如何总结两个不同的SQL查询的结果？

发布于 2025-01-21 09:20:12 字数 825 浏览 0 评论 0原文

我想概括一下这两个不同的查询。我加入了使用功能。但这显示了两个不同的细胞中的两个结果。

我需要一个结果来概括答案应为4069的这两个查询。

代码

SELECT COUNT(*) Active_Projects FROM
(SELECT ProjectID, ProjectStatusID, ClientID
FROM project) a
INNER JOIN 
(SELECT ProjectStatusID, ClientID, ProjectStatusName
FROM ProjectStatus) b
ON (a.ProjectStatusID = b.ProjectStatusID AND a.ClientID = b.ClientID)
WHERE ProjectStatusName LIKE '%active%'
AND a.ClientID = 4

UNION

SELECT COUNT(*) Total_Projects
FROM Project
WHERE ClientID=4

I want to SUM these two different queries. I joined both with USING function. but it shows the two results in two different cells.

I need a single result to SUM these two queries where the answer should be 4069.

The CODE

SELECT COUNT(*) Active_Projects FROM
(SELECT ProjectID, ProjectStatusID, ClientID
FROM project) a
INNER JOIN 
(SELECT ProjectStatusID, ClientID, ProjectStatusName
FROM ProjectStatus) b
ON (a.ProjectStatusID = b.ProjectStatusID AND a.ClientID = b.ClientID)
WHERE ProjectStatusName LIKE '%active%'
AND a.ClientID = 4

UNION

SELECT COUNT(*) Total_Projects
FROM Project
WHERE ClientID=4

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

甜尕妞 2025-01-28 09:20:12

尝试一下。 选择query1+ query2;

select (
    SELECT COUNT(*) Active_Projects FROM
(SELECT ProjectID, ProjectStatusID, ClientID
FROM project) a
INNER JOIN 
(SELECT ProjectStatusID, ClientID, ProjectStatusName
FROM ProjectStatus) b
ON (a.ProjectStatusID = b.ProjectStatusID AND a.ClientID = b.ClientID)
WHERE ProjectStatusName LIKE '%active%'
AND a.ClientID = 4
) + (
SELECT COUNT(*) Total_Projects
FROM Project
WHERE ClientID=4) ;

Try this. select query1+ query2;

select (
    SELECT COUNT(*) Active_Projects FROM
(SELECT ProjectID, ProjectStatusID, ClientID
FROM project) a
INNER JOIN 
(SELECT ProjectStatusID, ClientID, ProjectStatusName
FROM ProjectStatus) b
ON (a.ProjectStatusID = b.ProjectStatusID AND a.ClientID = b.ClientID)
WHERE ProjectStatusName LIKE '%active%'
AND a.ClientID = 4
) + (
SELECT COUNT(*) Total_Projects
FROM Project
WHERE ClientID=4) ;

回复收藏 0 原文

~没有更多了~