mysqli获取问题

Question

我有一个mysqli数据获取问题。我将尝试通过示例来解释我的问题：

我想从表中获取条目（由不同的人员）。现在我想在另一个表格中查找每个获取的人的姓名，并查看他是否有任何照片。 我的代码如下，但其没有工作，我收到以下错误：

mysqli::prepare() [mysqli.prepare]: All data must be fetched before a new statement prepare takes place 
    Call to a member function fetch() on a non-object in ... 

我的代码：

if ($stmt = $this->mysqli->prepare("SELECT entry, author, time FROM messages WHERE user = ?")) { 
      $stmt->bind_param("s", $user_name); 
      $stmt->execute(); 
      $stmt->bind_result($entry, $author, $time); 

     while ($stmt->fetch()) {    
       if ($stmt = $this->mysqli->prepare("SELECT photo_id FROM photos WHERE user = ?")) { 
        $stmt->bind_param("s", $author); 
        $stmt->execute(); 
        $stmt->bind_result($photo_id); 
       } 
      //echo $photo_id; 
     }  
    $stmt->close(); 
    } 

我会任何帮助非常感谢。

Answer 1

将第二条语句分配给新变量，以便它不会覆盖第一个变量并导致“所有数据必须被提取..”错误。

if ($stmt = $this->mysqli->prepare("SELECT entry, author, time FROM messages WHERE user = ?")) { 
     $stmt->bind_param("s", $user_name); 
     $stmt->execute(); 
     $stmt->bind_result($entry, $author, $time); 

     while ($stmt->fetch()) {    
      if ($st = $this->mysqli->prepare("SELECT photo_id FROM photos WHERE user = ?")) { 
       $st->bind_param("s", $author); 
       $st->execute(); 
       $st->bind_result($photo_id); 
      } 
      //echo $photo_id; 
      $st->close(); 
     }  
    $stmt->close(); 
}