为什么mysqli num_rows总是返回0？

Question

我一直在获取要使用mysqli返回的行数方面遇到麻烦。即使确实有一些结果，我每次都会得到0。if($stmt = $mysqli->prepare("SELECT id, title, visible, parent_id FROM content WHERE parent_id = ? ORDER BY page_order ASC;")){      $stmt->bind_param('s', $data->id);      $stmt->execute();    $num_of_rows = $stmt->num_rows;      $stmt->bind_result($child_id, $child_title, $child_visible, $child_parent);      while($stmt->fetch()){        //code    }    echo($num_of_rows);    $stmt->close();}为什么没有显示正确的数字？

Answer 1

您需要先调用MySqli_Stmt::store_result()num_rows查找：

if($stmt = $mysqli->prepare("SELECT id, title, visible, parent_id FROM content WHERE parent_id = ? ORDER BY page_order ASC;")){  

    $stmt->bind_param('s', $data->id);  

    $stmt->execute();

    $stmt->store_result(); <-- This needs to be called here!

    $num_of_rows = $stmt->num_rows;  

    $stmt->bind_result($child_id, $child_title, $child_visible, $child_parent);  

    while($stmt->fetch()){

        //code

    }

    echo($num_of_rows);

    $stmt->close();

}

请参阅上的文档MySQLi_Stmt->num_rows，该文档显示在页面顶部附近（在主要说明区域中）...

Answer 2

尝试$num_of_rows在if（statement）之后-在bind_param之前设置您的权限...除非那会改变您的结果。没有更多信息很难说。

Answer 3

您的查询似乎出了点问题。尝试进行直接查询，例如if($stmt = $mysqli->query("SELECT id, title, visible, parent_id FROM content WHERE parent_id = YOURIDHERE ORDER BY page_order ASC;")){，看看是否仍然可以解决0个结果