Question

我正在尝试从我的mysql数据库中获取一些信息，即用户unique_id。我一直在线跟踪教程并更改部件只是创建返回json对象的简单函数。但是，当我尝试启动该方法时，我遇到了错误：

No database selected
value no of java.lang.String cannot be converted to JSONObject

第二个问题是因为我猜想从数据库连接或检索数据不成功，但我无法找到问题所在。这是我的代码的config.php

<?php
define("DB_HOST", "localhost");
define("DB_USER", "root");
define("DB_PASSWORD", "");
define("DB_DATABASE", "test");
?>

DB_Connect.php

<?php
class DB_Connect {
    function __construct() { }
    function __destruct() { }
    // Connecting to database
    public function connect() {
        require_once 'config.php';
        // connecting to mysql
        $con = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
        // selecting database
        mysql_select_db(DB_DATABASE);
        // return database handler
        return $con;
    // Closing database connection
    public function close() {
        mysql_close();
    }
}
?>

DB_Functions.php

<?php
class DB_Functions {
    private $db;
    //put your code here
    // constructor
    function __construct() {
        require_once 'DB_Connect.php';
        // connecting to database
        $this->db = new DB_Connect();
        $this->db->connect();
         if(!$this->db){
          echo "Please try later.";
         }
    }
    // destructor
    function __destruct() {
    }
    // Check if user exists.
    public function isUserExisted($unique_id) {
        $result = mysql_query("SELECT unique_id from users WHERE unique_id = '$unique_id'");
        if($result === FALSE) {
         die(mysql_error());
         }
        $no_of_rows = mysql_num_rows($result);
        if ($no_of_rows > 0) {
            // user existed
            return true;
        } else {
            // user not existed
            return false;
        }
    }
}
?>

的index.php

<?php
if (isset($_POST['tag']) && $_POST['tag'] != '') {
    // get tag
    $tag = $_POST['tag'];

    // include db handler
    require_once 'include/DB_Functions.php';
    $db = new DB_Functions();

 $response = array("tag" => $tag, "success" => 0, "error" => 0);
 if ($tag == 'check') {
 // Request unique_id
        $unique_id = $_POST['unique_id'];

     // check if user is already existed
        if ($db->isUserExisted($unique_id)) {
            // user is already existed - error response
            $response["error"] = 2;
            $response["error_msg"] = "User already existed";
            echo json_encode($response);
        } else {
                $response["error"] = 1;
                $response["error_msg"] = "USER UNKNOWN!";
                echo json_encode($response);
}

        }
}
?>

在我的android JSONParses里面看起来像这样：

public class JSONParser {
    private static InputStream inputStream;
    private static JSONObject jSONObject;
    private static String json;

    JSONParser(){
        inputStream = null;
        jSONObject = null;
        json = " ";
    }


    public JSONObject getJSONFromUrl(String url, List<NameValuePair> parameters) throws ClientProtocolException{

        try {
            DefaultHttpClient defaultHttpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);
            httpPost.setEntity(new UrlEncodedFormEntity(parameters));

            HttpResponse httpResponse = defaultHttpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            inputStream = httpEntity.getContent();

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (IOException e) {
           e.printStackTrace();
       }



        try {
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream, "iso-8859-1"), 8);
            StringBuilder stringBuilder = new StringBuilder();
            String line = null;
            while((line = bufferedReader.readLine()) != null){
                stringBuilder.append(line + "n");
            }
            inputStream.close();
            json = stringBuilder.toString();
            Log.e("JSON", json);

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }


        try {
            jSONObject = new JSONObject(json);
        } catch (JSONException e) {
            Log.e("ERROR PARSING DATA: ",e.toString());
        }


        return jSONObject;
    }

}

最后一段代码，使用JSONParser的方法（它在异步调用中）

private JSONObject checkIfUserExists(String unique_id){
    List<NameValuePair> parameters = new ArrayList<NameValuePair>();
    parameters.add(new BasicNameValuePair("tag", CHECK_TAG));
    parameters.add(new BasicNameValuePair("unique_id", unique_id));
    JSONObject json=null;
    try {
        json = jsonParser.getJSONFromUrl(loginURL, parameters);
    } catch (ClientProtocolException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } 
    return json;
}

我收到的错误是我的JSONParser中没有选择数据库。我一直在查看代码，它似乎对我有好处，我知道这个问题很小，这就是为什么我找不到它。任何帮助将非常感激。提前谢谢。

@Edit。我刚检查了一下

   if($result === FALSE) {
          echo "DEAD";
         die(mysql_error());
         }

这打印死了，所以看来我的查询是错误的，或者有点没有实现，但我不知道它可能有什么问题......

Android从mysql获取信息 - php

0 个答案: