Question

让我解释一下我需要什么，我有2个名为A和B的表.B是A的子表。

这是Schema：

------------------------
Table B:
itemId version qty  AId
44       1      1   200
44       1      2   201
44       2      2   200
------------------------
Table A:
id  tId
200 100
201 100
------------------------

这就是我需要的：我需要所有具有相同tId的最新版本数量的总和。

这是我的查询：

select sum(qty) as sum from B
left join A on A.id=B.AId
where itemId=44 and tId=100 and 
version=(select max(version) from B where itemId=44 and tId=100)

当一个项目获得版本2且版本1被忽略时，结果会出错。

感谢。

修改

我到底需要的是：

itemId version qty  AId
44       2      2   200
44       1      2   201

Sum（qty）的结果必须为4，因为它们具有相同的tId并且在每个AId中都有Max版本。

Answer 1

使用窗口功能。

select itemid, version, qty, aid
from (
  select *, max(version) over (partition by AId) as latestVersion 
  from B
) as B
where version = latestVersion

总结

select tId, SUM(qty) AS qty_sum
from (
  select *, max(version) over (partition by AId) as latestVersion 
  from B
) as B
join A on B.AId = A.id
where version = latestVersion
group by tId

Answer 2

SELECT B.*
FROM B
INNER JOIN
(SELECT Aid,MAX(version) AS version FROM B WHERE itemId=44 GROUP BY AId) AS B1
ON B.Aid=B1.Aid 
AND B.version=B1.version
INNER JOIN
(SELECT * FROM A WHERE tId=100) AS A
ON A.id=B.Aid
Order BY B.aid

数量之和

SELECT SUM(B.qty)
FROM B
INNER JOIN
(SELECT Aid,MAX(version) AS version FROM B WHERE itemId=44 GROUP BY AId) AS B1
ON B.Aid=B1.Aid AND B.version=B1.version
INNER JOIN
(SELECT * FROM A WHERE tId=100) AS A
ON A.id=B.Aid
GROUP BY A.tid

<强>输出

itemid  version qty aid
44      2       2   200
44      1       2   201

<强>演示

http://sqlfiddle.com/#!17/092dd/5

Answer 3

试试这个

DECLARE @TA Table (id int,tid int)
DECLARE @TB Table (itemid int, version int,qty int,AID int)

INSERT INTO @TA
SELECT 200, 100
UNION ALL
SELECT 201, 100


INSERT INTO @TB
SELECT 44,1,1,201
UNION ALL
SELECT 44,1,2,200
UNION ALL
SELECT 44,2,3,200
UNION ALL
SELECT 44,2,5,201


DECLARE @tid int
SET @tid = 100

SELECT XB.* FROM @Tb XB INNER JOIN
(SELECT Version,Max(AID) Aid FROM @TA A INNER JOIN @TB B ON A.id = B.AID AND tid = @tid Group By Version) X
ON X.version = XB.version and XB.AID = X.Aid

Answer 4

工作解决方案

select b.* from B as b
inner join 
 (select AID,itemId,Max(version) as mVersion from B 
  group by AID,itemID) d 
on b.AID = d.AID and b.itemID = d.itemID and b.Version = d.mVersion
inner join A as a
on B.AID = a.id
where b.itemID = 44 --apply if you need

结果

itemid  version qty aid
44         2    2   200
44         1    2   201

这将为您提供数量总和

的结果

select itemID,sum(qty) from (
select b.* from B as b
inner join 
 (select AID,itemId,Max(version) as mVersion from B 
  group by AID,itemID) d 
on b.AID = d.AID and b.itemID = d.itemID and b.Version = d.mVersion
inner join A as a
on B.AID = a.id
where b.itemID = 44 --apply if you need 
) e group by itemID

结果

itemid  sum
44       4

Answer 5

我认为此查询可以帮助您解决问题

    SELECT itemId, version, qty , AId FROM (
      SELECT itemId, version, qty , AId FROM b 
      LEFT JOIN a ON (b.aid = a.id)
    ) temp
    WHERE version = (SELECT MAX(version) FROM b WHERE b.aid = temp.aid)
   and temp.tid = 100 and temp.itemId = 44

Answer 6

Postgres中最有效的解决方案greatest-n-per-group问题通常是使用（专有）运算符distinct on ()

因此，要获取每个a.id的最新版本，您可以使用：

select distinct on (a.id) b.*
from a 
  join b on a.id = b.aid
order by a.id, b.version desc;

以上回报：

itemid | version | qty | aid
-------+---------+-----+----
    44 |       2 |   2 | 200
    44 |       1 |   2 | 201

然后你可以总结结果：

select sum(qty) 
from (
  select distinct on (a.id) b.qty
  from a 
    join b on a.id = b.aid
  order by a.id, b.version desc
) t;

请注意，通常，派生表中的order by是无用的，但在这种情况下需要它，因为否则distinct on ()将无法正常工作。

在线示例：http://rextester.com/DRHK19268

sql join table，没有提供完美的结果

6 个答案: