我正在为似乎是我写过的最复杂的 SQL 查询而苦苦挣扎！
数据库结构：
frog_shared.staff
`ID`, `Firstname`, `Surname`
frog_observations.observations
`ID`, `Teacher_ID`, `Type`, `Main_Positive_Aspect`, `Main_Development_Aspect`, `Grade_for_Behaviour`, `Grade_for_Attainment`, `Grade_for_Teaching`
frog_observations.aspects
`ID`, `Observation_ID`, `Label_ID`, `Type`
frog_observations.aspect_labels
`ID`, `Title`
样本数据：
Member of staff:
`12345`, `Duncan`, `Wraight`
Observation:
`9888`, `12345`, `Formal`, `5`, `7`, `1`, `1`, `1`
Aspects:
`101`, `9888`, `2`, `P`
Aspect labels:
`2`, `Questioning`
我想要达到的目标：
我想列出一份我们在特定观察方面最好的老师的名单。例如，我想看看我的前 5 位老师的“提问”。
最重要的是，我想对数据进行一些过滤：
- 正式（
Type=Formal）和分享最佳实践（Type=SBP）观察的各个方面都应该被计算在内 - 仅当该观察的评分最多为 2 分（不高于，即不能有两个评分为 2 分，一个评分为 3 分）但共享最佳实践观察的评分均为 0 时，才应计算来自正式观察的方面
- 记录在观察表中的主要积极方面应在计数中加权
- 对于正式观察，主要积极方面的权重应为正常方面的 3*
- 为了分享最佳实践观察，主要积极方面的权重应为正常方面的 2*
我的尝试：
此声明按等级过滤观察结果，但不包括任何主要的积极方面（它们不包括在原始系统中）
SELECT
CONCAT(s.Firstname, " ", s.Surname) AS `Teacher`,
COUNT( * ) AS Total,
GREATEST(o.`Achievement_Grade`, o.`Behaviour_Grade`, o.`Teaching_Grade`) AS `Worst Grade`
FROM frog_observations.observations o
INNER JOIN frog_shared.staff s ON o.Teacher_ID = s.ID
INNER JOIN frog_observations.aspects a ON o.ID = a.Observation_ID
WHERE a.Aspect_ID = 4
AND ( GREATEST(o.`Achievement_Grade`, o.`Behaviour_Grade`, o.`Teaching_Grade`) BETWEEN 1 AND 2 ) AND o.Datetime > '2011-09-01'
AND a.Type = 'P'
GROUP BY s.ID
ORDER BY `Total` DESC
LIMIT 5
现在我想做的方式是这样的：
SELECT
CONCAT(s.Firstname, " ", s.Surname) AS `Teacher`,
COUNT(mp.*) AS `Number of Appraisal Main Positives`,
COUNT(sp.*) AS `Number of SBP Main Positives`,
COUNT(a.*) AS `Number of Other Positives`,
SUM(`Number of Appraisal Main Positives` + `Number of SBP Main Positives` `Number of Other Positives`) AS `Total`
FROM
`frog_observations`.`observations` o,
( `frog_observations`.`observations` WHERE `Type` = 'Formal') AS mp,
( `frog_observations`.`observations` WHERE `Type` = 'SBP') AS sp
INNER JOIN
`frog_observations`.`aspects` a ON a.Observation_ID = o.ID
GROUP BY s.ID
不幸的是，除了知道我想要的标题（即number of appraisal main positives、number of sharing best practice positives和number of standard positivesa weighted total）之外，我不知道如何编写我认为将是子查询的内容，以便从单个语句中获取所有这些信息。
任何指导表示赞赏。
编辑：示例输入/输出
输入
用户想要查看方面的前 5 名员工Questioning
流程
- 方面在表中Questioning有 ID 。- 然后，系统应为每位员工计算他们拥有的人数，并设置为。其中一些观察将是正式的，而一些将是SBP。- 系统还应计算表中每个员工所在的行数。4aspect_labelsobservationsMain_Positive_Aspect4TypeTypeaspectsAspect_ID4
输出
钥匙：
- FMP = Formal
observations表中的行数；; 和是（即在表中）TypeGREATEST(Achievement_Grade, Behaviour_Grade, Teaching_Grade) BETWEEN 1 AND 2Main_Positive_Aspect4Questioningaspects_label - SMP =
observations表中TypeSBP 和Main_Positive_Aspectis的行数4 - 其他=
aspects表中Aspect_ID每个4员工的行数（通过o.Observation_ID->o.Staff_ID链接） - 点数= 每项的加权总和 - 每个 FMP 值 3 分，每个 SMP 值 2 分，每个 Other 值 1。然后应
DESC按此列对查询进行排序。
示例输出：
-------------------------------------------------------------
| Staff Name FMP SMP Other Points |
|------------------------------------------------------------
| D Wraight 2 1 4 12 |
| A Nother 3 0 0 9 |
| J Bloggs 0 4 1 9 |
| J Arthur 1 1 1 6 |
| M Turner 0 1 0 2 |
-------------------------------------------------------------
以下是单独编写的查询，它们适用于我的数据库。基本上我需要将这些合并到一个查询中。如果可以的话！
#=======#
# FMP #
#=======#
SELECT
CONCAT(s.Firstname, " ", s.Surname) AS `Teacher`,
COUNT( * ) AS `FMP`
FROM frog_observations.observations o
INNER JOIN frog_shared.staff s ON o.Teacher_ID = s.ID
WHERE o.Main_Positive = 4
AND o.Type = 'F'
AND ( GREATEST(o.`Achievement_Grade`, o.`Behaviour_Grade`, o.`Teaching_Grade`) BETWEEN 1 AND 2 )
GROUP BY s.ID
#=======#
# SMP #
#=======#
SELECT
CONCAT(s.Firstname, " ", s.Surname) AS `Teacher`,
COUNT( * ) AS `SMP`
FROM frog_observations.observations o
INNER JOIN frog_shared.staff s ON o.Teacher_ID = s.ID
WHERE o.Main_Positive = 4
AND o.Type = 'S'
GROUP BY s.ID
#=========#
# Other #
#=========#
SELECT
CONCAT(s.Firstname, " ", s.Surname) AS `Teacher`,
COUNT( * ) AS `Other`
FROM observations o
INNER JOIN frog_shared.staff s ON o.Teacher_ID = s.ID
INNER JOIN aspects a ON o.ID = a.Observation_ID
WHERE a.Aspect_ID = 4
AND a.Type = 'P'
GROUP BY s.ID