Oracle SQL - 查询以计算来自多个表的值

Question

我试图通过加入多个表从 sql 查询中获得所需的结果。由于本人知识有限，望大家帮忙。

我正在尝试获取每个经理和员工的审核详细信息。我有 3 张桌子：

1. HR
2. REQUIRED_AUDITS
3. SCORE_ENTRY

以下是HR表的示例数据：

+----+---------------+------------------+
| id | manager_email |        VP        |
+----+---------------+------------------+
|  1 | john@com.com  | jake@com.com     |
|  2 | smith@com.com | kathleen@com.com |
|  3 | maria@com.com | james@com.com    |
|  4 | linda@com.com | david@com.com    |
|  5 | jess@com.com  | kim@com.com      |
+----+---------------+------------------+

以下是 REQUIRED_AUDITS 表的示例数据：

+----+---------------+----------------+-----------------+----------------+
| ID | manager_email | employee_email | audits_required | audit_eligible |
+----+---------------+----------------+-----------------+----------------+
|  1 | john@com.com  | brad@com.com   |               5 | Y              |
|  2 | linda@com.com | gloria@com.com |               2 | Y              |
|  3 | linda@com.com | susan@com.com  |               7 | Y              |
|  4 | john@com.com  | carmen@com.com |               5 | Y              |
|  5 | linda@com.com | aaron@com.com  |              25 | N              |
+----+---------------+----------------+-----------------+----------------+

以下是 SCORE_ENTRY 表的示例数据：

+----+---------------+----------------+-------+
| ID | manager_email | employee_email | Score |
+----+---------------+----------------+-------+
|  1 | linda@com.com | gloria@com.com | 85.04 |
|  2 | linda@com.com | susan@com.com  |   100 |
|  3 | john@com.com  | carmen@com.com | 80.50 |
+----+---------------+----------------+-------+

所以现在我想显示每位经理所需的审核次数、完成的审核次数以及完成的百分比。

+-------------------------------+----------------------------------------------------------------------------+--------------------------------------------+-----------------------+
| Manager(list from "HR" table) | Total Audits Required(Sum of audits required from "REQUIRED_AUDITS table") | Audits Performed(from "SCORE_ENTRY" table) | Percentage Completion |
+-------------------------------+----------------------------------------------------------------------------+--------------------------------------------+-----------------------+
| john@com.com                  |                                                                         10 |                                          1 | 10%                   |
| linda@com.com                 |                                                                          9 |                                          2 | 22.22%                |
| smith@com.com                 |                                                                          - |                                          - | -                     |
| maria@com.com                 |                                                                          - |                                          - | -                     |
| jess@com.com                  |                                                                          - |                                          - | -                     |
+-------------------------------+----------------------------------------------------------------------------+--------------------------------------------+-----------------------+

计算如下：
1. 计算审核次数，即从 REQUIRED_AUDITS 表：
考虑经理 linda@com.com:
所需审核总数：7+3，因为只有两名员工符合条件
2. 完成的审计总数：
audits_required（来自 REQUIRED AUDITS 表）- 已完成审计（来自 SCORE_ENTRY 表）

如前所述，我对 sql 的了解有限，这对我来说看起来真的很复杂，因此在 Stack Overflow 中伸出援手。

感谢任何帮助。

更新 为了进一步简化流程，我编写了不包括 HR 表的离散查询，并且在组合以下查询时需要帮助。

查询1：我称之为“DENOMINATOR”：

SELECT required_audits.manager_email, 
       SUM(audits_required) AS "TOTAL_AUDITS_REQUIRED" 
FROM   required_audits 
WHERE  Upper(required_audits.audit_eligible) = Upper('Y') 
GROUP  BY required_audits.manager_email 
ORDER  BY "total_audits_required" DESC 

+-------------------------------+-----------------------+
| required_audits.manager_email | TOTAL_AUDITS_REQUIRED |
+-------------------------------+-----------------------+
| john@com.com                  |                    10 |
| linda@com.com                 |                     9 |
+-------------------------------+-----------------------+

查询 2：我称之为 "NUMERATOR"：

SELECT score_entry.manager_email, 
       Count(id) 
FROM   score_entry 
GROUP  BY score_entry.manager_email 

+---------------------------+-----------+
| score_entry.manager_email | Count(id) |
+---------------------------+-----------+
| john@com.com              |         1 |
| linda@com.com             |         2 |
+---------------------------+-----------+

在最终或结果中，我需要的只是 NUMERATOR**/**DENOMINATOR * 100 的百分比。我对连接和在 where 子句中应用条件感到困惑。

+---------------+-----------------------+------------------+-----------------------+
|    Manager    | Total Audits Required | Audits Performed | Percentage Completion |
+---------------+-----------------------+------------------+-----------------------+
| john@com.com  |                    10 |                1 | 10%                   |
| linda@com.com |                     9 |                2 | 22.22%                |
+---------------+-----------------------+------------------+-----------------------+

在结果表中，仅应显示审核合格的数字。这是出错的另一个原因。

提前致谢。

谢谢，
里查

Answer 1

您可以使用多个公用表表达式，分别计算每个表达式并将它们连接在一起 - 只是为了让您了解发生了什么。

SQL Fiddle

查询：

WITH aud(manager_email,Total_audits) AS
  (SELECT manager_email,
    SUM (
    CASE
      WHEN audit_eligible = 'Y'
      THEN audits_required
    END )
  FROM REQUIRED_AUDITS
  GROUP BY manager_email
  ),  --Total_audits

  scores(manager_email,Audits_Performed) AS
  (SELECT manager_email,
    COUNT ( ID )
  FROM SCORE_ENTRY s
  GROUP BY manager_email
  )  --Audits_Performed

SELECT h.manager_email manager,
  a.Total_audits,
  s.Audits_Performed,
  100 * s.Audits_Performed / a.Total_audits percentage_complete
FROM HR h
LEFT OUTER JOIN aud a
ON h.manager_email = a.manager_email
LEFT OUTER JOIN scores s
ON h.manager_email = s.manager_email
ORDER BY 2 DESC NULLS LAST 

Results：

|       MANAGER | TOTAL_AUDITS | AUDITS_PERFORMED | PERCENTAGE_COMPLETE |
|---------------|--------------|------------------|---------------------|
|  john@com.com |           10 |                1 |                  10 |
| linda@com.com |            9 |                2 |   22.22222222222222 |
| smith@com.com |       (null) |           (null) |              (null) |
|  jess@com.com |       (null) |           (null) |              (null) |
| maria@com.com |       (null) |           (null) |              (null) |

Answer 2

鉴于已发布的数据，无需在此查询中使用HR 表，因为manager_email 也在required_audits 表上。在现实生活中，使用像 manager_email 这样的列作为外国将是一个非常糟糕的主意。它应该是hr.id 列。

尽管如此，我还是将它包含在查询中，因为您要求这样做。

算术相当简单。复杂之处在于，您需要使用left outer join 到 HR 到 REQUIRED_AUDITS，因为并非所有经理都有需要审核的员工，并且还需要加入 SCORE_ENTRY，因为并非所有员工都经过审核。

select hr.manager_email
       , sum(ra.audits_required) as tot_audits_required
       , count(se.score) as audits_performed
       , (count(se.score) / sum(ra.audits_required)) * 100 as pct_complete      
from hr
left outer join ( select * from  REQUIRED_AUDITS 
                  where audit_eligible = 'Y') ra 
    on ra.manager_email = hr.manager_email
left outer join SCORE_ENTRY se 
    on se.employee_email = ra.employee_email
group by hr.manager_email
order by hr.manager_email
/

Oracle LiveSQL 上的演示。

---

“在结果表上，只有审核合格的数字才会显示。”

要实现此版本的要求，只需将 REQUIRED_AUDITS 上的 left outer join 转换为 inner join。

Answer 3

假设从您上面的示例分数表中我们将查看得分行的数量而不是分数，这应该会有所帮助

select 
RA.manager_email, 
sum(RA.audits_requiered) as ReqQty,
(select count(manager_email) from SCORE_ENTRY as SE where SE.manager_email = RA.manager_email) as Compleated 

from REQUIRED_AUDITS as RA
group by manager_email
where audit_eligable='Y'

对于分数表，我不清楚这是否是您的意思。