Question

我是新来的，有人可以解决我无法用子查询解决的问题，任何想法如何解决问题？

基本上我需要所有患者＆＃34; pa_name＆＃34;，每个＆＃34;字段的最新检查：pe_d2＆＃34;喜欢＆＃34;预期结果：＆＃34;

我试图勾勒出结果，可能有助于理解问题......

＆＃34; pacient_exams＆＃34;表有很多记录，查询需要非常快。

提前感谢您寻求可能的解决方案！ []

patient_exams                                    
+-------+----------+----------+------------+------------+  
| pe_id | pe_pa_id | pe_ex_id |    pe_d1   |    pe_d2   |  
+-------+----------+----------+------------+------------+  
|   1   |     1    |     1    | 2014-05-19 | 2016-05-19 |
|   2   |     1    |     2    | 2014-05-19 | 2015-05-19 |
|   3   |     1    |     3    | 2014-05-26 | 2014-11-26 |
|   4   |     1    |     3    | 2014-05-19 | 2014-11-19 |
|   5   |     1    |     4    | 2013-05-19 | 2013-11-19 |
|   6   |     1    |     4    | 2014-05-19 | 2014-11-19 |
|   7   |     3    |     1    | 2013-08-19 | 2014-08-19 |
|   8   |     3    |     1    | 2014-05-01 | 2017-05-01 |
|   9   |     4    |     2    | 2013-05-02 | 2014-05-02 |
|   10  |     4    |     2    | 2013-11-01 | 2014-05-01 |
|   11  |     4    |     4    | 2013-05-02 | 2014-05-02 |
|   12  |     4    |     4    | 2013-11-01 | 2014-05-01 |
+-------+----------+----------+------------+------------+  

patient              exams 
+-------+---------+  +-------+---------+    
| pa_id | pa_name |  | ex_id | ex_name |
+-------+---------+  +-------+---------+    
|   1   | John M. |  |   1   |  Exam 1 | 
|   2   | Slater  |  |   2   |  Exam 2 | 
|   3   | Jonny   |  |   3   |  Exam 3 | 
|   4   | Jessy   |  |   4   |  Exam 4 | 
|  ...  | ...     |  |  ...  |  ...    |
+-------+---------+  +-------+---------+

预期结果：

+-------+---------+---------+------------+------------+  
| pe_id | pa_name | ex_name |    pe_d1   |    pe_d2   |  
+-------+---------+---------+------------+------------+
|   9   | Jessy   | Exam 2  | 2013-05-02 | 2014-05-02 |
|   11  | Jessy   | Exam 4  | 2013-05-02 | 2014-05-02 |
|   1   | John M. | Exam 1  | 2014-05-19 | 2016-05-19 |
|   2   | John M. | Exam 2  | 2014-05-19 | 2015-05-19 |
|   3   | John M. | Exam 3  | 2014-05-26 | 2014-11-26 |
|   6   | John M. | Exam 4  | 2014-05-26 | 2014-11-26 |
|   8   | Jonny   | Exam 1  | 2014-05-01 | 2017-05-01 |
+-------+---------+---------+------------+------------+

Answer 1

您可以在表格中使用联接，对于最长考试日期，您需要使用子查询进行patient_exams的额外自我加入，以获得考试日期的最大值，即max(pe_d2)

select 
pe.pe_id,
p.pa_name ,
e.ex_name ,
pe.pe_d1   ,
pe.pe_d2
from exams e
join patient_exams pe on(e.ex_id = pe.pe_ex_id)
join patient p on(p.pa_id= pe.pe_pa_id)
join (select `pe_pa_id`, `pe_ex_id` ,max(pe_d2) pe_d2 
      from patient_exams
     group by `pe_pa_id`, `pe_ex_id`) pee
on (pe.`pe_pa_id`= pee.`pe_pa_id` and
    pe.`pe_ex_id` = pee.`pe_ex_id` and
    pe.pe_d2 = pee.pe_d2
   )
order by p.pa_name  ,pee.pe_d2 desc

Demo

Answer 2

您需要先从patient_exams表中获取最新记录，然后将所有3个表与过滤后的结果一起加入，如下所示：

SELECT pe_id, pa_name, ex_name, pe_d1, pe_d2
FROM patient_exams pe
JOIN patient p
ON pe.pe_pa_id = p.pa_id
JOIN exams e
ON pe.pe_ex_id = e.ex_id
JOIN (
  SELECT pe_pa_id, pe_ex_id, MAX(pe_d2) AS max_pe_d2
  FROM patient_exams
  GROUP BY pe_pa_id, pe_ex_id
) AS t
ON pe.pe_pa_id = t.pe_pa_id 
AND pe.pe_ex_id = t.pe_ex_id 
AND pe.pe_d2 = t.max_pe_d2
ORDER BY pa_name, ex_name

Demo/Solution

感谢大家，工作正常！

在MySQL中选择和分组子查询 - 返回最新记录

2 个答案:

Demo

Demo/Solution