个人中心
gpt4 book ai didi

php - mysqli 信号到 php 错误

转载 作者：行者123 更新时间：2023-11-30 22:15:53 25 4
gpt4 key购买 nike

<分区>

我有 Mysql 存储过程，在某些情况下会给出这样的信号错误:

    CREATE DEFINER=`root`@`localhost` PROCEDURE `InsertAction`(IN `EmployerID` INT, IN `StoreFromID` INT, IN `StoreToID` INT, IN `StoreID` INT, IN `ProductID` INT, IN `Quantity` DECIMAL(10,2), IN `DualOperation` TINYINT, IN `inOrOut` TINYINT)
    LANGUAGE SQL
    NOT DETERMINISTIC
    CONTAINS SQL
    SQL SECURITY DEFINER
    COMMENT ''
BEGIN
    START TRANSACTION;
        SELECT @lastActionID;
        SELECT @lastTransferID;
        SELECT @retval;
        SELECT SUM(ad.Quantity) INTO @retVal FROM productin pri JOIN actiondetails ad ON ad.ID=pri.ID;
        IF DualOperation = 1
            THEN
                IF @retVal>Quantity
                    THEN
                        INSERT INTO Actions (EmployerID, StorehouseID, `Date`)
                            VALUES (EmployerID, StoreFromID, CURDATE());
                        SET @lastActionID = (SELECT ID FROM Actions ORDER BY ID DESC LIMIT 1);
                        INSERT INTO ProductTransfer (ID, TransferType)
                            VALUES (@lastActionID, 0);

                        INSERT INTO ActionDetails (ID,ProductID, Quantity)
                            VALUES (@lastActionID, ProductID, Quantity);

                        SET @lastTransferID = (SELECT ID FROM ProductTransfer ORDER BY ID DESC LIMIT 1);
                        INSERT INTO Actions (EmployerID, StorehouseID, `Date`) VALUES (EmployerID, StoreToID, CURDATE());
                        SET @lastActionID = (SELECT ID FROM Actions ORDER BY ID DESC LIMIT 1);
                        INSERT INTO ProductTransfer (ID, TransferType, ParentID) VALUES (@lastActionID, 1, @lastTransferID);

                        INSERT INTO ActionDetails (ID,ProductID, Quantity)
                            VALUES (@lastActionID, ProductID, Quantity);
                    ELSE
                        SIGNAL SQLSTATE '45000'
                    SET MESSAGE_TEXT = 'Not enough materials';
            END IF;
        ELSE
                INSERT INTO Actions (EmployerID, StorehouseID, `Date`)
                    VALUES (EmployerID, StoreID, CURDATE());
                SET @lastActionID = (SELECT ID FROM Actions ORDER BY ID DESC LIMIT 1);
                INSERT INTO ActionDetails (ID, ProductID, Quantity)
                    VALUES (@lastActionID, ProductID, Quantity);
                IF InOrOut = 0
                    THEN
                        INSERT INTO ProductIn (ID, OrganizationID) values (@lastActionID, NULL);
                    ELSE
                        IF @retVal>Quantity
                            THEN
                                INSERT INTO ProductOut (ID, OrganizationID) values (@lastActionID, NULL);
                            ELSE
                                SIGNAL SQLSTATE '45000'
                            SET MESSAGE_TEXT = 'Not enough materials';
                    END IF;
                END IF;
        END IF;
    COMMIT;
END

当我通过 Mysql 查询运行此代码时，一切似乎都运行良好。它发出“ Material 不足”的信号 IF @retVal<=Quantity并且没有插入任何记录(按原样工作)...但是当我从 PHP 调用这个过程时，它根本不会给出任何错误。没有插入任何行，但我无法收到通知失败的通知...这是 php 代码:

$mysqli->query("CALL `InsertAction`('6', '1', '2', '0', '13', '431243241', '1', '0')");

the $mysqli->sqlstate给出 0000 .我应该如何理解程序已完成或收到信号？

所以我真正想要的是，如果 @retVal<=Quantity然后给出 php 异常。此代码打印 "string"出:

try {
$mysqli->query("CALL `InsertAction`(6, 1, 2, 0, 13, 431243241, 1, 0)");
}
catch (Exception $e){
    echo "string";
}

25 4 0
文章推荐： mysql - 在 1 亿+ 字符串中有效地搜索存在 5000 次
文章推荐： python - 如何删除已存在列平均值的列
文章推荐： python:动态类型转换 - unicode 对象到 python 对象转换
文章推荐： Mysql查询计数器x/y
行者123
个人简介

我是一名优秀的程序员,十分优秀！

滴滴打车优惠券免费领取
滴滴打车优惠券
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com