尝试获取准确的信息(递归CTE)

时间:2019-11-18 04:23:34

标签: mysql sql common-table-expression recursive-query

我有不同的表,目标是为每个客户获取批准工作流程,并以此方式显示该信息:

>客户| APPROVER1 | APPROVER2 | APPROVER3 | APPROVER4

首先,我有一个称为实体的表

(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被分配给数学安德鲁... 308是表示    Matt Andrew是客户

     

13被分配给John Connor ... 308是表示    John Connor是客户

由于Math Andrew和John Connor是客户(也称为客户),因此必须将他们链接到一个或多个批准者

一个客户可能有1个批准人,2个批准人,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个或多个批准时,我是说

  

客户端-批准人4(这是1-1的关系)PS:客户端将    总是以某种方式与批准人有关

     

客户-审批人1-审批人4(在这种情况下,将为2    关系。一个:CLIENT-APPROVER1和另一个APPROVER1-APPROVER4)

     

客户-审批人1-审批人2-审批人4(在这种情况下,    3关系。一个:CLIENT-APPROVER1,APPROVER1- APPROVER2和    APPROVER2-APPROVER4)

等等...(希望您能想到这个主意)

表type_entities

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

表type_relation

(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-审批人1:(459,“客户审批人1”)

客户-审批人2:(471,“客户审批2”)

客户端-APPROVER3:(461,“ APPROVER1-APPROVER3”)

客户端-批准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')


  

这很重要:当客户链接到一个批准人时,    RELATION是在关系表中创建的。

表关系:

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

787是排成一行时分配的数字
  459关系:客户-批准人
  CHAIN1-MathAndre是客户端
  18是批准人

遵循以下想法:

APPROVER1已链接到APPROVER2

(788,460,18,20)

APPROVER2已链接到APPROVER3

(789,463,20,21)

APPROVER3已链接到APPROVER4

(790,467,21,26)

因此,我想在屏幕上显示此内容:

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

最后2行只是一个示例


这是我到目前为止(正在运行)的内容:

LINK_sample_SQL

但它显示信息时未显示列名(CLIENT,APPROVER1,APPROVER2,APPROVER3,APPROVER4)。这显示为:

  

CHAIN1-MathAndrew-ZATCH-Ger-Mar-John

我想以这种方式显示数据:

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

我很迷路,请你帮我一下吗?

编辑:

  

批准人的最大数量为:4

1 个答案:

答案 0 :(得分:1)

您应该使用条件聚合来根据需要格式化数据。尝试以下解决方案,假设您拥有MySQL ver.8并且窗口功能可用:

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查询并添加用于表格式设置的必要信息:(1)row_id和(2)col_id。然后将这些值用于条件放大以创建表。

相关问题