是否可以获得前两个查询的insert_id?我可以得到最后一个,但我想要前两个。在下面的示例中,处理完表单后,我想向地址表添加一个新角色,向角色表添加新行,然后使用前两个查询中插入的查询的id向客户添加一行。这有可能吗。
if(isset($_POST["submit"])) :
$username = $_POST["username"];
$password = $_POST["password"];
$role = $_POST["role"];
$permission1 = intval($_POST["permission1"]);
$permission2 = intval($_POST["permission2"]);
$city = $_POST["city"];
$state = $_POST["state"];
mysqli_autocommit($connection,FALSE);
mysqli_query($connection,"INSERT INTO Address(city, state) VALUES('{$city}', '{$state}')");
mysqli_query($connection,"INSERT INTO Roles(roleName, permission1, permission2) VALUES('{$role}', '{$permission1}', '{$permission2}')");
mysqli_query($connection,"INSERT INTO Customers(username, password, roleId, addressId) VALUES ('{$username}', '{$password}'," . mysqli_insert_id($connection) . " , " . mysqli_insert_id($connection) . ")");
if(mysqli_error($connection)):
echo mysqli_error($connection);
endif;
mysqli_commit($connection);
endif;
最后一个查询中的mysqli_insert_id($connection)
将从Roles
插入中提取Id两次。有什么方法可以让它从前两个查询中获取两个Id值吗?
if(isset($_POST["submit"])) :
$username = $_POST["username"];
$password = $_POST["password"];
$role = $_POST["role"];
$permission1 = intval($_POST["permission1"]);
$permission2 = intval($_POST["permission2"]);
$city = $_POST["city"];
$state = $_POST["state"];
mysqli_autocommit($connection,FALSE);
mysqli_query($connection,"INSERT INTO Address(city, state) VALUES('{$city}', '{$state}')");
//this will give you first id//
$first_inserted_id = mysqli_insert_id($connection);
mysqli_query($connection,"INSERT INTO Roles(roleName, permission1, permission2) VALUES('{$role}', '{$permission1}', '{$permission2}')");
//this will give you second id//
$second_inserted_id = mysqli_insert_id($connection);
mysqli_query($connection,"INSERT INTO Customers(username, password, roleId, addressId) VALUES ('{$username}', '{$password}'," . $first_inserted_id . " , " . $second_inserted_id . ")");
//this will give you third id//
$third_inserted_id = mysqli_insert_id($connection);
if(mysqli_error($connection)):
echo mysqli_error($connection);
endif;
mysqli_commit($connection);
endif;