📜  通知 ajax php mysql - PHP 代码示例

📅  最后修改于: 2022-03-11 14:54:26.556000             🧑  作者: Mango

代码示例2
$sql = "update tbl_noti set status = 'read'";
       $result = $conn->query($sql);
       $row = $result->fetch_assoc();
       $count = $result->num_rows;
       echo $count;
       $conn->close();