Question

我有一张桌子

date      doctor   patient
20-1-18   xx        11
20-1-18   xx        22
20-1-18   yy        33
21-1-18   zz        44
21-1-18   xx        55
21-1-18   yy        66
22-1-18   zz        77

我想展示每天看过不止一次患者的医生。连同患者姓名

预期产量

date      doctor patient
20-1-18   xx        11
20-1-18   xx        22

但是我只得到这个

date      doctor patient
20-1-18   xx        11

如何使其同时显示两个值

Answer 1

如果使用的是SQL Server，这可能是解决方案：

select [date], doctor, patient from (
    select [date], doctor, pateint,
           count(*) over (partition by [date], doctor) cnt
    from my_table
) a where cnt > 1

Answer 2

您需要一个带有having子句的子查询：

select *
from tab
where (date, doctor) in
(
  select date, doctor
    from tab
   group by date, doctor
  having count(1)>1
);
+-----------+-------+---------+
|   date    | doctor| patient |
+-----------+-------+---------+
| 2018-01-20|   xx  |    11   |
+-----------+-------+---------+
| 2018-01-20|   xx  |    22   |
+-----------+-------+---------+

SQL Fiddle Demo

Answer 3

如果您的dbms不支持窗口功能，则可以编写一个子查询以通过COUNT然后date,doctor来获得self join

select t2.*
from 
(
SELECT date,doctor,count(*) cn   
FROM T
GROUP BY date,doctor 
) t1 INNER JOIN T t2 on t1.date = t2.date and t1.doctor = t2.doctor
WHERE cn > 1

sqlifddle

[结果] ：

|    date | doctor | patient |
|---------|--------|---------|
| 20-1-18 |     xx |      11 |
| 20-1-18 |     xx |      22 |

如何显示与计数相同次数的记录

3 个答案: