我有以下数据表：
1.Table follow_up as：
mysql> select * from follow_up;
+--------------+----------------+--------------------------------------------------+-------------------+---------+---------------+-----------+---------------+----------+
| follow_up_id | feedback_close | feedback_open | is_email_required | is_Open | reminder_date | client_id | conclusion_id | stage_id |
+--------------+----------------+--------------------------------------------------+-------------------+---------+---------------+-----------+---------------+----------+
| 1 | NULL | dsffsdfsdfsd | 1 | 1 | 2017-09-20 | 101 | 96 | 72 |
| 2 | NULL | FSGDFHFGHFG | 1 | 1 | 2017-09-28 | 101 | 251 | 72 |
| 3 | NULL | Tender stage fb | 0 | 1 | NULL | 101 | 98 | 163 |
| 4 | NULL | Call back tender stage update date from 28 to 30 | 1 | 1 | 2017-09-28 | 101 | 96 | 163 |
| 5 | NULL | Metting follow up for next meeting | 1 | 1 | 2017-10-02 | 101 | 96 | 73 |
+--------------+----------------+--------------------------------------------------+-------------------+---------+---------------+-----------+---------------+----------+
2。表logs表示为：
mysql> SELECT * from logs where transaction = 'FLWUP';
+---------+---------+---------------------+---------+-------------+
| user_id | menu_id | logs_time | tran_id | transaction |
+---------+---------+---------------------+---------+-------------+
| 84 | 69 | 2017-09-19 19:31:04 | 1 | FLWUP |
| 84 | 69 | 2017-09-19 19:31:25 | 2 | FLWUP |
| 84 | 69 | 2017-09-20 19:10:41 | 2 | FLWUP |
| 84 | 69 | 2017-09-21 12:35:01 | 3 | FLWUP |
| 84 | 69 | 2017-09-21 12:35:26 | 4 | FLWUP |
| 84 | 69 | 2017-09-21 12:36:16 | 4 | FLWUP |
| 84 | 69 | 2017-09-21 12:38:30 | 5 | FLWUP |
+---------+---------+---------------------+---------+-------------+
7 rows in set (0.00 sec)
第3。表allcode表格为：
mysql> select * from allcode where code_type like 'MARK%';
+------------------+---------+------+----------------------+
| code_type | code_id | srno | code_name |
+------------------+---------+------+----------------------+
| MARKETING_STAGES | 72 | 1 | Enquiry |
| MARKETING_STAGES | 73 | 3 | Meeting |
| MARKETING_STAGES | 74 | 4 | Presentation |
| MARKETING_STAGES | 163 | 2 | Tender |
+------------------+---------+------+----------------------+
11 rows in set (0.00 sec)
我调用了一个查询并得到了结果：
mysql> select f.follow_up_id,f.feedback_open, f.feedback_close, f.reminder_date,
ast.code_name as stage, ac.code_name as conclusion, max(l.logs_time)
from follow_up f
join logs l on l.tran_id = f.follow_up_id
join allcode ast on ast.code_id = f.stage_id
join allcode ac on ac.code_id = f.conclusion_id
where l.transaction='FLWUP' and f.client_id = 101
group by ast.code_name order by ast.srno;
+--------------+------------------------------------+----------------+---------------+---------+------------+---------------------+
| follow_up_id | feedback_open | feedback_close | reminder_date | stage | conclusion | max(l.logs_time) |
+--------------+------------------------------------+----------------+---------------+---------+------------+---------------------+
| 1 | dsffsdfsdfsd | NULL | 2017-09-20 | Enquiry | Call Back | 2017-09-20 19:10:41 |
| 3 | Tender stage fb | NULL | NULL | Tender | Next | 2017-09-21 12:36:16 |
| 5 | Metting follow up for next meeting | NULL | 2017-10-02 | Meeting | Call Back | 2017-09-21 12:38:30 |
+--------------+------------------------------------+----------------+---------------+---------+------------+---------------------+
3 rows in set (0.00 sec)
但我希望结果为：
+--------------+-----------------------------------------------------+----------------+---------------+---------+------------+---------------------+
| follow_up_id | feedback_open | feedback_close | reminder_date | stage | conclusion | max(l.logs_time) |
+--------------+-----------------------------------------------------+----------------+---------------+---------+------------+---------------------+
| 2 | FSGDFHFGHFG | NULL | 2017-09-20 | Enquiry | Call Back | 2017-09-20 19:10:41 |
| 4 | Call back tender stage update date from 28 to 30 | NULL | NULL | Tender | Next | 2017-09-21 12:36:16 |
| 5 | Metting follow up for next meeting | NULL | 2017-10-02 | Meeting | Call Back | 2017-09-21 12:38:30 |
+--------------+-----------------------------------------------------+----------------+---------------+---------+------------+---------------------+
3 rows in set (0.00 sec)
我无法加入并分组以获得所需的结果。
表conclusion_id的列stage_id和follow_up指的是表code_id的{{1}}。
问题：
我想要的结果是
allcode，stage_id和srno
allcode表的最后/最近follow_up_id 答案 0 :(得分：1)
DEMO包括我的答案，根据需要提供完整组的原始问题，以及Reupal在演示中的答案。您在结果ID中缺少示例数据中的值，因此我只是根据ID创建它们（现在更新为ISO，回调但缺少98.）
我的结果与您的结果不符;但我相信你的预期结果是错误的。
当多个stage_ID存在时，好像你想要每个stage_ID的最大follow_up_ID
这可以通过派生表/内联视图来处理，获取由stage_ID分组的最大follow_UP_ID并将其连接回您的集合。将结果限制为仅包括stage_Id的最大follow_Up_ID。
我也不是mySQL的扩展组的粉丝，并且更喜欢包括未在组中的select中聚合的所有列。使用扩展组倾向于隐藏潜在问题。在这种情况下，仅通过ast.code_name进行分组允许引擎从其他列中选择非不同的值。您最终没有获得所需的结果，而且它隐藏了您在查询中获得多条记录的事实，如果不是通过使用/误用扩展组。
SELECT f.follow_up_id,f.feedback_open, f.feedback_close, f.reminder_date,
ast.code_name as stage, ac.code_name as conclusion, max(l.logs_time)
from follow_up f
join logs l on l.tran_id = f.follow_up_id
join allcode ast on ast.code_id = f.stage_id
join allcode ac on ac.code_id = f.conclusion_id
JOIN SELECT max(follow_up_ID) MFID, stage_ID
FROM follow_up
GROUP BY stage_ID) Z
on f.follow_up_ID = Z.MFID
and F.Stage_ID = Z.Stage_ID
WHERE l.transaction='FLWUP' and f.client_id = 101
GROUP BY f.follow_up_id,f.feedback_open, f.feedback_close, f.reminder_date,
ast.code_name , ac.code_name
ORDER BY ast.srno;
答案 1 :(得分：0)
请尝试以下操作，注意顺序和按顺序分组。
select f.follow_up_id,f.feedback_open, f.feedback_close, f.reminder_date,
ast.code_name as stage, ac.code_name as conclusion, max(l.logs_time)
from follow_up f
join logs l on l.tran_id = f.follow_up_id
join allcode ast on ast.code_id = f.stage_id
join allcode ac on ac.code_id = f.conclusion_id
where l.transaction='FLWUP' and f.client_id = 101
group by follow_up.stage_id order by ast.srno, follow_up.follow_up_id DESC;
这应该有效，如果不是，那么你应该搜索如何在多列上设置排序。
参考。 article- SQL multiple column ordering