我有一个非常简单的查询,不确定我在这里做错了什么。
我的数据库调用没有收到我期望的插入id
。
表格:
存储过程:CREATE DEFINER=`root`@`localhost` PROCEDURE `addCustomerProduct`(IN in_customerID INT, in_productID INT)
BEGIN
INSERT INTO order_customer_product (customerID, productID, retailAmountAtPurchase, faceValue)
SELECT
in_customerID,
in_productID,
p.retail,
p.faceValue
FROM
products as p
WHERE
p.productID = in_productID;
END
PHP:
public function addProduct($data, $userID)
{
// Do we already have a pending order for this user?
$orderID = $this->doesOrderExist($userID);
// We had no order, lets create one
if (!$orderID) {
$orderID = $this->createOrder($userID);
}
/**
* Insert the customer product.
* This relates a denomination to a customer.
*/
$customerProductID = $this->addCustomerProduct($data);
// Add this customer product to the order
$this->addProductToOrder(array("customerProductID" => $customerProductID, "orderID" => $orderID));
// Return
return $customerProductID;
}
/**
* Description: Add a customer product / reward
* Page: client/add_reward
*/
public function addCustomerProduct($data){
$procedure = "CALL addCustomerProduct(?,?)";
$result = $this->db->query($procedure, $data);
return $this->db->insert_id();
}
这个有问题的行是:
$customerProductID = $this->addCustomerProduct($data);
。正在向表中插入新记录,该表具有PK/AI。数据插入正常,但
$customerProductID
返回0
。也许从选择语句插入不会返回
insert ID
吗?
@Ravi 更新-
更新2:
我创建了一个单独的方法,并硬编码查询和发送的数据。
它成功添加记录,AI增加,0
作为last id
返回。
public function test(){
$procedure = "CALL addCustomerProduct(?,?)";
$result = $this->db->query($procedure, array("customerID" => 1, "productID" => 20));
echo $this->db->insert_id();
}
我还重新启动了MySQL服务器,以确保没有任何奇怪的情况发生。
另外,更新了存储过程,只需将随机数据插入表中,而不使用select。
CREATE DEFINER=`root`@`localhost` PROCEDURE `addCustomerProduct`(IN in_customerID INT, in_productID INT)
BEGIN
INSERT INTO order_customer_product (customerID, productID, retailAmountAtPurchase, faceValue)
VALUES(8,2,'4.55',25);
END
更新3:
在插入后,我会打印出最后一次运行的查询以及结果。您会注意到有1行受影响(插入正在发生),但 insert_id
仍为0。
CALL addCustomerProduct('8','33')
CI_DB_mysqli_result Object
(
[conn_id] => mysqli Object
(
[affected_rows] => 1
[client_info] => mysqlnd 5.0.12-dev - 20150407 - $Id: b396954eeb2d1d9ed7902b8bae237b287f21ad9e $
[client_version] => 50012
[connect_errno] => 0
[connect_error] =>
[errno] => 0
[error] =>
[error_list] => Array
(
)
[field_count] => 0
[host_info] => Localhost via UNIX socket
[info] =>
[insert_id] => 0
[server_info] => 5.6.35
[server_version] => 50635
[stat] => Uptime: 1637 Threads: 3 Questions: 508 Slow queries: 0 Opens: 113 Flush tables: 1 Open tables: 106 Queries per second avg: 0.310
[sqlstate] => 00000
[protocol_version] => 10
[thread_id] => 25
[warning_count] => 0
)
[result_id] => 1
[result_array] => Array
(
)
[result_object] => Array
(
)
[custom_result_object] => Array
(
)
[current_row] => 0
[num_rows] =>
[row_data] =>
)
更新4:
根据我所做的一些研究,除非您使用mysqli方法,例如$this->db->insert()
,否则它不会向您返回最后插入的id。
我将尝试弄清楚Ravi的建议,但似乎Code Igniter不允许显示的示例。至少我现在知道,除非您使用“insert”方法而不是存储过程,否则这不是正常行为。
mysqli
方法就像上面那样。值得一试! - Professor Abronsiusinsert_id()
,但可能会包含一些 Codeigniter 插入并干扰的语句。 - Rick JamesaddCustomerProduct()
函数中执行存储过程中的查询呢? - Paul Spiegeltest()
函数中尝试以下代码:$this->db->conn_id->query("CALL addCustomerProduct(1,20)"); echo $this->db->conn_id->insert_id;
它仍然是0
吗? - Paul Spiegel