Question

我有

users
------------------------
id | name | other_stuff.....

engagement
------------------------
user_id | type_code |

type_code是varchar，但是A，B，C或NULL

[为了清晰起见编辑：用户可以拥有每种类型代码的许多约定。所以我想要计算他们每人的数量。 ]

我想返回所有用户行，但是要计算A，B和C类型的约定。 E.g。

users_result
------------------------
user_id | user_name | other_stuff..... | count_A | count_B | count_C |

我已经做了很多搜索，但在其他解决方案中发现了以下问题：

“other_stuff ...”实际上是来自十几个其他联接的分组/连接结果，所以它已经是一个怪物了。所以我需要能够将其他字段添加到预先存在的“SELECT ...... FROM users ...”查询中。
所需的三个数据位均来自同一个参与表，每个参与表都有自己的条件。我没有找到任何允许我在同一张相关桌子上使用这三个条件的内容。

由于

[编辑]

我试图简化这个问题，所以人们不必查看大量不必要的东西，但似乎我可能没有提供足够的信息。这是原始查询的“大多数”。我已经取出了很多选中的字段，因为有负载，但我已经离开了大部分连接，所以你可以基本看到实际发生的事情。

SELECT 
  user.id, 
  user.first_name, 
  user.second_name, 
  GROUP_CONCAT(DISTINCT illness.id ORDER BY illness.id SEPARATOR ',' ) AS reason_for_treatment, 
  IF(ww_id=1000003, 1,'') as user_refused_program, 
  Group_CONCAT(DISTINCT physical_activity.name SEPARATOR ', ') AS programme_options, 
  COUNT(CASE WHEN engagement_item.type_code LIKE 'wm6%' THEN 1 ELSE NULL END) as count_A, 
  COUNT(CASE WHEN engagement_item.type_code LIKE 'wm12%' THEN 1 ELSE NULL END) as count_B, 
  COUNT(CASE WHEN engagement_item.type_code LIKE 'wm6%' THEN 1 ELSE NULL END) as count_C 
FROM `user` 
LEFT JOIN session AS session_induction ON (user.id = session_induction.user_id AND session_induction.session_type_id = 3) 
LEFT JOIN stats AS stats_induction ON session_induction.id = stats_induction.session_id 
LEFT JOIN session AS session_interim ON (user.id = session_interim.user_id AND session_interim.session_type_id = 4) 
LEFT JOIN stats AS stats_interim ON session_interim.id = stats_interim.session_id 
LEFT JOIN session AS session_final ON (user.id = session_final.user_id AND session_final.session_type_id = 5) 
LEFT JOIN stats AS stats_final ON session_final.id = stats_final.session_id 
LEFT JOIN user_has_illness ON user.ID = user_has_illness.user_id 
LEFT JOIN illness ON user_has_illness.illness_id = illness.id 
LEFT JOIN user_has_physical_activity ON user.ID = user_has_physical_activity.user_id 
LEFT JOIN physical_activity ON user_has_physical_activity.physical_activity_id = physical_activity.id 
LEFT JOIN engagement_item ON user.ID = engagement_item.user_ID 
WHERE (user.INDUCTION_DATE>='2010-06-09' AND user.INDUCTION_DATE<='2011-06-09' AND user.archive!='1' ) 
GROUP BY user.id, engagement_item.user_id

值得一提的是它工作正常 - 返回所有需要所有详细信息的用户。除了count_A B和C cols。

[编辑后面添加了稍微简化的查询]

删除不相关的连接并选择。

SELECT 
  user.id, 
  user.first_name, 
  COUNT(CASE WHEN engagement_item.type_code LIKE 'wm6%' THEN 1 ELSE NULL END) as count_A, 
  COUNT(CASE WHEN engagement_item.type_code LIKE 'wm12%' THEN 1 ELSE NULL END) as count_B, 
  COUNT(CASE WHEN engagement_item.type_code LIKE 'wm6%' THEN 1 ELSE NULL END) as count_C 
FROM `user` 
LEFT JOIN engagement_item ON user.ID = engagement_item.user_ID 
GROUP BY user.id, engagement_item.user_id

Answer 1

或许这样的事情？

select e.user_id, u.name,
       sum(case e.type_code when 'A' then 1 else 0 end) as count_A,
       sum(case e.type_code when 'B' then 1 else 0 end) as count_B,
       sum(case e.type_code when 'C' then 1 else 0 end) as count_C
from engagement e join users u on (e.user_id = u.id)
group by e.user_id, u.name

有趣的是在CASE中使用SUM将计数分成三个块。

Answer 2

SELECT e.user_id, u.name,
       COUNT(CASE type_code WHEN 'A' THEN 1 ELSE NULL END) as count_A,
       COUNT(CASE type_code WHEN 'B' THEN 1 ELSE NULL END) as count_B,
       COUNT(CASE type_code WHEN 'C' THEN 1 ELSE NULL END) as count_C
FROM engagement e join users u on (e.user_id = u.id)
GROUP BY e.user_id, u.name

我会使用COUNT代替SUM，因为这就是它的用途，在非NULL时计算事物。

SELECT 
  user.id, 
  user.first_name, 
  user.second_name, 
  GROUP_CONCAT(DISTINCT illness.id ORDER BY illness.id SEPARATOR ',' ) AS reason_for_treatment, 
  IF(ww_id=1000003, 1,'') as user_refused_program, 
  Group_CONCAT(DISTINCT physical_activity.name SEPARATOR ', ') AS programme_options,
  ei.count_A, ei.count_B, ei.count_C
FROM `user` 
LEFT JOIN ( SELECT user_id
    , COUNT(CASE WHEN engagement_item.type_code LIKE 'wm6%' THEN 1 ELSE NULL END) as count_A
    , COUNT(CASE WHEN engagement_item.type_code LIKE 'wm12%' THEN 1 ELSE NULL END) as count_B 
    , COUNT(CASE WHEN engagement_item.type_code LIKE 'wm6%' THEN 1 ELSE NULL END) as count_C 
    FROM engagement_item
    GROUP BY userid ) ei
LEFT JOIN session AS session_induction ON (user.id = session_induction.user_id AND session_induction.session_type_id = 3) 
LEFT JOIN stats AS stats_induction ON session_induction.id = stats_induction.session_id 
LEFT JOIN session AS session_interim ON (user.id = session_interim.user_id AND session_interim.session_type_id = 4) 
LEFT JOIN stats AS stats_interim ON session_interim.id = stats_interim.session_id 
LEFT JOIN session AS session_final ON (user.id = session_final.user_id AND session_final.session_type_id = 5) 
LEFT JOIN stats AS stats_final ON session_final.id = stats_final.session_id 
LEFT JOIN user_has_illness ON user.ID = user_has_illness.user_id 
LEFT JOIN illness ON user_has_illness.illness_id = illness.id 
LEFT JOIN user_has_physical_activity ON user.ID = user_has_physical_activity.user_id 
LEFT JOIN physical_activity ON user_has_physical_activity.physical_activity_id = physical_activity.id 
LEFT JOIN engagement_item ON user.ID = engagement_item.user_ID 
WHERE (user.INDUCTION_DATE>='2010-06-09' AND user.INDUCTION_DATE<='2011-06-09' AND user.archive!='1' ) 
GROUP BY user.id, engagement_item.user_id, ei.count_A, ei.count_B, ei.count_C

相关表的Mysql COUNT结果行

[编辑后面添加了稍微简化的查询]

2 个答案: