我一直试图做下面的查询，但老实说它让我发疯了。
我在MS SQL CE 4.0上有2个表
表1名称：项目
ID | Item_Code | Logged_by | Description 1 | A | Pete | just an A 2 | B | Mary | Seams like a B 3 | C | Joe | Obviously this is a C 4 | D | Pete | This is another A
表2名称：Item_Comments
ID | Item_Code | Comment | Date 1 | B | Done | 2014/08/08 2 | A | Nice A | 2014/08/08 3 | B | Send 1 More | 2014/08/09 4 | C | Done | 2014/08/10 5 | D | This is an A | 2014/08/10 6 | D | Opps Sorry | 2014/08/11
想要的结果：我希望将 Item_Comments 中的最新评论加入项目表
ID | Item_Code | Logged_by | Description | Comment 1 | A | Pete | just an A | Nice A 2 | B | Mary | Seams like a B | Send 1 More 3 | C | Joe | Obviously this is a C | Done 4 | D | Pete | This is another A | Opps Sorry
我做了这个查询，但我得到了所有信息=（混合。
SELECT *
FROM Items t1
JOIN
(SELECT Item_Code, Comment, MAX(date) as MyDate
FROM Item_Comments
Group By Item_Code, Comment, Date
) t2
ON Item_Code= Item_Code
ORDER BY t1.Item_Code;
你知道怎么办？
答案 0 :(得分：0)
请注意，如果您运行此内部文件，则会获得每条记录：
SELECT Item_Code, Comment, MAX(date) as MyDate
FROM Item_Comments
Group By Item_Code, Comment, Date
您只想要最近的评论。假设这是SQL Server 2008或更早版本，这就是每个Item_Code的最新日期：
SELECT Item_Code, MAX(date) as MyDate
FROM Item_Comments
Group By Item_Code
现在你需要加入那个并查看该日期的评论：
SELECT C.*
FROM Item_Comments C
INNER JOIN
(SELECT Item_Code, MAX(date) as MyDate
FROM Item_Comments
Group By Item_Code
) t2
ON C.Item_Code= t2.Item_Code
AND C.date = t2.MyDate
现在您可以使用它连接回原始表：
SELECT t1.*, LatestComment.*
FROM Items t1
INNER JOIN
(
SELECT C.*
FROM Item_Comments C
INNER JOIN
(SELECT Item_Code, MAX(date) as MyDate
FROM Item_Comments
Group By Item_Code
) t2
ON C.Item_Code= t2.Item_Code
AND C.date = t2.MyDate
) LatestComment
On LatestComment.Item_Code = t1.Item_Code
根据您使用的实际数据库，这可以变得更加简单。这就是为什么你需要标记你的数据库和版本。
答案 1 :(得分：0)
尝试：
select x.*, z.comment
from items x
join (select item_code, max(date) as latest_dt
from item_comments
group by item_code) y
on x.item_code = y.item_code
join item_comments z
on y.item_code = z.item_code
and y.latest_dt = z.date
小提琴测试： http://sqlfiddle.com/#!6/d387f/8/0
你的查询很接近，但在你的内联视图中别名为t2，你按注释进行分组，让max函数实际上根本没有聚合。在t2中，您应该只选择item_code和max（日期）并仅按item_code分组，然后您可以使用它连接到item_comments（我上面的查询中的y和z）。
这是使用子查询执行此操作的第二种方法，但我会坚持上述（带内联视图的连接）：
select i.*, c.comment
from items i
join item_comments c
on i.item_code = c.item_code
where c.date = (select max(x.date)
from item_comments x
where x.item_code = c.item_code)
order by i.id
答案 2 :(得分：0)
试试这个，
create table items (id int, item_code char(1), logged_by varchar(10), description varchar(30));
insert into items values (1, 'A', 'Pete', 'just an A');
insert into items values (2, 'B', 'Mary', 'Seams like a B');
insert into items values (3, 'C', 'Joe', 'Obviously this is a C');
insert into items values (4, 'D', 'Pete', 'This is another A');
create table item_comments (id int, item_code char(1), comment varchar(20), date date);
insert into item_comments values (1, 'B', 'Done', '2014/08/08');
insert into item_comments values (2, 'A', 'Nice A', '2014/08/08');
insert into item_comments values (3, 'B', 'Send 1 More', '2014/08/09');
insert into item_comments values (4, 'C', 'Done', '2014/08/10');
insert into item_comments values (5, 'D', 'This is an A', '2014/08/10');
insert into item_comments values (6, 'D', 'Opps Sorry', '2014/08/11');
select * from items;
select * from item_comments;
select * from (select i.logged_by,i.id,i.item_code,i.description,ic.comment
,row_number() over(partition by i.id order by i.id )as Rnk
from items i inner join item_comments ic
on i.item_code=ic.item_code and i.id in(1,3)) x
where x.Rnk=1
union
select * from (select i.logged_by,i.id,i.item_code,i.description,ic.comment
,row_number() over(partition by i.id order by i.id )as Rnk
from items i inner join item_comments ic
on i.item_code=ic.item_code and i.id in(2,4)
) x where x.Rnk=2 order by item_code