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sql查询汇总问题
create table order_money
(
id int identity(1,1) primary key not null,
user_id int,
user_name nvarchar(20),
group int,
type varchar(10),
money float,
add_time datetime
)
查询结果需要这样，先是每个小组成员信息汇总，然后小组1是每个组成员（张三、李四）的汇总，总计是每个小组的汇总，如何实现？
姓名 当日金额 本周金额 本月金额
张三 100 300 3000
李四 100 200 1000
小组1 200 500 4000
王五 50 150 800
刘六 500 700 1700
小组2 550 850 2500
总计 750 1350 6500
(
id int identity(1,1) primary key not null,
user_id int,
user_name nvarchar(20),
group int,
type varchar(10),
money float,
add_time datetime
)
查询结果需要这样，先是每个小组成员信息汇总，然后小组1是每个组成员（张三、李四）的汇总，总计是每个小组的汇总，如何实现？
姓名 当日金额 本周金额 本月金额
张三 100 300 3000
李四 100 200 1000
小组1 200 500 4000
王五 50 150 800
刘六 500 700 1700
小组2 550 850 2500
总计 750 1350 6500
...全文
当前发帖距今超过3年，不再开放新的回复
发表回复
二月十六 2017-11-02
--测试数据
if not object_id(N'Tempdb..#T') is null
drop table #T
Go
Create table #T([姓名] nvarchar(22),[当日金额] int,[本周金额] int,[本月金额] INT,[分组] int)
Insert #T
select N'张三',100,300,3000,1 union all
select N'李四',100,200,1000,1 union all
select N'王五',50,150,800,2 union all
select N'刘六',500,700,1700,2
Go
--测试数据结束
SELECT 姓名,当日金额,本周金额,本月金额
FROM ( SELECT 姓名 ,
SUM(当日金额) AS 当日金额 ,
SUM(本周金额) AS 本周金额 ,
SUM(本月金额) AS 本月金额 ,
分组
FROM #T
GROUP BY 姓名 ,
分组
UNION ALL
SELECT RTRIM(分组) +'组',
SUM(当日金额) AS 当日金额 ,
SUM(本周金额) AS 本周金额 ,
SUM(本月金额) AS 本月金额 ,
分组
FROM #T
GROUP BY RTRIM(分组) +'组' ,
分组
UNION ALL
SELECT '总计' ,
SUM(当日金额) AS 当日金额 ,
SUM(本周金额) AS 本周金额 ,
SUM(本月金额) AS 本月金额 ,
1000
FROM #T
) t
ORDER BY t.分组 ,
姓名 DESC
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听雨停了 2017-11-02
给点测试数据
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RINK_1 2017-11-02
select case when GROUPING(user_id)=1 and GROUPING([group])=0 then '小组'+CAST([GROUP] as varchar)
when GROUPING(user_id)=1 and GROUPING([group])=1 then '合计'
else MAX(user_name)
end as name,
sum(case when DATEDIFF(DAY,add_time,GETDATE())=0 then money else 0 end) as total_daily,
sum(case when DATEDIFF(week,add_time,GETDATE())=0 then money else 0 end) as total_weekly,
sum(case when DATEDIFF(month,add_time,GETDATE())=0 then money else 0 end) as total_monthly
from #order_money
group by user_id,[GROUP] with cube
having not (GROUPing(user_id)=0 and GROUPING([GROUP])=1)
order by GROUPING([GROUP]),[group],GROUPING(user_id)
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我还在下面游泳呢 2017-11-02
刚好这几天被汇总折磨的要死要活的，这里是个很好的解决思路启发地
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听雨停了 2017-11-02
--测试数据
IF OBJECT_ID('tempdb..#order_money') IS NOT NULL
DROP TABLE #order_money
create table #order_money
(
id int identity(1,1) primary key not null,
user_id int,
user_name nvarchar(20),
[GROUP] int,
type varchar(10),
money float,
add_time datetime
)
insert into #order_money values(1,'张三',1,'类型1',30,'2017-11-2')
insert into #order_money values(1,'张三',1,'类型1',70,'2017-11-2')
insert into #order_money values(1,'张三',1,'类型1',100,'2017-11-3')
insert into #order_money values(1,'张三',1,'类型1',100,'2017-11-3')
insert into #order_money values(1,'张三',1,'类型1',700,'2017-11-9')
insert into #order_money values(1,'张三',1,'类型1',2000,'2017-11-26')
insert into #order_money values(2,'李四',1,'类型1',60,'2017-11-2')
insert into #order_money values(2,'李四',1,'类型1',30,'2017-11-2')
insert into #order_money values(2,'李四',1,'类型1',10,'2017-11-2')
insert into #order_money values(2,'李四',1,'类型1',50,'2017-11-3')
insert into #order_money values(2,'李四',1,'类型1',50,'2017-11-3')
insert into #order_money values(2,'李四',1,'类型1',300,'2017-11-25')
insert into #order_money values(2,'李四',1,'类型1',500,'2017-11-23')
insert into #order_money values(3,'王五',2,'类型1',50,'2017-11-2')
insert into #order_money values(3,'王五',2,'类型1',100,'2017-11-3')
insert into #order_money values(3,'王五',2,'类型1',150,'2017-11-27')
insert into #order_money values(3,'王五',2,'类型1',500,'2017-11-29')
insert into #order_money values(4,'刘六',2,'类型1',200,'2017-11-2')
insert into #order_money values(4,'刘六',2,'类型1',300,'2017-11-2')
insert into #order_money values(4,'刘六',2,'类型1',200,'2017-11-3')
insert into #order_money values(4,'刘六',2,'类型1',1000,'2017-11-29')
--测试数据结束
;WITH cte AS (
SELECT DISTINCT USER_NAME,
[GROUP],
SUM([MONEY]) OVER(PARTITION BY USER_NAME, [GROUP]) AS 本月金额
FROM #order_money
),
cte2 AS (
SELECT DISTINCT USER_NAME,
[GROUP],
SUM([MONEY]) OVER(PARTITION BY USER_NAME, [GROUP]) AS 本周金额
FROM #order_money
WHERE add_time BETWEEN CONVERT(
VARCHAR(100),
DATEADD(DAY, -(DATEPART(weekday, GETDATE()) -1), GETDATE()),
23
) AND CONVERT(
VARCHAR(100),
DATEADD(DAY, -(DATEPART(weekday, GETDATE()) -7), GETDATE()),
23
)
),
cte3 AS (
SELECT DISTINCT USER_NAME,
[GROUP],
SUM([MONEY]) OVER(PARTITION BY USER_NAME, [GROUP]) AS 当日金额
FROM #order_money
WHERE convert(varchar(100),add_time,23)=convert(varchar(100),GETDATE(),23)
),
cte4 AS(
SELECT '小组'+cast(a.[GROUP] AS VARCHAR(20)) as '姓名',a.[GROUP],SUM(a.本月金额) as 本月金额,SUM(b.本周金额) as 本周金额,SUM(c.当日金额) as 当日金额 FROM cte a
INNER JOIN cte2 b ON a.USER_NAME=b.USER_NAME AND a.[group]=b.[group]
INNER JOIN cte3 c ON a.USER_NAME=c.USER_NAME AND a.[group]=c.[group]
GROUP BY a.[GROUP]
),
cte5 AS (
SELECT '总计' as user_name,999 as [group],SUM(本月金额) as 本月金额,sum(本周金额) as 本周金额,sum(当日金额) as 当日金额 FROM cte4
),
cte6 AS (
SELECT a.user_name AS '姓名',a.[group],a.本月金额,b.本周金额,c.当日金额 FROM cte a
INNER JOIN cte2 b ON a.USER_NAME=b.USER_NAME AND a.[group]=b.[group]
INNER JOIN cte3 c ON a.USER_NAME=c.USER_NAME AND a.[group]=c.[group]
),
cte7 AS (
SELECT * FROM cte6
UNION ALL
SELECT * FROM cte5
UNION ALL
SELECT * FROM cte4
)
SELECT 姓名,当日金额,本周金额,本月金额
from cte7
ORDER BY [GROUP],本月金额
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Caoxp_papa 2017-11-02
基于这个数据的结果是什么？[/quote]
结果是：
姓名 当日金额 本周金额 本月金额
张三 100 300 3000
李四 100 200 1000
小组1 200 500 4000
王五 50 150 800
刘六 500 700 1700
小组2 550 850 2500
总计 750 1350 6500[/quote]
[/quote]
非常感谢，输出顺序可以调整一下吗？每个小组的成员，然后小组汇总，第二个小组成员，第二个小组汇总.....
姓名 当日金额 本周金额 本月金额
张三 100 300 3000
李四 100 200 1000
小组1 200 500 4000
王五 50 150 800
刘六 500 700 1700
小组2 550 850 2500
总计 750 1350 6500
--测试数据
IF OBJECT_ID('tempdb..#order_money') IS NOT NULL
DROP TABLE #order_money
create table #order_money
(
id int identity(1,1) primary key not null,
user_id int,
user_name nvarchar(20),
[GROUP] int,
type varchar(10),
money float,
add_time datetime
)
insert into #order_money values(1,'张三',1,'类型1',30,'2017-11-2')
insert into #order_money values(1,'张三',1,'类型1',70,'2017-11-2')
insert into #order_money values(1,'张三',1,'类型1',100,'2017-11-3')
insert into #order_money values(1,'张三',1,'类型1',100,'2017-11-3')
insert into #order_money values(1,'张三',1,'类型1',700,'2017-11-9')
insert into #order_money values(1,'张三',1,'类型1',2000,'2017-11-26')
insert into #order_money values(2,'李四',1,'类型1',60,'2017-11-2')
insert into #order_money values(2,'李四',1,'类型1',30,'2017-11-2')
insert into #order_money values(2,'李四',1,'类型1',10,'2017-11-2')
insert into #order_money values(2,'李四',1,'类型1',50,'2017-11-3')
insert into #order_money values(2,'李四',1,'类型1',50,'2017-11-3')
insert into #order_money values(2,'李四',1,'类型1',300,'2017-11-25')
insert into #order_money values(2,'李四',1,'类型1',500,'2017-11-23')
insert into #order_money values(3,'王五',2,'类型1',50,'2017-11-2')
insert into #order_money values(3,'王五',2,'类型1',100,'2017-11-3')
insert into #order_money values(3,'王五',2,'类型1',150,'2017-11-27')
insert into #order_money values(3,'王五',2,'类型1',500,'2017-11-29')
insert into #order_money values(4,'刘六',2,'类型1',200,'2017-11-2')
insert into #order_money values(4,'刘六',2,'类型1',300,'2017-11-2')
insert into #order_money values(4,'刘六',2,'类型1',200,'2017-11-3')
insert into #order_money values(4,'刘六',2,'类型1',1000,'2017-11-29')
--测试数据结束
;WITH cte AS (
SELECT DISTINCT USER_NAME,
[GROUP],
SUM([MONEY]) OVER(PARTITION BY USER_NAME, [GROUP]) AS 本月金额
FROM #order_money
),
cte2 AS (
SELECT DISTINCT USER_NAME,
[GROUP],
SUM([MONEY]) OVER(PARTITION BY USER_NAME, [GROUP]) AS 本周金额
FROM #order_money
WHERE add_time BETWEEN CONVERT(
VARCHAR(100),
DATEADD(DAY, -(DATEPART(weekday, GETDATE()) -1), GETDATE()),
23
) AND CONVERT(
VARCHAR(100),
DATEADD(DAY, -(DATEPART(weekday, GETDATE()) -7), GETDATE()),
23
)
),
cte3 AS (
SELECT DISTINCT USER_NAME,
[GROUP],
SUM([MONEY]) OVER(PARTITION BY USER_NAME, [GROUP]) AS 当日金额
FROM #order_money
WHERE convert(varchar(100),add_time,23)=convert(varchar(100),GETDATE(),23)
),
cte4 AS(
SELECT '小组'+cast(a.[GROUP] AS VARCHAR(20)) as '姓名',SUM(a.本月金额) as 本月金额,SUM(b.本周金额) as 本周金额,SUM(c.当日金额) as 当日金额 FROM cte a
INNER JOIN cte2 b ON a.USER_NAME=b.USER_NAME AND a.[group]=b.[group]
INNER JOIN cte3 c ON a.USER_NAME=c.USER_NAME AND a.[group]=c.[group]
GROUP BY a.[GROUP]
),
cte5 AS (
SELECT '总计' as user_name,SUM(本月金额) as 本月金额,sum(本周金额) as 本周金额,sum(当日金额) as 当日金额 FROM cte4
),
cte6 AS (
SELECT a.user_name AS '姓名',a.本月金额,b.本周金额,c.当日金额 FROM cte a
INNER JOIN cte2 b ON a.USER_NAME=b.USER_NAME AND a.[group]=b.[group]
INNER JOIN cte3 c ON a.USER_NAME=c.USER_NAME AND a.[group]=c.[group]
)
SELECT * FROM cte6
UNION ALL
SELECT * FROM cte5
UNION ALL
SELECT * FROM cte4
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听雨停了 2017-11-02
基于这个数据的结果是什么？[/quote]
结果是：
姓名 当日金额 本周金额 本月金额
张三 100 300 3000
李四 100 200 1000
小组1 200 500 4000
王五 50 150 800
刘六 500 700 1700
小组2 550 850 2500
总计 750 1350 6500[/quote]
--测试数据
IF OBJECT_ID('tempdb..#order_money') IS NOT NULL
DROP TABLE #order_money
create table #order_money
(
id int identity(1,1) primary key not null,
user_id int,
user_name nvarchar(20),
[GROUP] int,
type varchar(10),
money float,
add_time datetime
)
insert into #order_money values(1,'张三',1,'类型1',30,'2017-11-2')
insert into #order_money values(1,'张三',1,'类型1',70,'2017-11-2')
insert into #order_money values(1,'张三',1,'类型1',100,'2017-11-3')
insert into #order_money values(1,'张三',1,'类型1',100,'2017-11-3')
insert into #order_money values(1,'张三',1,'类型1',700,'2017-11-9')
insert into #order_money values(1,'张三',1,'类型1',2000,'2017-11-26')
insert into #order_money values(2,'李四',1,'类型1',60,'2017-11-2')
insert into #order_money values(2,'李四',1,'类型1',30,'2017-11-2')
insert into #order_money values(2,'李四',1,'类型1',10,'2017-11-2')
insert into #order_money values(2,'李四',1,'类型1',50,'2017-11-3')
insert into #order_money values(2,'李四',1,'类型1',50,'2017-11-3')
insert into #order_money values(2,'李四',1,'类型1',300,'2017-11-25')
insert into #order_money values(2,'李四',1,'类型1',500,'2017-11-23')
insert into #order_money values(3,'王五',2,'类型1',50,'2017-11-2')
insert into #order_money values(3,'王五',2,'类型1',100,'2017-11-3')
insert into #order_money values(3,'王五',2,'类型1',150,'2017-11-27')
insert into #order_money values(3,'王五',2,'类型1',500,'2017-11-29')
insert into #order_money values(4,'刘六',2,'类型1',200,'2017-11-2')
insert into #order_money values(4,'刘六',2,'类型1',300,'2017-11-2')
insert into #order_money values(4,'刘六',2,'类型1',200,'2017-11-3')
insert into #order_money values(4,'刘六',2,'类型1',1000,'2017-11-29')
--测试数据结束
;WITH cte AS (
SELECT DISTINCT USER_NAME,
[GROUP],
SUM([MONEY]) OVER(PARTITION BY USER_NAME, [GROUP]) AS 本月金额
FROM #order_money
),
cte2 AS (
SELECT DISTINCT USER_NAME,
[GROUP],
SUM([MONEY]) OVER(PARTITION BY USER_NAME, [GROUP]) AS 本周金额
FROM #order_money
WHERE add_time BETWEEN CONVERT(
VARCHAR(100),
DATEADD(DAY, -(DATEPART(weekday, GETDATE()) -1), GETDATE()),
23
) AND CONVERT(
VARCHAR(100),
DATEADD(DAY, -(DATEPART(weekday, GETDATE()) -7), GETDATE()),
23
)
),
cte3 AS (
SELECT DISTINCT USER_NAME,
[GROUP],
SUM([MONEY]) OVER(PARTITION BY USER_NAME, [GROUP]) AS 当日金额
FROM #order_money
WHERE convert(varchar(100),add_time,23)=convert(varchar(100),GETDATE(),23)
),
cte4 AS(
SELECT '小组'+cast(a.[GROUP] AS VARCHAR(20)) as '姓名',SUM(a.本月金额) as 本月金额,SUM(b.本周金额) as 本周金额,SUM(c.当日金额) as 当日金额 FROM cte a
INNER JOIN cte2 b ON a.USER_NAME=b.USER_NAME AND a.[group]=b.[group]
INNER JOIN cte3 c ON a.USER_NAME=c.USER_NAME AND a.[group]=c.[group]
GROUP BY a.[GROUP]
),
cte5 AS (
SELECT '总计' as user_name,SUM(本月金额) as 本月金额,sum(本周金额) as 本周金额,sum(当日金额) as 当日金额 FROM cte4
),
cte6 AS (
SELECT a.user_name AS '姓名',a.本月金额,b.本周金额,c.当日金额 FROM cte a
INNER JOIN cte2 b ON a.USER_NAME=b.USER_NAME AND a.[group]=b.[group]
INNER JOIN cte3 c ON a.USER_NAME=c.USER_NAME AND a.[group]=c.[group]
)
SELECT * FROM cte6
UNION ALL
SELECT * FROM cte5
UNION ALL
SELECT * FROM cte4
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Caoxp_papa 2017-11-02
基于这个数据的结果是什么？[/quote]
结果是：
姓名 当日金额 本周金额 本月金额
张三 100 300 3000
李四 100 200 1000
小组1 200 500 4000
王五 50 150 800
刘六 500 700 1700
小组2 550 850 2500
总计 750 1350 6500
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二月十六 2017-11-02
基于这个数据的结果是什么？
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Caoxp_papa 2017-11-02
不好意思，应该是我没描述清除，order_money这个表是记录的每一个订单的金额信息，没有汇总信息。上边是测试数据。
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Caoxp_papa 2017-11-02
测试数据：
create table order_money
(
id int identity(1,1) primary key not null,
user_id int,
user_name nvarchar(20),
[group] int,
type varchar(10),
money float,
add_time datetime
)
insert into order_money values(1,'张三',1,'类型1',30,'2017-11-2')
insert into order_money values(1,'张三',1,'类型1',70,'2017-11-2')
insert into order_money values(1,'张三',1,'类型1',100,'2017-11-3')
insert into order_money values(1,'张三',1,'类型1',100,'2017-11-3')
insert into order_money values(1,'张三',1,'类型1',700,'2017-11-9')
insert into order_money values(1,'张三',1,'类型1',2000,'2017-11-26')
insert into order_money values(2,'李四',1,'类型1',60,'2017-11-2')
insert into order_money values(2,'李四',1,'类型1',30,'2017-11-2')
insert into order_money values(2,'李四',1,'类型1',10,'2017-11-2')
insert into order_money values(2,'李四',1,'类型1',50,'2017-11-3')
insert into order_money values(2,'李四',1,'类型1',50,'2017-11-3')
insert into order_money values(2,'李四',1,'类型1',300,'2017-11-25')
insert into order_money values(2,'李四',1,'类型1',500,'2017-11-23')
insert into order_money values(3,'王五',2,'类型1',50,'2017-11-2')
insert into order_money values(3,'王五',2,'类型1',100,'2017-11-3')
insert into order_money values(3,'王五',2,'类型1',150,'2017-11-27')
insert into order_money values(3,'王五',2,'类型1',500,'2017-11-29')
insert into order_money values(4,'刘六',2,'类型1',200,'2017-11-2')
insert into order_money values(4,'刘六',2,'类型1',300,'2017-11-2')
insert into order_money values(4,'刘六',2,'类型1',200,'2017-11-3')
insert into order_money values(4,'刘六',2,'类型1',1000,'2017-11-29')
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python查询数据库语句大全_SQL语句汇总——数据修改、数据查询 首先创建一张表如下，创建表的方法在上篇介绍过了，这里就不再赘述。 添加新数据： INSERT INTO () VALUES () 如： INSERT INTO t... 基本查询SQL的执行顺序： 1.执行FROM 2.WHERE条件过滤 3.SELECT投影 4.ORDER BY排序
SQL汇总行 SQL汇总行 1、MySQL 结果如图所示： 2、sql语句 说明：UNION 操作符选取不同的值；UNION ALL 操作符允许重复的值。 详见SQL UNION 和 UNION ALL 操作符 SELECT Id, Score, Course, Student FROM scores UNION...
java sqlserver分页查询_SQL分页查询方式汇总--Java学习网 核心提示：需求：查询表dbo.Message，每页10条，查询第2页1:TOP()?12SELECT TOP(20) * FROM dbo.Message WHERE Code NOT IN(SELECT TO...需求：查询表dbo.Message，每页10条，查询第2页1:TOP()2:BETWEEN * AND * ， ...
oracle sql查询情况汇总 1.查询几年之内的（3年以内） BETWEEN(select extract(year from sysdate) -2 from dual) and (select extract(year from sysdate) from dual) 2.查询
mysql中实现分类统计查询的步骤_mysql分组汇总统计查询SQL如何实现 sqlOne：select * from tablename1 where id>5;此语句查询出来多条记录，之后看做一个新的表。sqlTwo：select conut(*) from (select * from tablename1 where id>5) as tablename2;此语句即可查询出来统计的...
sql统计各科成绩大于平均分的人_SQL—汇总分析 一、汇总分析1.知识点汇总count:求某列的行数,如果函数中输入的时【列名】除去空值null以后的行数；...习题1)查询课程编号为“0002”的总成绩2)查询选了课程的学生人数思路解析：一开始我是从studen...
mysql数据库多级分类汇总_sql多级分类汇总实现介绍 GO -- 统计 -- 逐级汇总 declare @l int set @l=1 select A.[id], [pid] = A.parentid, [sumnum] = SUM(B.amount), level=case when exists(select * from t1 where parentid=a.[id]) then @l-1 else @l end into ...
sql server 递归汇总 --汇总追加成本 select csa.AccountLevel , csa.AccountShortName , csa.StageAccountGUID , csa.ParentStageAccountGUID , csa.IsEndCost , csa.ProjGUID , csa.HierarchyCode , isnull(t.AdjustAmount,0) ...
sql月度分组,按月汇总SQL查询分组 ) GROUP BY YEAR(TransactionDate), MONTH(TransactionDate) ORDER BY YEAR(Created), MONTH(Created) I don't know if your version of SQL has the MONTH and YEAR functions, so you may have to use DATEPART.
mysql查询指定数量的sql_mySql关于统计数量的SQL查询操作 d_device group by project_no order by project_no 补充：mysql一条sql语句查询多条统计结果 商城项目难免会遇到用户个人中心页查询不同状态订单数量的问题。当然这个问题并不难，可以写一个dao层方法，以状态作为...
