使用 PHP 通过 Android 插入和接收来自 MySQL 的数据

Posted 2023-04-19

【中文标题】使用 PHP 通过 Android 插入和接收来自 MySQL 的数据

【英文标题】：insert and receive the data from MySQL through Android using PHP

【发布时间】：2013-08-27 08:46:18

【问题描述】：

我在运行此应用程序时遇到错误。 “解析数据时出错 org.json.JSONException：值

这是我的 register_user.java

package com.iwantnew.www;
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.os.AsyncTask;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;

public class register_user extends Activity
    // Progress Dialog
        private ProgressDialog pDialog;
    JSONParser jsonParser = new JSONParser();
    EditText signup_username;
    EditText signup_email;
    EditText signup_password;
    Button registerBtn;
    TextView registerErrorMsg;

    // url to create new product
        private static String url_create_users = "http://10.0.2.2/http://10.0.2.2/android_iwant/signup_success.php";
        // JSON Node names
        private static final String TAG_SUCCESS = "success";

    public void onCreate(Bundle savedInstanceState) 
        super.onCreate(savedInstanceState);
        setContentView(R.layout.signup_form);
        // Edit Text
                signup_username = (EditText) findViewById(R.id.signup_username);
                signup_email = (EditText) findViewById(R.id.signup_email);
                signup_password = (EditText) findViewById(R.id.signup_password);                
                registerBtn = (Button) findViewById(R.id.regiserBtn);

                // button click event
                registerBtn.setOnClickListener(new View.OnClickListener() 

                    @Override
                    public void onClick(View view) 
                        // creating new product in background thread
                        new CreateNewUser().execute();
                    
                );

    
    class CreateNewUser extends AsyncTask<String, String, String> 

        /**
         * Before starting background thread Show Progress Dialog
         * */
        @Override
        protected void onPreExecute() 
            super.onPreExecute();
            pDialog = new ProgressDialog(register_user.this);
            pDialog.setMessage("Creating Users..");
            pDialog.setIndeterminate(false);
            pDialog.setCancelable(true);
            pDialog.show();
        

        /**
         * Creating product
         * */
        protected String doInBackground(String... args) 
            /*String username = signup_username.getText().toString();
            String email =  signup_email.getText().toString();
            String password = signup_password.getText().toString();*/

            //Building Parameters
            List<NameValuePair> params = new ArrayList<NameValuePair>();
            params.add(new BasicNameValuePair("username", signup_username.getText().toString()));
            params.add(new BasicNameValuePair("email",signup_email.getText().toString()));
            params.add(new BasicNameValuePair("password", signup_password.getText().toString()));

            // getting JSON Object
            // Note that create product url accepts POST method
            JSONObject json = jsonParser.makeHttpRequest(url_create_users,
                    "POST", params);

            // check log cat fro response
            Log.d("Create Response", json.toString());

            // check for success tag
            try 
                int success = json.getInt(TAG_SUCCESS);

                if (success == 1) 
                    // successfully created users
                    Intent i = new Intent(getApplicationContext(), MainActivity.class);
                    startActivity(i);

                    // closing this screen
                    finish();
                 else 
                    // failed to create users

                
             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 done
            pDialog.dismiss();
        

    

这是 JSONParser.java。这是android的数据解析类

package com.iwantnew.www;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.util.List;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONException;
import org.json.JSONObject;

import android.util.Log;

public class JSONParser 

    static InputStream is = null;
    static JSONObject jObj = null;
    static String json = "";

    // constructor
    public JSONParser() 

    

    // function get json from url
    // by making HTTP POST or GET method
    public JSONObject makeHttpRequest(String url, String method,
            List<NameValuePair> params) 

        // Making HTTP request
        try 

            // check for request method
            if(method == "POST")
                // request method is POST
                // defaultHttpClient
                DefaultHttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(url);
                httpPost.setEntity(new UrlEncodedFormEntity(params));

                HttpResponse httpResponse = httpClient.execute(httpPost);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();

            else if(method == "GET")
                // request method is GET
                DefaultHttpClient httpClient = new DefaultHttpClient();
                String paramString = URLEncodedUtils.format(params, "utf-8");
                url += "?" + paramString;
                HttpGet httpGet = new HttpGet(url);

                HttpResponse httpResponse = httpClient.execute(httpGet);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();
                      

         catch (UnsupportedEncodingException e) 
            e.printStackTrace();
         catch (ClientProtocolException e) 
            e.printStackTrace();
         catch (IOException e) 
            e.printStackTrace();
        

        try 
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) 
                sb.append(line + "\n");
            
            is.close();
            json = sb.toString();
         catch (Exception e) 
            Log.e("Buffer Error", "Error converting result " + e.toString());
        

        // try parse the string to a JSON object
        try 
            jObj = new JSONObject(json);
         catch (JSONException e) 
            Log.e("JSON Parser", "Error parsing data " + e.toString());
        

        // return JSON String
        return jObj;

    

这是 PHP 代码（signup_success.php）

<?php
// array for JSON response
$response = array();
if(isset($_POST['username']) && isset($_POST['email']) && isset($_POST['password']))
    $username = $_POST['username'];
    $email = $_POST['email'];
    $password = $_POST['password'];

    // include db connect class
    require_once __DIR__ . '/db_connect.php';


    // connecting to db
    $db = new DB_CONNECT();

     $uuid = uniqid('', true);
        $hash = $this->hashSSHA($password);
        $encrypted_password = $hash["encrypted"]; // encrypted password
        $salt = $hash["salt"]; // salt
        $result = mysql_query("INSERT INTO users(unique_id, name, email, encrypted_password, salt, created_at) VALUES('$uuid', '$username', '$email', '$encrypted_password', '$salt', NOW())");
        // check for successful store
        if ($result) 
            // get user details 
            $uid = mysql_insert_id(); // last inserted id
            $result = mysql_query("SELECT * FROM users WHERE uid = $uid");
            // return user details
            return mysql_fetch_array($result);
         else 
            return false;
        

        $result =  mysql_query("INSERT INTO users(name,email,password) VALUES ('$username','$email,$password')");
     // check if row inserted or not
    if ($result) 
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = "successfully signed Up.";

        // echoing JSON response
        echo json_encode($response);
        else 
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";

        // echoing JSON response
        echo json_encode($response);
    
 else 
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);


?>

【问题讨论】：

调试并发布此sb.toString();

请发布您从BufferReader 获得的信息作为回复。

【参考方案1】：

解析数据时出错 org.json.JSONException

这个错误是因为解析json响应失败，因为来自服务器端的响应不是预期的。

在你的 PHP 脚本中我发现了一个问题

// check for successful store
        if ($result) 
            // get user details 
            $uid = mysql_insert_id(); // last inserted id
            $result = mysql_query("SELECT * FROM users WHERE uid = $uid");
            // return user details
            return mysql_fetch_array($result);
         else 
            return false;
        

在此块中，if 和 else 都有 return 语句，因此无法访问之后编写的代码。 我不明白波纹管代码的作用，但是如果插入数据，则必须使用 json_encode 以 json 的形式进行响应。 因此，您必须从 if 块中删除 return

// check for successful store
        if ($result) 
            // get user details 
            $uid = mysql_insert_id(); // last inserted id
            $result = mysql_query("SELECT * FROM users WHERE uid = $uid");
            // return user details
           echo json_encode(mysql_fetch_array($result));
           exit;               
        

一些使用 api 的建议 1.首先由邮递员从网上检查您的结果，并确认输出符合您的要求。 2.在android中使用debugging找出错误。

调试代码非常有用。

【参考方案2】：

是的，问题转换为 json 格式。让我举个例子来说明如何创建 json。

$dt=array('key' => 'value'); 响应/输出：

键：值

$dt=array(array('key' => 'value'),array('key2' => 'value2')) 响应/输出：

[key:value,key2:values2]