mysql - 试图获得准确的信息(CTE - 递归)

Question

我有不同的表，目标是为每个客户获取批准工作流程，并以这种方式显示该信息:

> 客户 |批准人1 |批准人2 |批准人3 |批准人4

首先，我有一个名为实体的表

(12, 'Math Andrew', 308, 'CHAIN1-MathAndrew')
(13, 'John Connor', 308, 'CHAIN2-JohnConnor')
(18, 'ZATCH', 309, null),
(19, 'MAX', 309, null),
(20, 'Ger',310, null),
(21, 'Mar',310, null),
(22, 'Maxwell',311, null),
(23, 'Ryan',312, null),
(24, 'Juy',313, null),
(25, 'Angel',314, null),
(26, 'John',315, null);

请注意:

12 was assigned to Math Andrew... 308 is the number that says that Matt Andrew is a CLIENT

13 was assigned to John Connor... 308 is the number that says that John Connor is a CLIENT

因为 Math Andrew 和 John Connor 是客户(也称为客户)，所以他们必须链接到一个或多个批准者

一个客户可能有 1 个批准人，或 2 个批准人，或 3 个批准人或 4 个批准人，实体表中存在不同的批准人。

当我说客户“可能有”1 个或多个批准者时，我的意思是这个

CLIENT - APPROVER4 (this is a 1-1 relationship) PS: A CLIENT WILL ALWAYS BE RELATED TO the APPROVER4 IN SOME WAY OR ANOTHER

CLIENT - APPROVER1 - APPROVER4 (in this case there Will be 2 relations.. ONE: CLIENT-APPROVER1 and another APPROVER1-APPROVER4)

CLIENT - APPROVER1 - APPROVER2 - APPROVER4 (in this case there Will be 3 relations.. ONE: CLIENT-APPROVER1, APPROVER1- APPROVER2 AND APPROVER2 - APPROVER4)

等等......(希望你明白了)

表类型实体

(308,'CLIENT'),
(309,'APPROVER1'),
(310,'APPROVER2'),
(311,'APPROVER3'),
(312,'J3 APPROVER4'),
(313,'J4 APPROVER4'),
(314,'J5 APPROVER4'),
(315, 'J6 APPROVER4'),
(316,'J7 APPROVER4');

表类型关系

(444,'J6 CLIENT-APPROVER4'),
(445,'J3 CLIENT-APPROVER4'),
(446,'J4 CLIENT-APPROVER4'),
(447,'J10 CLIENT-APPROVER4'),
(449,'J5 CLIENT-APPROVER4'),
(453,'J5 CLIENT-APPROVER4'),
(456,'J7 CLIENT-APPROVER4'),
(457,'J8 CLIENT-APPROVER4'),
(458,'CLIENT-APPROVER3'),
(459,'CLIENT-APPROVER1'),
(460,'APPROVER1-APPROVER2'),
(461,'APPROVER1-APPROVER3'),
(462,'J3 APPROVER1-APPROVER4'),
(463,'APPROVER2-APPROVER3'),
(464,'J3 APPROVER3-APPROVER4'),
(465,'J4 APPROVER3-APPROVER4'),
(466,'J5 APPROVER3-APPROVER4'),
(467,'J6 APPROVER3-APPROVER4'),
(468,'J7 APPROVER3-APPROVER4'),
(469,'J8 APPROVER3-APPROVER4'),
(470,'J10 APPROVER3-APPROVER4'),
(471,'CLIENT-APPROVER2');

---

关系类型:

客户 - 批准人 1 : (459,'客户批准人 1')

客户 - 批准人 2 : (471,'客户批准人 2')

客户 - APPROVER3 : (461,'APPROVER1-APPROVER3')

客户 - 批准人4:

(445,'J3 CLIENT-APPROVER4')

(446,'J4 CLIENT-APPROVER4')

(449,'J5 CLIENT-APPROVER4')

(444,'J6 CLIENT-APPROVER4')

(456,'J7 CLIENT-APPROVER4')

(457,'J8 CLIENT-APPROVER4')

(447,'J10 CLIENT-APPROVER4')

批准人 1 -批准人 2:

(460,'APPROVER1-APPROVER2')

批准人 2 - 批准人 3:

(463,'APPROVER2-APPROVER3')

批准人 3 - 批准人 4:

(464,'J3 APPROVER3-APPROVER4')

(465,'J4 APPROVER3-APPROVER4')

(466,'J5 APPROVER3-APPROVER4')

(467,'J6 APPROVER3-APPROVER4')

(468,'J7 APPROVER3-APPROVER4')

(469,'J8 APPROVER3-APPROVER4')

(470,'J10 APPROVER3-APPROVER4')

---

THIS IS IMPORTANT: when a client is linked to one approver, a NEW RELATION is created inside relationships table.

表关系:

(787,459,12,18)
(788,460,18,20)
(789,463,20,21)
(790,467,21,26)

787 IS THE NUMBER THAT WAS ASSIGNED WHEN THAT ROW WAS CREATED
459 REPRESENTS THE RELATION: CLIENT - APPROVER
CHAIN1-MathAndre is theclient
18 is the approver

按照思路:

APPROVER1 链接到 APPROVER2

(788,460,18,20)

APPROVER2 链接到 APPROVER3

(789,463,20,21)

APPROVER3 链接到 APPROVER4

(790,467,21,26)

所以，我想在屏幕上显示这个:

|CLIENT               | APPROVER1 | APPROVER2 | APPROVER3 | APPROVER4|
|CHAIN1-MathAndrew    |   ZATCH   |   Ger     |    Mar    |    John  |
|CHAIN2-JohnConnor    |    MAX    |           |    Mario  |    Steven|
|CHAIN3-MarioShapiro  |    IVAN   |           |           |    John  |

最后两行只是一个例子

---

这是我目前所拥有的(它正在工作):

LINK_sample_SQL

但它显示信息而不显示列名称(CLIENT、APPROVER1、APPROVER2、APPROVER3、APPROVER4)。这是显示这个:

CHAIN1-MathAndrew-ZATCH-Ger-Mar-John

我想这样显示数据:

|CLIENT               | APPROVER1 | APPROVER2 | APPROVER3 | APPROVER4|
|CHAIN1-MathAndrew    |   ZATCH   |   Ger     |    Mar    |    John  |
|CHAIN2-JohnConnor    |    MAX    |           |    Mario  |    Steven|
|CHAIN3-MarioShapiro  |    IVAN   |           |           |    John  |

我迷路了，你能帮帮我吗？

编辑:

The maximum amount of approvers is: 4

Answer 1

您应该根据需要使用条件聚合来格式化数据。尝试以下解决方案，我假设您有 MySQL ver.8 并且窗口函数可用:

WITH recursive relationships_CTE as (
  select e.id, e.description AS name, 1 col_id, 
    row_number() over (order by e.id) row_id
  from entities e
  where e.description like 'CHAIN%'
    UNION ALL
  select r.description_entitiy_2, e.name, col_id+ 1, row_id
  from relationships_CTE cte
  left join relationships r
    on r.description_entitiy_1 = cte.id
  join entities e 
    on r.description_entitiy_2 = e.id
)
select 
  max(case when col_id = 1 then name end) client,
  max(case when col_id = 2 then name end) approver1,
  max(case when col_id = 3 then name end) approver2,
  max(case when col_id = 4 then name end) approver3,
  max(case when col_id = 5 then name end) approver4
from relationships_CTE
group by row_id

DB-FIDDLE DEMO

该解决方案使用您的 SQL 查询并添加表格格式化所需的信息:(1) row_id 和 (2) col_id。然后将这些值用于条件聚合以创建表。