mysql 將多個SUM查詢合併為乙個結果

標籤：multiple-tables

sumsql

mysql

sql-server

所以…我們有3個與比賽有關的**,其中db跟蹤他們為每個比賽獲得的積分.競賽1、2和3.每當使用者獲得成就時,就會為該使用者建立乙個新行,並附加得分.因此,為了計算使用者獲得的所有積分,我使用選擇總和

select userid,sum(amount1)as"contest 1 points"from [company].[dbo].[contest1]where userid notin(0,1)group by userid
order by userid

因為我還有另外兩個比賽,所以我也要查詢每個比賽…

select userid,sum(amount2)/.65as "category 2 points"from [company].[dbo].[contest2]where dategiven >=201301and dategiven <=201305group by userid
order by userid
select userid,sum(amount3)as "category 3 points"from [company].[dbo].[contest3]whereuserid notin(1,2)group by userid
order by userid

我基本上需要將每個使用者從每個競賽中獲得的所有積分加到1個基本上顯示結果的列中

userid,共totals(競賽1競賽2競賽3)

或至少像

user,共1個競賽,共2個競賽,共3個競賽

到目前為止,我這樣做的方法是將每個結果複製/貼上到excel中,然後我使用vlookup將它們相互匹配,這有點麻煩,並且肯定有一種方法可以在sql中完成.我對sql來說還很陌生,我曾嘗試加入並使用usig

on來匹配userid,但是我的語法以及我了解到查詢都可以插入其中的方式出了點問題.

解決方法:

您需要union結果：

select userid,sum(points)as total
from
(select userid,sum(amount1)as "points"from [company].[dbo].[contest1]where userid not in (0,1)group by userid
union all       
select userid,sum(amount2)/.65as "category 2 points"from [company].[dbo].[contest2]where dategiven >=201301and dategiven <=201305group by userid
union all       
select userid,sum(amount3)as "category 3 points"from [company].[dbo].[contest3]where userid not in (1,2)group by userid
)as dt
group by userid
order by 2desc;

select userid,sum("category 1 points"),sum("category 2 points"),sum("category 3 points")from
(select userid,sum(amount1)as "category 1 points"from [company].[dbo].[contest1]where userid not in (0,1)group by userid
union all       
select userid,sum(amount2)/.65as "category 2 points"from [company].[dbo].[contest2]where dategiven >=201301and dategiven <=201305group by userid
union all       
select userid,sum(amount3)as "category 3 points"from [company].[dbo].[contest3]where userid not in (1,2)group by userid
)as dt
group by userid
order by 2desc;

當然,每個userdi /類別只有一行,因此min或max將返回相同的結果.

如果想要0,則將為不存在的資料返回null,而應使用coalesce(「 category x points」,0).

您也可以加入結果集,但是除非保證每個使用者都參加了每個比賽,否則您需要使用coalesce進行full outer join：

select userid,"category 1 points","category 2 points","category 3 points"from
(select userid,sum(amount1)as "category 1 points"from [company].[dbo].[contest1]where userid not in (0,1)group by userid
)as t1
full join
on t1.userid =t2.userid
(select userid,sum(amount2)/.65as "category 2 points"from [company].[dbo].[contest2]where dategiven >=201301and dategiven <=201305group by userid
)as t2
full join
(select userid,sum(amount3)as "category 3 points"from [company].[dbo].[contest3]where userid not in (1,2)group by userid
)as t3
on coalesce(t1.userid,t2.userid)=t3.userid
order by 2desc;

標籤：multiple-tables,sum,sql,mysql,sql-server

mysql 將多個SUM查詢合併為乙個結果

mysql將查詢到的多個id合併成id字串

mysql 合併查詢

如何將多個字典合併？

mysql 將多個SUM查詢合併為乙個結果

mysql將查詢到的多個id合併成id字串

mysql 合併查詢

如何將多個字典合併？

相關推薦