mysqli返回的行数不正确

Question

我正在为我的网站和我的生活写一个'喜欢'的功能，我无法使用此功能！

假设这是一个简化版本，我的数据库结构：

ID --- POST_ID --- USER_ID

1 --- 1 ----------- 1

2 --- 2 ----------- 1

这是我写的代码：

public function userHasLikedPost($post_id,$user_id){ 
     $this->post_id_clean = sanitize($post_id); 
     $this->user_id_clean = sanitize($user_id); 

     global $mysqli,$db_table_prefix; 
     $stmt = $mysqli->prepare("SELECT like_type FROM ".$db_table_prefix."post_likes WHERE user_id = ? AND post_id = ?"); 

     $stmt->bind_param("ii", $this->post_id_clean, $this->user_id_clean); 

     $stmt->execute(); 
     $stmt->store_result(); 

     $stmt->bind_result($like_type); 

     $stmt->fetch(); 
     $result; 

     if($stmt->num_rows >0){ 
      $result = array("liked" => 1, "like_type" => $like_type); 
     }else{ 
      $result = array("liked" => 0); 
     } 

     return print_r($stmt); 

     $stmt->close(); 
    } 

当我打电话userHasLikedPost（1,1）我的代码将返回正确行数

mysqli_stmt对象（[affected_rows] => 1 [INSERT_ID] => 0 [NUM_ROWS] = > [param_count] => 2 [field_count] => 1 [errno] => 0 [error] => [error_list] => Array（）[sqlstate] => 00000 [id] => 5）1

但是，如果我呼叫userHasLikedPost（2,1），我会得到不正确的行数。

mysqli_stmt对象（[affected_rows] => 0 [INSERT_ID] => 0 [NUM_ROWS] => [param_count] => 2 [FIELD_COUNT] => 1 [错误] => 0 [错误] => [error_list] => Array（）[sqlstate] => 00000 [id] => 5）1

这是怎么回事？我正确使用mysqli吗？数据存在于数据库中，如果我在PHPMyAdmin中手动运行查询，我将得到正确的输出。我花了2个小时试图让这一个功能工作......预先感谢任何回复:)

Answer 1

我解决了我的问题。我的绑定是错误的顺序：/多么尴尬哈哈！

解决方案：

替换这行代码：

$stmt->bind_param("ii", $this->post_id_clean, $this->user_id_clean); 

有了：

$stmt->bind_param("ii", $this->user_id_clean, $this->post_id_clean);