{closure}（）缺少參數2;

我正在使用Slim Framework處理我的REST風格的Web服務，並且當我嘗試從3個表中獲取數據時出現此錯誤。{closure}（）缺少參數2;

這是在我的課的功能：

public function getCard($card_id, $card_type_id, $user_id) { 
    $stmt = $this->conn->prepare("SELECT uc.id, c.card_name, ct.category, c.card_desc, c.card_picture ,c.status, c.created_at from cards c, user_cards uc, cards_type ct WHERE c.card_type_id = ct.card_type_id AND c.card_id = uc.card_id AND uc.user_id = ?"); 

    if ($stmt == FALSE) { 
     die($this->conn->error); 
    } else { 
     $stmt->bind_param("iii", $card_id, $card_type_id, $user_id); 
     if ($stmt->execute()) { 
      $res = array(); 
      $stmt->bind_result($id, $card_name, $category, $card_desc, $card_picture, $status, $created_at); 

      $stmt->fetch(); 
      $res["id"]   = $id; 
      $res["card_name"] = $card_name; 
      $res["category"]  = $category; 
      $res["card_desc"] = $card_desc; 
      $res["card_picture"] = $card_picture; 
      $res["status"]  = $status; 
      $res["created_at"] = $created_at; 
      $stmt->close(); 
      return $res; 
     } else { 
      return NULL; 
     } 
    } 
}

而且這是在index.php文件的代碼：

$app->get('/cards/users/:card_id', 'authenticate', function($card_id, $card_type_id) { 
    global $user_id; 
    $response = array(); 
    $db  = new Card(); 

    // fetch card 
    $result = $db->getCard($card_id, $card_type_id, $user_id); 

    if ($result != NULL) { 
     $response["error"]  = false; 
     $response["id"]   = $result["id"]; 
     $response["card_name"] = $result["card_name"]; 
     $response["category"]  = $result["category"]; 
     $response["card_desc"] = $result["card_desc"]; 
     $response["card_picture"] = $result["card_picture"]; 
     $response["status"]  = $result["status"]; 
     $response["createdAt"] = $result["created_at"]; 
     echoRespnse(200, $response); 
    } else { 
     $response["error"] = true; 
     $response["message"] = "The requested resource doesn't exists"; 
     echoRespnse(404, $response); 
    } 
});

難道我寫的東西錯了嗎？

來源

2014-11-01 Rendy Setya Pratama

退房這一行：

$stmt = $this->conn->prepare("SELECT uc.id, c.card_name, ct.category, c.card_desc, c.card_picture ,c.status, c.created_at from cards c, user_cards uc, cards_type ct WHERE c.card_type_id = ct.card_type_id AND c.card_id = uc.card_id AND uc.user_id = ?");

查詢可能很難看到，所以我只是格式化查詢：

SELECT uc.id, 
     c.card_name, 
     ct.category, 
     c.card_desc, 
     c.card_picture, 
     c.status, 
     c.created_at 
FROM cards c, 
     user_cards uc, 
     cards_type ct 
WHERE c.card_type_id = ct.card_type_id 
     AND c.card_id = uc.card_id 
     AND uc.user_id = ?

注意，它只有一個問號，表明您應該只將一個參數綁定到您的準備好的語句。

所以，你應該改變這一行：

$stmt->bind_param("i", $user_id); //have only one argument

來源

2014-11-01 05:44:14

我已經嘗試過的方式，但我得到了錯誤太多，因爲我寫上我的課這樣的代碼：公共職能getCard（$ card_id的，$ card_type_id，$ USER_ID）{ – 2014-11-01 05:51:53

或者我的查詢出錯了？ – 2014-11-01 05:53:05

查看查詢，只有一個問號，所以你只能綁定一個參數。看看這是否有效。 – 2014-11-01 05:58:39

{closure}（）缺少參數2;

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