Получение определенных строк данных

Question

Я пытаюсь получить несколько строк данных. Однако, я получаю сообщение об ошибке.

Я хочу получить строки, имя которых (под столбцом имя) = имя пользователя. Какую часть моего кода мне нужно изменить?

protected String doInBackground(String... args) { 
     // Building Parameters 
     try { 
     List<NameValuePair> params = new ArrayList<NameValuePair>(); 
     params.add(new BasicNameValuePair("name", name)); 
     // getting JSON string from URL 
     JSONObject json = jParser.makeHttpRequest(url_all_coupons, "GET", params); 

     // Check your log cat for JSON reponse 
     Log.d("All Products: ", json.toString()); 


      // Checking for SUCCESS TAG 
      int success = json.getInt(TAG_SUCCESS); 

      if (success == 1) { 
       // products found 
       // Getting Array of Products 
       products = json.getJSONArray(TAG_PRODUCTS); 

       // looping through All Products 
       for (int i = 0; i < products.length(); i++) { 
        JSONObject c = products.getJSONObject(i); 

        // Storing each json item in variable 
        String couponexpires = c.getString(TAG_COUPONEXPIRES); 
        String coupondetails = c.getString(TAG_COUPONDETAILS); 

        // creating new HashMap 
        HashMap<String, String> map = new HashMap<String, String>(); 

        // adding each child node to HashMap key => value 
        map.put(TAG_COUPONEXPIRES, couponexpires); 
        map.put(TAG_COUPONEXPIRES, coupondetails); 

        // adding HashList to ArrayList 
        couponsList.add(map); 

Мой PHP код

<?php 



// array for JSON response 
$response = array(); 


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

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

if (isset($_GET["name"])) { 
    $name= $_GET['name']; 


$result = mysql_query("SELECT * FROM coupons WHERE name = '$name'") or die(mysql_error()); 

// check for empty result 
if (mysql_num_rows($result) > 0) { 
    $response["products"] = array(); 

    while ($row = mysql_fetch_array($result)) { 
     // temp user array 
     $product = array(); 
     $product["couponcreated"] = $row["couponcreated"]; 
     $product["couponexpires"] = $row["couponexpires"]; 
     $product["coupondetails"] = $row["coupondetails"]; 

     array_push($response["products"], $product); 
    } 
    // success 
    $response["success"] = 1; 

    // echoing JSON response 
    echo json_encode($response); 
} else { 
    // no products found 
    $response["success"] = 0; 
    $response["message"] = "No products found"; 

    // echo no users JSON 
    echo json_encode($response); 
} 
?> 

Мой LogCat

5202-5251/info.androidhive.loginandregistration E/JSON Parser﹕ Error parsing data org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject 
12-21 11:24:50.750 5202-5251/info.androidhive.loginandregistration W/dalvikvm﹕ threadid=11: thread exiting with uncaught exception (group=0x41b9f700) 
12-21 11:24:50.755 5202-5251/info.androidhive.loginandregistration E/AndroidRuntime﹕ FATAL EXCEPTION: AsyncTask #1 
    java.lang.RuntimeException: An error occured while executing doInBackground() 
      at android.os.AsyncTask$3.done(AsyncTask.java:299) 
      at java.util.concurrent.FutureTask.finishCompletion(FutureTask.java:352) 
      at java.util.concurrent.FutureTask.setException(FutureTask.java:219) 
      at java.util.concurrent.FutureTask.run(FutureTask.java:239) 
      at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230) 
      at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1080) 
      at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:573) 
      at java.lang.Thread.run(Thread.java:841) 
    Caused by: java.lang.NullPointerException 
      at info.androidhive.loginandregistration.CouponPageActivity$LoadAllProducts.doInBackground(CouponPageActivity.java:103) 
      at info.androidhive.loginandregistration.CouponPageActivity$LoadAllProducts.doInBackground(CouponPageActivity.java:76) 
      at android.os.AsyncTask$2.call(AsyncTask.java:287) 

Answer 1

Существует ошибка синтаксиса в скрипте (если условие не закрыто). используйте ниже сценарий:

<?php 
// array for JSON response 
$response = array(); 
$response["success"] = 0; 
$response["message"] = "No products found"; 

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

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

if (isset($_GET["name"])) { 
    $name = $_GET['name']; 


    $result = mysql_query("SELECT * FROM coupons WHERE name = '$name'") or die(mysql_error()); 

// check for empty result 
    if (mysql_num_rows($result) > 0) { 
     $response["products"] = array(); 

     while ($row = mysql_fetch_array($result)) { 
      // temp user array 
      $product = array(); 
      $product["couponcreated"] = $row["couponcreated"]; 
      $product["couponexpires"] = $row["couponexpires"]; 
      $product["coupondetails"] = $row["coupondetails"]; 

      array_push($response["products"], $product); 
     } 
     // success 
     $response["success"] = 1; 
     $response["message"] = ''; 
    } else { 
     // no products found 


     // echo no users JSON 

    } 
} 
echo json_encode($response); 
?>