Android/php：從MySqli數據庫表中填充JSONArray行的AutoCompleteTextView列表

Question

我試圖在我的應用程序中實現數據庫搜索功能，該功能搜索用戶表並返回之類的用戶數組，如搜索串。我需要這個數組來填充下降到acTextView下面的列表，但是我不能讓JSONArray正確傳遞，而我不知道如何將數據傳遞給acTextView。任何幫助，將不勝感激！

應用端的搜索活動：

public class PopupAddContact extends AppCompatActivity { 

@Override 
protected void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.popup_add_contact); 

    getWindow().setLayout(ConstraintLayout.LayoutParams.WRAP_CONTENT, ConstraintLayout.LayoutParams.WRAP_CONTENT); 
    getWindow().setBackgroundDrawable(new ColorDrawable(Color.TRANSPARENT)); 

    final AutoCompleteTextView searchInput = (AutoCompleteTextView) findViewById(R.id.searchText); 

    searchInput.addTextChangedListener(new TextWatcher() { 
     @Override 
     public void beforeTextChanged(CharSequence charSequence, int i, int i1, int i2) { 

     } 

     @Override 
     public void onTextChanged(CharSequence charSequence, int i, int i1, int i2) { 
      final String name = searchInput.getText().toString(); 

      Response.Listener<String> responseListener = new Response.Listener<String>() { 
       @Override 
       public void onResponse(String response) { 
        try { 
         System.out.println(response); 
         //ArrayList<String> array = new ArrayList<String>(); 
         JSONArray jsonResponse = new JSONArray(response); 

         System.out.println(jsonResponse); 
         System.out.println("working"); 


         /* 
         if (jsonResponse != null) { 

          Integer contact_user_id = jsonResponse.getInt(); 
          String contact_name = jsonResponse.getString("name"); 
          String contact_email = jsonResponse.getString("email"); 



         } else if(response != null){ 
          Toast.makeText(PopupAddContact.this, "No users found", Toast.LENGTH_SHORT).show(); 
         } */ 

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

      PopupContactRequest AddContactRequest = new PopupContactRequest(name, responseListener); 
      RequestQueue queue = Volley.newRequestQueue(PopupAddContact.this); 
      queue.add(AddContactRequest); 

     } 

     @Override 
     public void afterTextChanged(Editable editable) { 

     } 
    }); 
} 

PHP腳本中檢索用戶的數組：

if(!empty($_POST["name"])){ // check post data and then start 
$mysqli = new mysqli("db creds"); 

$name = $_POST["name"]; 
$response=array(); 

if (mysqli_connect_errno()) { 
    printf("Connect failed: %s\n", mysqli_connect_error()); 
    exit(); 
} 

$query = "SELECT user_id, name, email FROM users WHERE name LIKE %?%"; 

if ($stmt = $mysqli->prepare($query)) { // prepare query 

    $stmt->bind_param("s", $name); // bind parameters 

    $stmt->execute(); 

    $stmt->bind_result($data); 

    while ($stmt->fetch()) { 
     $response[] = $data; //assign each data to response array 
    } 

} 

echo json_encode($response); 

$mysqli->close(); 

}else{ 
    echo "no users found"; 
} 

Answer 1

我想你的PHP文件是不是呼應一個有效的JSON，嘗試改變這行：

if ($stmt = $mysqli->prepare($query)) { // prepare query 

    $stmt->bind_param("s", $name); // bind parameters 

    $stmt->execute(); 

    $stmt->bind_result($user_id, $name, $email); 
    while ($stmt->fetch()) { 
     $response[] = [ 
      "user_id" => $user_id, 
      "name" => $name, 
      "email" => $email, 
     ]; //assign each data to response array 
    } 
} 

echo json_encode(array_values($response)); 

首先，你應該打電話給PHP網站，並驗證結果在json驗證器像https://jsonlint.com/然後你可以繼續使用java代碼。

這也是一個很好的做法呼應的JSON響應之前寫的JSON頭：

header('Content-Type: application/json'); 
echo json_encode($response);