Question

我有以下mysql tabels

tbl_users

user_id     email               db_add_date
1           steve@gmail.com     2014-04-22 19:32:35
2           mark@gmail.com      2014-05-02 10:12:01
3           allan@yahoo.com     2014-03-15 05:14:56
4           jim_ross@gmail.com  2014-02-12 14:10:08

tbl_points

id  user_id     points
1   1           10  
2   3           1   
3   2           15
4   1           16
5   4           46
6   3           24

我想要获取这些电子邮件＆amp;单点

Expected Result

email               userid      points
mark@gmail.com      2           15
jim_ross@gmail.com  4           46

请帮忙......

更新说明

我想获取点数表中只有1个条目的电子邮件记录

Answer 1

SQL Fiddle

MySQL 5.5.32架构设置：

create table tbl_users
(
  user_id int,
  email varchar(20),
  db_add_date datetime
);

insert into tbl_users values
(1,           'steve@gmail.com   ',  '2014-04-22 19:32:35'),
(2,           'mark@gmail.com    ',  '2014-05-02 10:12:01'),
(3,           'allan@yahoo.com   ',  '2014-03-15 05:14:56'),
(4,           'jim_ross@gmail.com',  '2014-02-12 14:10:08');

create table tbl_points
(
  id int,
  user_id int,
  points int
);

insert into tbl_points values
(1,   1,           10),  
(2,   3,           1 ),  
(3,   2,           15),
(4,   1,           16),
(5,   4,           46),
(6,   3,           24);

查询1 ：

select u.email,
       u.user_id,
       p.points
from tbl_users as u
  inner join (
             select user_id, 
                    sum(points) as points
             from tbl_points
             group by user_id
             having count(*) = 1
             ) as p
    on u.user_id = p.user_id

<强> Results ：

|              EMAIL | USER_ID | POINTS |
|--------------------|---------|--------|
| mark@gmail.com     |       2 |     15 |
| jim_ross@gmail.com |       4 |     46 |

Answer 2

<强> your Expected Result...!!! （简单和排序查询）

SELECT u.email,u.user_id as userid,p.points 
FROM tbl_users  u 
LEFT JOIN tbl_points p 
ON u.user_id = p.user_id 
GROUP BY u.user_id 
HAVING COUNT(u.user_id) = 1

<强>结果： -

>     email               userid      points
>     mark@gmail.com      2           15
>     jim_ross@gmail.com  4           46

查询仅获取单点记录

2 个答案: