java.lang.String類型的值無法轉換爲JSONObject

-1

我在這裏掙扎了一點，我在我的應用程序中有一個研究按鈕，它將調用一個PHP文件來執行SELECT（並獲取幾行代碼）。這裏是我的PHP文件：java.lang.String類型的值無法轉換爲JSONObject

$con = mysqli_connect("***", "***", "***", "***"); 
$city = $_POST["city"]; 
$statement = mysqli_prepare($con, "SELECT name FROM Restaurants WHERE city = ? "); 
mysqli_stmt_bind_param($statement, "s", $city); 
mysqli_stmt_execute($statement); 

mysqli_stmt_store_result($statement); 
mysqli_stmt_bind_result($statement, $city); 

$response = array(); 
$response["success"] = false; 
$i=-1 

while(mysqli_stmt_fetch($statement)){ 
    $response["success"] = true; 
    while($row = mysqli_fetch_array($statement)){ 
     $i++; 

     $response["name"][$i]=$row[$i]; 
     $response["name"][$i]=$row[$i]; 

    }  
} 
echo json_encode($response);

我從.java文件中收到錯誤。這裏是按鈕聽衆在我的.java文件：

public void bSearchRestaurantClicked(View v) { 

    final EditText etCity = (EditText) findViewById(R.id.etSearchRestaurantsCity); 
    final String city = etCity.getText().toString(); 

    Response.Listener<String> responseListener = new Response.Listener<String>() { 
     public void onResponse(String response) { 
      try { 
       JSONObject jsonResponse = new JSONObject(response); 
       boolean success = jsonResponse.getBoolean("success"); 

       if (success) { 

        String nameCities = jsonResponse.getString("name"); 

        System.out.println("=======> "+nameCities); 

        Intent intent = new Intent(AreaActivityClient.this, ResultSearchActivity.class); 

        intent.putExtra("city", city); 

        intent.putExtra("cities", nameCities); 

        AreaActivityClient.this.startActivity(intent); 

       } else { 
        AlertDialog.Builder builder = new AlertDialog.Builder(AreaActivityClient.this); 
        builder.setMessage("Login Failed !").setNegativeButton("Retry", null).create().show(); 
       } 
      } catch (JSONException e) { 
       e.printStackTrace(); 
      } 
     } 
    }; 
    SearchRequest searchRequest = new SearchRequest(city, responseListener); 
    RequestQueue queue = Volley.newRequestQueue(AreaActivityClient.this); 
    queue.add(searchRequest);

的錯誤是在這一行：

   JSONObject jsonResponse = new JSONObject(response);

這裏是我的SearchRequest.java文件：

private static final String SEARCH_REQUEST_URL="file.php"; 
private Map<String,String> params; 
public SearchRequest(String city, Response.Listener<String> listener){ 
    super(Request.Method.POST,SEARCH_REQUEST_URL,listener,null); 
    params = new HashMap<>(); 
    params.put("city",city); 
}

我很確定錯誤來自.php文件，但我找不到它... 預先感謝您的答案。

來源

2016-11-07 Lie Hili

也許u能

log.v(response);

然後檢查響應是從簡單的JSON字符串

調試複雜

來源

2016-11-07 06:41:02

java.lang.String類型的值無法轉換爲JSONObject

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