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)…