Question

我试图在android中学习远程数据库连接。我在互联网上找到了一个使用PHP和JSON的教程。它工作正常。但PHP脚本在同一网页中回显编码的JSON数据。只有这样才能将数据传递给android应用程序。当我仅将PHP脚本用于移动应用程序时，它可能没问题。

但是，如果我也希望将我的想法作为一个网站提出来呢？我不希望json数据在网页上可见。我遇到的所有PHP脚本似乎都以相同的方式编码JSON数据。

是否可以将JSON数据传递给Android应用程序而不在网页上回显？或者我应该为网站和移动应用程序创建单独的PHP脚本吗？我已经在下面添加了我迄今为止所使用的代码。

的login.php：

<?php

//load and connect to MySQL database stuff
require("config.inc.php");

if (!empty($_POST)) {
//gets user's info based off of a username.
$query = " 
        SELECT 
            id, 
            username, 
            password
        FROM users 
        WHERE 
            username = :username 
    ";

$query_params = array(
    ':username' => $_POST['username']
);

try {
    $stmt   = $db->prepare($query);
    $result = $stmt->execute($query_params);
}
catch (PDOException $ex) {
    // For testing, you could use a die and message. 
    //die("Failed to run query: " . $ex->getMessage());

    //or just use this use this one to product JSON data:
    $response["success"] = 0;
    $response["message"] = "Database Error1. Please Try Again!";
    die(json_encode($response));

}

//This will be the variable to determine whether or not the user's information is correct.
//we initialize it as false.
$login_ok = false;

//fetching all the rows from the query
$row = $stmt->fetch();
if ($row) {
    //if we encrypted the password, we would unencrypt it here, but in our case we just
    //compare the two passwords
    if ($_POST['password'] === $row['password']) {
        $login_ok = true;
    }
}

// If the user logged in successfully, then we send them to the private members-only page 
// Otherwise, we display a login failed message and show the login form again 
if ($login_ok) {
    $response["success"] = 1;
    $response["message"] = "Login successful!";
    die(json_encode($response));

} else {
    $response["success"] = 0;
    $response["message"] = "Invalid Credentials!";
    die(json_encode($response));
}
} else {
?>
    <h1>Login</h1> 
    <form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" method="post"> 
        Username:<br /> 
        <input type="text" name="username" placeholder="username" /> 
        <br /><br /> 
        Password:<br /> 
        <input type="password" name="password" placeholder="password" value="" /> 
        <br /><br /> 
        <input type="submit" value="Login" /> 
    </form> 
    <a href="register.php">Register</a>
<?php
}

?>

Android java文件 - Login.java

package com.example.mysqltest;

import java.util.ArrayList;
import java.util.List;

import org.apache.http.NameValuePair;
import org.apache.http.message.BasicNameValuePair;
import org.json.JSONException;
import org.json.JSONObject;

import android.app.Activity;
import android.app.ProgressDialog;
import android.content.Intent;
import android.content.SharedPreferences;
import android.content.SharedPreferences.Editor;
import android.os.AsyncTask;
import android.os.Bundle;
import android.preference.PreferenceManager;
import android.util.Log;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;

public class Login extends Activity implements OnClickListener{
private EditText user, pass;
private Button mSubmit, mRegister;

 // Progress Dialog
private ProgressDialog pDialog;

// JSON parser class
JSONParser jsonParser = new JSONParser();

//php login script location:

//localhost :
//testing on your device
//put your local ip instead,  on windows, run CMD > ipconfig
//or in mac's terminal type ifconfig and look for the ip under en0 or en1




private static final String LOGIN_URL =     "http://192.168.1.2:8080/webservice/login.php";

//JSON element ids from repsonse of php script:
private static final String TAG_SUCCESS = "success";
private static final String TAG_MESSAGE = "message";

@Override
protected void onCreate(Bundle savedInstanceState) {
    // TODO Auto-generated method stub
    super.onCreate(savedInstanceState);
    setContentView(R.layout.login);

    //setup input fields
    user = (EditText)findViewById(R.id.username);
    pass = (EditText)findViewById(R.id.password);

    //setup buttons
    mSubmit = (Button)findViewById(R.id.login);
    mRegister = (Button)findViewById(R.id.register);

    //register listeners
    mSubmit.setOnClickListener(this);
    mRegister.setOnClickListener(this);

}

@Override
public void onClick(View v) {
    // TODO Auto-generated method stub
    switch (v.getId()) {
    case R.id.login:
            new AttemptLogin().execute();
        break;
    case R.id.register:
            Intent i = new Intent(this, Register.class);
            startActivity(i);
        break;

    default:
        break;
    }
}

class AttemptLogin extends AsyncTask<String, String, String> {

     /**
     * Before starting background thread Show Progress Dialog
     * */
    boolean failure = false;

    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        pDialog = new ProgressDialog(Login.this);
        pDialog.setMessage("Attempting login...");
        pDialog.setIndeterminate(false);
        pDialog.setCancelable(true);
        pDialog.show();
    }

    @Override
    protected String doInBackground(String... args) {
        // TODO Auto-generated method stub
         // Check for success tag
        int success;
        String username = user.getText().toString();
        String password = pass.getText().toString();
        try {
            // Building Parameters
            List<NameValuePair> params = new ArrayList<NameValuePair>();
            params.add(new BasicNameValuePair("username", username));
            params.add(new BasicNameValuePair("password", password));

            Log.d("request!", "starting");
            // getting product details by making HTTP request
            JSONObject json = jsonParser.makeHttpRequest(
                   LOGIN_URL, "POST", params);

            // check your log for json response
            Log.d("Login attempt", json.toString());

            // json success tag
            success = json.getInt(TAG_SUCCESS);
            if (success == 1) {
                Log.d("Login Successful!", json.toString());

                SharedPreferences sp=PreferenceManager.getDefaultSharedPreferences(Login.this);
                Editor edit=sp.edit();
                edit.putString("username", username);
                edit.commit();

                Intent i = new Intent(Login.this, ReadComments.class);
                finish();
                startActivity(i);
                return json.getString(TAG_MESSAGE);
            }else{
                Log.d("Login Failure!", json.getString(TAG_MESSAGE));
                return json.getString(TAG_MESSAGE);

            }
        } catch (JSONException e) {
            e.printStackTrace();
        }

        return null;

    }
    /**
     * After completing background task Dismiss the progress dialog
     * **/
    protected void onPostExecute(String file_url) {
        // dismiss the dialog once product deleted
        pDialog.dismiss();
        if (file_url != null){
            Toast.makeText(Login.this, file_url, Toast.LENGTH_LONG).show();
        }

    }

}

}

Answer 1

是否可以将json数据传递给Android应用程序而不在网页上回显？

不，因为这是Web服务器与客户端通信的方式。但是，您可以为网站和移动应用的API使用相同的代码。只需将您的逻辑与演示文稿分开。在一个地方进行处理，在另一个地方进行表示。也许看一下可以帮助你做到这一点的框架，比如Laravel。您可以在模型中进行处理，并为您的api和您的网站提供单独的控制器。

Answer 2

我写的方式是，

有一个php脚本，它接受用户名/ pass并返回JSON输出，成功或失败。
我将从Android应用程序调用该PHP脚本。
对于网页，我将编写一个带有用户名/密码表单字段的简单html页面，并使用AJAX调用php脚本。

你的php脚本试图成为一个网页以及一个服务，所以我认为需要解耦。

将您的php视为Android和Android的服务提供商HTML网页。

使用PHP和JSON的Android远程数据库连接

2 个答案: