如何取出time最早的记录tb id name time
1 001 2010-07-1
2 007 2010-07-2
3 002 2010-07-3
4 003 2010-07-6
5 004 2010-06-1
6 005 2010-07-7
7 006 2010-07-18output
--------------------------
id name time
5 004 2010-06-1
1 001 2010-07-1
2 007 2010-07-2
3 002 2010-07-3
4 003 2010-07-6
5 004 2010-06-1
6 005 2010-07-7
7 006 2010-07-18output
--------------------------
id name time
5 004 2010-06-1
TOP 1
ID,NAME,TIME
FROM TB
ORDER BY TIME
where time=(select min(time) from tb)
IF OBJECT_ID('TB') IS NOT NULL DROP TABLE TBCREATE TABLE TB(ID CHAR(1),NAME CHAR(3),TIME DATETIME)INSERT INTO TB
SELECT '1','001','2010-07-1' UNION ALL
SELECT '2','007','2010-07-2' UNION ALL
SELECT '3','002','2010-07-3' UNION ALL
SELECT '4','003','2010-07-6' UNION ALL
SELECT '5','004','2010-06-1' UNION ALL
SELECT '6','005','2010-07-7' UNION ALL
SELECT '7','006','2010-07-18'-----方法一
SELECT TOP 1 * FROM TB ORDER BY TIME ASC-----方法二
SELECT * FROM TB WHERE TIME=(SELECT MIN(TIME) FROM TB)DROP TABLE TBOUTPUT
----------------------------------
id name time
5 004 2010-06-01 00:00:00.000
这两条都没问题了,再加个select * from tb a where not exists (select 1 from tb b where b.time<a.time)
TOP 1
ID,NAME,TIME
FROM TB
ORDER BY TIME