面的链接

How to connect Android with PHP, MySQL for Basic CRUD

Android Login and Registration with PHP, MySQL and SQLite

来源

2017-03-07 16:37:11 ken

如何从Android Studio的XAMPP数据库获取用户ID

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