使用
mysqli获取返回的行数一直很困难.我每次都得到0回,尽管肯定有一些结果.
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); $stmt->close(); }
为什么不显示正确的数字？
您需要在num_rows查找之前调用
MySqli_Stmt::store_result()：
if($stmt = $MysqLi->prepare("SELECT id,$data->id); $stmt->execute(); $stmt->store_result(); <-- This needs to be called here! $num_of_rows = $stmt->num_rows; $stmt->bind_result($child_id,$child_parent); while($stmt->fetch()){ //code } echo($num_of_rows); $stmt->close(); }
请参阅the docs on MySQLi_Stmt->num_rows,它说它在页面顶部附近(在主要描述块)…
