MySQL : Using MIN and GROUP BY

Question

Table have the following data Each same ID has the same datetime.

ID |     sendtime  |  gcm  | type
1, 2010-07-14 12:12:12, 1111111,  a
1, 2010-07-14 12:12:12, 2222222,  b
1, 2010-07-14 12:12:12, 3333333,  c
2, 2010-07-10 11:11:11, 4444444,  d
2, 2010-07-10 11:11:11, 5555555,  c
3, 2010-07-15 13:13:13, 6666666,  b
4, 2010-07-14 14:14:14, 7777777,  a

I want to get values following

ID |     sendtime  |  gcm  | type
2, 2010-07-10 11:11:11, 4444444,  d
2, 2010-07-10 11:11:11, 5555555,  c

Result means to get all rows having same ID and COUNT(*) > 1 and MIN(datetime).

I tried the following query.

SELECT *
FROM USER a INNER JOIN
     (SELECT ID, MIN(sendtime)
      FROM USER
      GROUP BY ID
      HAVING COUNT(*) > 1
     ) b
     ON a.ID = b.ID;

But result is

1, 2010-07-14 12:12:12, 1111111,  a
1, 2010-07-14 12:12:12, 2222222,  b
1, 2010-07-14 12:12:12, 3333333,  c
2, 2010-07-10 11:11:11, 4444444,  d
2, 2010-07-10 11:11:11, 5555555,  c

How to get right values?

Answer 1

Here is one way to express your logic:

select u.*
from user u
where u.sendtime = (select min(u2.sendtime) from user u2 where u2.id = u.id
                   ) and
      (select count(*)
       from user u2
       where u2.id = u.id and u2.sendtime = u.sendtime
      ) >= 2;

Answer 2

I solved this work.

SELECT a.* FROM USER a INNER JOIN (SELECT ID FROM USER GROUP BY ID HAVING COUNT(*) > 1 ORDER BY SENDTIME ASC LIMIT 1 ) b ON a.ID = b.ID;