如何使用辅助函数创建更新两个字段的存储过程 mySQL
How to use helper functions to create a stored procedure that updates two fields mySQL
所以我在 mySQL 上有几个 table:
CREATE TABLE IF NOT EXISTS `salarygrade` (
`GRADE` INT(11) NOT NULL,
`HOURLYRATE` FLOAT NOT NULL,
PRIMARY KEY (`GRADE`));
============================================= ==============================
CREATE TABLE IF NOT EXISTS `staffongrade` (
`STAFFNO` INT(11) NOT NULL,
`GRADE` INT(11) NOT NULL,
`STARTDATE` DATE NULL DEFAULT NULL,
`FINISHDATE` DATE NULL DEFAULT NULL,
INDEX `STAFFONGRADE_FK` (`STAFFNO` ASC),
INDEX `STAFFONGRADE2_FK` (`GRADE` ASC),
PRIMARY KEY (`GRADE`, `STAFFNO`),
CONSTRAINT `FK_STAFFONG_STAFFONGR_SALARYGR`
FOREIGN KEY (`GRADE`)
REFERENCES `salarygrade` (`GRADE`),
CONSTRAINT `FK_STAFFONG_STAFFONGR_STAFF`
FOREIGN KEY (`STAFFNO`)
REFERENCES `staff` (`STAFFNO`));
============================================= ==============================
CREATE TABLE IF NOT EXISTS `campaign` (
`CAMPAIGN_NO` INT(11) NOT NULL,
`TITLE` VARCHAR(30) NOT NULL,
`CUSTOMER_ID` INT(11) NOT NULL,
`THEME` VARCHAR(40) NULL DEFAULT NULL,
`CAMPAIGNSTARTDATE` DATE NULL DEFAULT NULL,
`CAMPAIGNFINISHDATE` DATE NULL DEFAULT NULL,
`ESTIMATEDCOST` INT(11) NULL DEFAULT NULL,
`ACTUALCOST` FLOAT NULL DEFAULT NULL,
PRIMARY KEY (`CAMPAIGN_NO`),
INDEX `OWNS_FK` (`CUSTOMER_ID` ASC),
CONSTRAINT `FK_CAMPAIGN_OWNS_CUSTOMER`
FOREIGN KEY (`CUSTOMER_ID`)
REFERENCES `customer` (`CUSTOMER_ID`)
ON DELETE RESTRICT
ON UPDATE RESTRICT);
============================================= ==============================
CREATE TABLE IF NOT EXISTS `workson` (
`STAFFNO` INT(11) NOT NULL,
`CAMPAIGN_NO` INT(11) NOT NULL,
`WDATE` DATE NOT NULL,
`HOUR` FLOAT NULL DEFAULT NULL,
PRIMARY KEY (`STAFFNO`, `CAMPAIGN_NO`, `WDATE`),
INDEX `WORKSON_FK` (`STAFFNO` ASC),
INDEX `FK_WORKSON_WORKSON2_CAMPAIGN_idx` (`CAMPAIGN_NO` ASC),
CONSTRAINT `FK_WORKSON_WORKSON2_CAMPAIGN`
FOREIGN KEY (`CAMPAIGN_NO`)
REFERENCES `campaign` (`CAMPAIGN_NO`)
ON DELETE RESTRICT
ON UPDATE RESTRICT,
CONSTRAINT `FK_WORKSON_WORKSON_STAFF`
FOREIGN KEY (`STAFFNO`)
REFERENCES `staff` (`STAFFNO`));
我想创建一个名为 sp_finish_campaign (in c_title varchar(30)) 的存储过程，它获取活动的标题并通过将 CAMPAIGNFINISHDATE 更新为当前日期并将 ACTUALCOST 更新为活动成本，根据不同员工在不同日期投入的小时数计算，以及薪资等级（这会根据 staffID 和基于 STARTDATE 和 FINISHDATE 的时间范围而变化staffongrade table.
为了计算 ACTUALCOST，我创建了一个辅助函数：
DELIMITER //
CREATE FUNCTION rate_on_date(staff_id int, given_date date)
RETURNS int
DETERMINISTIC
BEGIN
DECLARE salaryGrade int;
SET salaryGrade = (select grade from staffongrade
where staffno = staff_id AND (given_date BETWEEN STARTDATE AND FINISHDATE));
RETURN salaryGrade;
END //
DELIMITER ;
其中returns支付grade根据staff_id和given_date参数我给它：
select rate_on_date(1, "2018-02-02") as Grade_On_Date;
对于这个参数，我假设我必须从 workson table 中获取它，如下所示：
我试过使用 select 语句来获得工资等级：
select hourlyrate as 'grade' from salarygrade
where rate_on_date(1, "2018-02-02") = grade;
要计算 ACTUALCOST 我假设我必须通过将 HOUR 列与成绩成本相乘来进行计算，并使用 WDATE 和 STAFFNO 列workson table 作为我的存储过程的参数，它将通过输入活动标题来计算和更新活动的 CAMPAIGNFINISHDATE 和 ACTUALCOST。
但是我该怎么做呢？
我只是对如何创建这个过程感到困惑，也对如何在我的存储过程中正确使用这些辅助函数感到困惑。
我觉得这个问题很长，但我真的不知道要问什么或应该采取什么方向来解决这个问题。
你真的不需要函数。 mysql 可以进行多次 table 更新（参见 https://dev.mysql.com/doc/refman/8.0/en/update.html）在你的情况下它可能看起来像这样
update campaign c
join
(select c.campaign_no,
sum(hour * hourlyrate) cost
from campaign c
join workson w on w.campaign_no = c.campaign_no
join staffongrade s on s .staffno = w.staffno and w.wdate between s.startdate and s.finishdate
join salarygrade g on g.grade = s.grade
group by c.campaign_no
) s
on s.campaign_no = c.campaign_no
set actualcost = s.cost
where c.campaign_no = 1
;
子查询需要的地方
如果你简化你的数据，这应该很容易证明；
drop table if exists salarygrade,campaign,workson,staffongrade;
CREATE TABLE `salarygrade`
( GRADE INT NOT NULL,
hOURLYRATE decimal(10,2) NOT NULL
);
insert into salarygrade values(1,10),(2,20);
cREATE TABLE IF NOT EXISTS `staffongrade` (
`STAFFNO` INT(11) NOT NULL,
`GRADE` INT(11) NOT NULL,
`STARTDATE` DATE NULL DEFAULT NULL,
`FINISHDATE` DATE NULL DEFAULT NULL
);
insert into staffongrade values
(1,1,'2019-01-01','2019-06-30'),(1,2,'2019-06-01','2019-12-31'),(2,1,'2019-01-01','2019-01-31');
CREATE TABLE IF NOT EXISTS `campaign` (
`CAMPAIGN_NO` INT(11) NOT NULL,
`CAMPAIGNSTARTDATE` DATE NULL DEFAULT NULL,
`CAMPAIGNFINISHDATE` DATE NULL DEFAULT NULL,
`ESTIMATEDCOST` INT(11) NULL DEFAULT NULL,
`ACTUALCOST` FLOAT NULL DEFAULT NULL
);
insert into campaign values (1,'2019-01-01','2019-12-31',null,null);
CREATE TABLE IF NOT EXISTS `workson` (
`STAFFNO` INT(11) NOT NULL,
`CAMPAIGN_NO` INT(11) NOT NULL,
`WDATE` DATE NOT NULL,
`HOUR` FLOAT NULL DEFAULT NULL
);
insert into workson values
(1,1,'2019-01-01',1),(1,1,'2019-12-01',1),(2,1,'2019-01-01',1);
select * from campaign;
+-------------+-------------------+--------------------+---------------+------------+
| CAMPAIGN_NO | CAMPAIGNSTARTDATE | CAMPAIGNFINISHDATE | ESTIMATEDCOST | ACTUALCOST |
+-------------+-------------------+--------------------+---------------+------------+
| 1 | 2019-01-01 | 2019-12-31 | NULL | 40 |
+-------------+-------------------+--------------------+---------------+------------+
1 row in set (0.00 sec)
必须冲刺所以我会让你把更新放到一个过程中。
如果 staffongrade 的完成日期为 NULL，则需要进行一些数据清理。为简单起见，我会创建一个临时的 table 来填补空白并更改更新语句以使用 projectfinishdate（如果不知道，则替换为 suitable 未来日期）。此代码将在更新之前插入到您的程序中
所以
insert into staffongrade values
(1,1,'2019-01-01',null),(1,2,'2019-07-01',null),(2,1,'2019-01-01',null);
drop temporary table if exists staffongradetemp;
create temporary table staffongradetemp like staffongrade;
insert into staffongradetemp
select s.STAFFNO,s.GRADE,s.STARTDATE,
case when s.FINISHDATE is not null then s.finishdate
else date_sub((select s1.startdate
from staffongrade s1
where s1.STAFFNO = s.STAFFNO and s1.startdate > s.STARTDATE
order by startdate limit 1), interval 1 day)
end
from staffongrade s
;
select * from staffongradetemp;
+---------+-------+------------+------------+
| STAFFNO | GRADE | STARTDATE | FINISHDATE |
+---------+-------+------------+------------+
| 1 | 1 | 2019-01-01 | 2019-06-30 |
| 1 | 2 | 2019-07-01 | NULL |
| 2 | 1 | 2019-01-01 | NULL |
+---------+-------+------------+------------+
3 rows in set (0.00 sec)
这会将所有最后的 finshdates 保留为 null，我们可以使用 coalesce
将其捕获在更新语句中
update campaign c
join
(select c.campaign_no,
sum(hour * hourlyrate) cost
from campaign c
join workson w on w.campaign_no = c.campaign_no
join **staffongradetemp s** on s .staffno = w.staffno and w.wdate between s.startdate and **coalesce(s.finishdate,c.CAMPAIGNFINISHDATE)**
join salarygrade g on g.grade = s.grade
where c.campaign_no = 1
group by c.campaign_no
) s
on s.campaign_no = c.campaign_no
set actualcost = s.cost
where 1 = 1;
所以我在 mySQL 上有几个 table:
CREATE TABLE IF NOT EXISTS `salarygrade` (
`GRADE` INT(11) NOT NULL,
`HOURLYRATE` FLOAT NOT NULL,
PRIMARY KEY (`GRADE`));
============================================= ==============================
CREATE TABLE IF NOT EXISTS `staffongrade` (
`STAFFNO` INT(11) NOT NULL,
`GRADE` INT(11) NOT NULL,
`STARTDATE` DATE NULL DEFAULT NULL,
`FINISHDATE` DATE NULL DEFAULT NULL,
INDEX `STAFFONGRADE_FK` (`STAFFNO` ASC),
INDEX `STAFFONGRADE2_FK` (`GRADE` ASC),
PRIMARY KEY (`GRADE`, `STAFFNO`),
CONSTRAINT `FK_STAFFONG_STAFFONGR_SALARYGR`
FOREIGN KEY (`GRADE`)
REFERENCES `salarygrade` (`GRADE`),
CONSTRAINT `FK_STAFFONG_STAFFONGR_STAFF`
FOREIGN KEY (`STAFFNO`)
REFERENCES `staff` (`STAFFNO`));
============================================= ==============================
CREATE TABLE IF NOT EXISTS `campaign` (
`CAMPAIGN_NO` INT(11) NOT NULL,
`TITLE` VARCHAR(30) NOT NULL,
`CUSTOMER_ID` INT(11) NOT NULL,
`THEME` VARCHAR(40) NULL DEFAULT NULL,
`CAMPAIGNSTARTDATE` DATE NULL DEFAULT NULL,
`CAMPAIGNFINISHDATE` DATE NULL DEFAULT NULL,
`ESTIMATEDCOST` INT(11) NULL DEFAULT NULL,
`ACTUALCOST` FLOAT NULL DEFAULT NULL,
PRIMARY KEY (`CAMPAIGN_NO`),
INDEX `OWNS_FK` (`CUSTOMER_ID` ASC),
CONSTRAINT `FK_CAMPAIGN_OWNS_CUSTOMER`
FOREIGN KEY (`CUSTOMER_ID`)
REFERENCES `customer` (`CUSTOMER_ID`)
ON DELETE RESTRICT
ON UPDATE RESTRICT);
============================================= ==============================
CREATE TABLE IF NOT EXISTS `workson` (
`STAFFNO` INT(11) NOT NULL,
`CAMPAIGN_NO` INT(11) NOT NULL,
`WDATE` DATE NOT NULL,
`HOUR` FLOAT NULL DEFAULT NULL,
PRIMARY KEY (`STAFFNO`, `CAMPAIGN_NO`, `WDATE`),
INDEX `WORKSON_FK` (`STAFFNO` ASC),
INDEX `FK_WORKSON_WORKSON2_CAMPAIGN_idx` (`CAMPAIGN_NO` ASC),
CONSTRAINT `FK_WORKSON_WORKSON2_CAMPAIGN`
FOREIGN KEY (`CAMPAIGN_NO`)
REFERENCES `campaign` (`CAMPAIGN_NO`)
ON DELETE RESTRICT
ON UPDATE RESTRICT,
CONSTRAINT `FK_WORKSON_WORKSON_STAFF`
FOREIGN KEY (`STAFFNO`)
REFERENCES `staff` (`STAFFNO`));
我想创建一个名为 sp_finish_campaign (in c_title varchar(30)) 的存储过程，它获取活动的标题并通过将 CAMPAIGNFINISHDATE 更新为当前日期并将 ACTUALCOST 更新为活动成本，根据不同员工在不同日期投入的小时数计算，以及薪资等级（这会根据 staffID 和基于 STARTDATE 和 FINISHDATE 的时间范围而变化staffongrade table.
为了计算 ACTUALCOST，我创建了一个辅助函数：
DELIMITER //
CREATE FUNCTION rate_on_date(staff_id int, given_date date)
RETURNS int
DETERMINISTIC
BEGIN
DECLARE salaryGrade int;
SET salaryGrade = (select grade from staffongrade
where staffno = staff_id AND (given_date BETWEEN STARTDATE AND FINISHDATE));
RETURN salaryGrade;
END //
DELIMITER ;
其中returns支付grade根据staff_id和given_date参数我给它：
select rate_on_date(1, "2018-02-02") as Grade_On_Date;
对于这个参数，我假设我必须从 workson table 中获取它，如下所示：
我试过使用 select 语句来获得工资等级：
select hourlyrate as 'grade' from salarygrade
where rate_on_date(1, "2018-02-02") = grade;
要计算 ACTUALCOST 我假设我必须通过将 HOUR 列与成绩成本相乘来进行计算，并使用 WDATE 和 STAFFNO 列workson table 作为我的存储过程的参数，它将通过输入活动标题来计算和更新活动的 CAMPAIGNFINISHDATE 和 ACTUALCOST。
但是我该怎么做呢？
我只是对如何创建这个过程感到困惑，也对如何在我的存储过程中正确使用这些辅助函数感到困惑。 我觉得这个问题很长，但我真的不知道要问什么或应该采取什么方向来解决这个问题。
你真的不需要函数。 mysql 可以进行多次 table 更新（参见 https://dev.mysql.com/doc/refman/8.0/en/update.html）在你的情况下它可能看起来像这样
update campaign c
join
(select c.campaign_no,
sum(hour * hourlyrate) cost
from campaign c
join workson w on w.campaign_no = c.campaign_no
join staffongrade s on s .staffno = w.staffno and w.wdate between s.startdate and s.finishdate
join salarygrade g on g.grade = s.grade
group by c.campaign_no
) s
on s.campaign_no = c.campaign_no
set actualcost = s.cost
where c.campaign_no = 1
;
子查询需要的地方
如果你简化你的数据，这应该很容易证明；
drop table if exists salarygrade,campaign,workson,staffongrade;
CREATE TABLE `salarygrade`
( GRADE INT NOT NULL,
hOURLYRATE decimal(10,2) NOT NULL
);
insert into salarygrade values(1,10),(2,20);
cREATE TABLE IF NOT EXISTS `staffongrade` (
`STAFFNO` INT(11) NOT NULL,
`GRADE` INT(11) NOT NULL,
`STARTDATE` DATE NULL DEFAULT NULL,
`FINISHDATE` DATE NULL DEFAULT NULL
);
insert into staffongrade values
(1,1,'2019-01-01','2019-06-30'),(1,2,'2019-06-01','2019-12-31'),(2,1,'2019-01-01','2019-01-31');
CREATE TABLE IF NOT EXISTS `campaign` (
`CAMPAIGN_NO` INT(11) NOT NULL,
`CAMPAIGNSTARTDATE` DATE NULL DEFAULT NULL,
`CAMPAIGNFINISHDATE` DATE NULL DEFAULT NULL,
`ESTIMATEDCOST` INT(11) NULL DEFAULT NULL,
`ACTUALCOST` FLOAT NULL DEFAULT NULL
);
insert into campaign values (1,'2019-01-01','2019-12-31',null,null);
CREATE TABLE IF NOT EXISTS `workson` (
`STAFFNO` INT(11) NOT NULL,
`CAMPAIGN_NO` INT(11) NOT NULL,
`WDATE` DATE NOT NULL,
`HOUR` FLOAT NULL DEFAULT NULL
);
insert into workson values
(1,1,'2019-01-01',1),(1,1,'2019-12-01',1),(2,1,'2019-01-01',1);
select * from campaign;
+-------------+-------------------+--------------------+---------------+------------+
| CAMPAIGN_NO | CAMPAIGNSTARTDATE | CAMPAIGNFINISHDATE | ESTIMATEDCOST | ACTUALCOST |
+-------------+-------------------+--------------------+---------------+------------+
| 1 | 2019-01-01 | 2019-12-31 | NULL | 40 |
+-------------+-------------------+--------------------+---------------+------------+
1 row in set (0.00 sec)
必须冲刺所以我会让你把更新放到一个过程中。
如果 staffongrade 的完成日期为 NULL，则需要进行一些数据清理。为简单起见，我会创建一个临时的 table 来填补空白并更改更新语句以使用 projectfinishdate（如果不知道，则替换为 suitable 未来日期）。此代码将在更新之前插入到您的程序中
所以
insert into staffongrade values
(1,1,'2019-01-01',null),(1,2,'2019-07-01',null),(2,1,'2019-01-01',null);
drop temporary table if exists staffongradetemp;
create temporary table staffongradetemp like staffongrade;
insert into staffongradetemp
select s.STAFFNO,s.GRADE,s.STARTDATE,
case when s.FINISHDATE is not null then s.finishdate
else date_sub((select s1.startdate
from staffongrade s1
where s1.STAFFNO = s.STAFFNO and s1.startdate > s.STARTDATE
order by startdate limit 1), interval 1 day)
end
from staffongrade s
;
select * from staffongradetemp;
+---------+-------+------------+------------+
| STAFFNO | GRADE | STARTDATE | FINISHDATE |
+---------+-------+------------+------------+
| 1 | 1 | 2019-01-01 | 2019-06-30 |
| 1 | 2 | 2019-07-01 | NULL |
| 2 | 1 | 2019-01-01 | NULL |
+---------+-------+------------+------------+
3 rows in set (0.00 sec)
这会将所有最后的 finshdates 保留为 null，我们可以使用 coalesce
将其捕获在更新语句中update campaign c
join
(select c.campaign_no,
sum(hour * hourlyrate) cost
from campaign c
join workson w on w.campaign_no = c.campaign_no
join **staffongradetemp s** on s .staffno = w.staffno and w.wdate between s.startdate and **coalesce(s.finishdate,c.CAMPAIGNFINISHDATE)**
join salarygrade g on g.grade = s.grade
where c.campaign_no = 1
group by c.campaign_no
) s
on s.campaign_no = c.campaign_no
set actualcost = s.cost
where 1 = 1;