我需要您的帮助,使用vb6和access db构建一个sql查询。下面是一个场景:2个表,给出并拥有Tb1字段Id、Name、Tb2 Id、Name、column,我需要有两个表中每个名称的总金额,这样就有了总计给列和总数有列,但我的查询不起作用。
Select tb1.id,tb1.name,sum(tb1.amount) as TG, tb2.id,tb2.name,sum(tb2.amount) as TH
from tb1 inner join
tb2
on tb1.id=tb2.id
group by... Etc如果我有10条记录,其中id =1在tb1上,3条记录在tb上,那么tb2上的总数是错误的(它重复了tb1上每条记录在tb2上的和)
我也尝试过使用Union在行中获得一个正确的结果,但是我应该想获得类似的结果
Id Name Have Give
1 John Doe 200,00 76,00我希望用照片来解释得更好



Triyng @Parfait建议,获得的结果与我之前编写的查询非常相似。
提前感谢您的帮助
发布于 2020-12-17 23:24:23
考虑通过id分别连接两个表的聚合
聚合查询(另存为存储访问查询)
SELECT tb1.idF
, tb1.[name]
, SUM(tb1.Give) AS TG
FROM tblGive tb1
GROUP BY tb1.idF
, tb1.[name] SELECT tb2.IDB
, tb2.[name]
, SUM(tb2.Have) AS TH
FROM tblHave tb2
GROUP BY tb2.IDB
, tb2.name最终查询(运行以返回两个表中的所有不同名称)
SELECT NZ(agg1.idF, agg2.idB) AS [id]
, NZ(agg1.name, agg2.name) AS [name]
, NZ(agg2.TH, 0) AS [Have]
, NZ(agg1.TG, 0) AS [Give]
FROM tblGiveAgg agg1
LEFT JOIN tblHaveAgg agg2
ON agg1.idF = agg2.idB
UNION
SELECT NZ(agg1.idF, agg2.idB) AS [id]
, NZ(agg1.name, agg2.name) AS [name]
, NZ(agg2.TH, 0) AS [Have]
, NZ(agg1.TG, 0) AS [Give]
FROM tblGiveAgg agg1
RIGHT JOIN tblHaveAgg agg2
ON agg1.idF = agg2.idB;用下面的数据演示
CREATE TABLE tblGive (
ID AUTOINCREMENT,
IdF INTEGER,
[Name] TEXT(10),
Give INTEGER
);
INSERT INTO tblGive (IdF, [Name], [Give]) VALUES (1, 'JOHN', 37);
INSERT INTO tblGive (IdF, [Name], [Give]) VALUES (2, 'ANNA', 10);
INSERT INTO tblGive (IdF, [Name], [Give]) VALUES (3, 'BILL', -37);
INSERT INTO tblGive (IdF, [Name], [Give]) VALUES (2, 'ANNA', 116);
INSERT INTO tblGive (IdF, [Name], [Give]) VALUES (1, 'JOHN', 120);
CREATE TABLE tblHave (
ID AUTOINCREMENT,
IDB INTEGER,
[Name] TEXT(10),
Have INTEGER
);
INSERT INTO tblHave (IDB, [Name], [Have]) VALUES (1, 'JOHN', 200);
INSERT INTO tblHave (IDB, [Name], [Have]) VALUES (2, 'ANNA', 400);
INSERT INTO tblHave (IDB, [Name], [Have]) VALUES (3, 'BILL', 150);
INSERT INTO tblHave (IDB, [Name], [Have]) VALUES (1, 'JOHN', 25);
INSERT INTO tblHave (IDB, [Name], [Have]) VALUES (1, 'JOHN', 70);

最后的完全连接查询返回以下结果:

发布于 2020-12-17 22:39:39
尝试使用union all,然后聚合:
Select id, name, sum(tg) as tg, sum(th) as th
from (select id, name, amount as tg, 0 as th from tb1
union all
select id, name, 0, amount from tbl2
) as t
group by id, name;我不确定在union all子句中是否所有版本的MS Access都支持from。如果没有,则需要将该片段封装在视图中。
https://stackoverflow.com/questions/65349339
复制相似问题