📜  mysqli 为更新记录做准备 - 任何代码示例

📅  最后修改于: 2022-03-11 14:57:12.690000             🧑  作者: Mango

代码示例1
$q = "UPDATE table1 SET title=?, description=?, date=? WHERE id=?";
$stmt = mysqli_prepare($dbc, $q);
mysqli_stmt_bind_param($stmt, 'sssi', $title, $descrip, $date1, $id);
mysqli_stmt_execute($stmt);