两个 Select 语句合二为一答案

【问题标题】：Two Select statements into one两个 Select 语句合二为一
【发布时间】：2021-07-09 11:59:05
【问题描述】：

我使用的是 MySql 8.0.23 并且有两个 select 语句。第一个是这样的：

SELECT w.userid,w.reportdate, t.teamname, sum(w.worked)/60 as Worked FROM apticproject.view_workedtime as w
inner join user u on u.userid=w.userid
left outer join teammembers m on m.member=w.userid
left outer join teams t on t.teamid=m.team
where reportdate >= '2021-03-01' and reportdate <'2021-04-01'
group by w.userid
order by fullname ;

并给出以下输出：

第二个 select 语句如下所示：

SELECT r.userid, sum(r.invoicetime)/60 as invoicetime FROM apticproject.view_registeredtime as r
inner join user u on u.userid=r.userid 
left join teammembers m on m.member=r.userid
left join teams t on t.teamid=m.team
where fromdate >= '2021-03-01' and fromdate <'2021-04-01' and r.invoiced=1 and r.approved=1 and charge_customer=1 and r.zeroinvoice=0 
group by userid
order by fullname;

并给出以下输出：

我希望能够将 invoicetime 列从第二个 select 语句连接到第一个 select 语句，以便 invoicetime 列显示在第一个 select 语句中的 Worked 列之后。如您所见，我有可用作连接点的用户 ID。最有效的方法是什么？非常感谢您的帮助。

【问题讨论】：

将您的查询用作子查询，按公共列连接。
请edit 提及您使用的 MySQL / MariaDB 版本。使用 v8+，这些查询更容易编写，因为它们有 common table expressions。
@O.Jones，它已更新

标签： mysql sql join select subquery

【解决方案1】：

我设法找到了如下所示的解决方案。在 MySQL 中你没有 FULL JOIN，只有 Union 存在，这是我使用的语法：

select x.userid, x.username, x.fullname,x.reportdate, x.teamname, x.Worked,y.invoicetime
from
(SELECT w.userid, u.username, u.fullname,w.reportdate, t.teamname, sum(w.worked)/60 as Worked FROM apticproject.view_workedtime as w
inner join user u on u.userid=w.userid
left outer join teammembers m on m.member=w.userid
left outer join teams t on t.teamid=m.team
where reportdate >= '2021-03-01' and reportdate <'2021-04-01'
group by w.userid) as x

left join

(SELECT r.userid, sum(r.invoicetime)/60 as invoicetime FROM apticproject.view_registeredtime as r
inner join user u on u.userid=r.userid 
left join teammembers m on m.member=r.userid
left join teams t on t.teamid=m.team
where fromdate >= '2021-03-01' and fromdate <'2021-04-01' and r.invoiced=1 and r.approved=1 and charge_customer=1 and r.zeroinvoice=0 
group by userid) as y on y.userid=x.userid

Union

select x.userid, x.username, x.fullname,x.reportdate, x.teamname, x.Worked,y.invoicetime
from
(SELECT w.userid, u.username, u.fullname,w.reportdate, t.teamname, sum(w.worked)/60 as Worked FROM apticproject.view_workedtime as w
inner join user u on u.userid=w.userid
left outer join teammembers m on m.member=w.userid
left outer join teams t on t.teamid=m.team
where reportdate >= '2021-03-01' and reportdate <'2021-04-01'
group by w.userid) as x

right join

(SELECT r.userid, sum(r.invoicetime)/60 as invoicetime FROM apticproject.view_registeredtime as r
inner join user u on u.userid=r.userid 
left join teammembers m on m.member=r.userid
left join teams t on t.teamid=m.team
where fromdate >= '2021-03-01' and fromdate <'2021-04-01' and r.invoiced=1 and r.approved=1 and charge_customer=1 and r.zeroinvoice=0 
group by userid) as y on y.userid=x.userid
order by fullname;

【讨论】：

【解决方案2】：

如果您不希望在第二个查询中看到第一个没有用户记录的项目，您可以在 Darko86 答案的后半部分删除联合和整个右连接。

【讨论】：