2012-05-04 47 views
14

我有一个表,看起来像这样:SQL显示GROUP BY中的最新记录?

id | SubjectCode | Grade | DateApproved | StudentId 

1   SUB123   1.25   1/4/2012   2012-12345 

2   SUB123   2.00   1/5/2012   2012-12345 

3   SUB123   3.00   1/5/2012   2012-98765 

我想GROUP BY SubjectCode,但我想它显示最近DateApproved所以它看起来像:

id | SubjectCode | Grade | DateApproved | StudentId 

2   SUB123   2.00   1/5/2012   2012-12345 

3   SUB123   3.00   1/5/2012   2012-98765 

如何做到我有点失落?

编辑:

玉家伙现在即时通讯我真正的PC上,比较遗憾的是结构不佳的问题。

这里就是我真正想要做的事:

SELECT GD.GradebookDetailId, G.SubjectCode, G.Description, G.UnitsAcademic, G.UnitsNonAcademic, 
GD.Grade, GD.Remarks, G.FacultyName, STR_TO_DATE(G.DateApproved, '%m/%d/%Y %h:%i:%s') AS 'DateAproved' 
FROM gradebookdetail GD INNER JOIN gradebook G ON GD.GradebookId=G.GradebookId 
WHERE G.DateApproved IS NOT NULL AND G.GradebookType='final' AND StudentIdNumber='2012-12345' 

GROUP BY <?????> 
ORDER BY G.SubjectCode ASC 

基本上,我只是想显示“SubjectCode”的最近“DateApprove”,所以我没有得到多个条目。

+0

你真的想按StudentId。做到这一点的方法是选择max(DateApproved)。 (假设DateApproved是一个日期字段) –

+0

Ok,DateApproved是一个VARCHAR,这个工作:MAX(STR_TO_DATE(DateApproved,'%d%m%y'))? –

+0

嗯。我猜想值得一试。你必须尝试看看。我不太熟悉mysql语法。 –

回答

0
SELECT * 
FROM TheTable a 
WHERE NOT EXISTS(SELECT * 
       FROM TheTable b 
       WHERE b.StudentCode = a.StudentCode AND b.DateApproved > a.DateApproved) 
25

开始与此:

select StudentId, max(DateApproved) 
from tbl 
group by StudentId 

然后集成了主查询:

select * 
from tbl 
where (StudentId, DateApproved) in 

(
    select StudentId, max(DateApproved) 
    from tbl 
    group by StudentId 
) 

您也可以使用此:

select * 
from tbl 
join (select StudentId, max(DateApproved) as DateApproved 
     from tbl 
     group by StudentId) 
using (StudentId, DateApproved) 

但我更喜欢的元组测试,它的方式neater

现场测试:http://www.sqlfiddle.com/#!2/771b8/5

7
SELECT t2.* 
FROM temp t2 
INNER JOIN(
    SELECT MAX(DateApproved) as MaxDate, StudentId 
    FROM temp 
    GROUP BY StudentId 
    ) t1 ON t1.MaxDate = t2.DateApproved and t1.StudentId = t2.StudentId 
+0

H可能需要'GROUP BY StudentId,SubjectCode'(并适当地改变'ON'条件)。但这是要走的路。 –