试图获得准确的信息（CTE - 递归）
Trying to obtain the accurate information (CTE - recursive)
我有不同的 table，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 table 称为实体
(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，实体内部存在不同的批准人 table。
当我说客户 "could have" 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...（希望您明白）
table 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');
table 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 - 批准人 1 : (459,'CLIENT-APPROVER1')
客户 - 批准人 2 : (471,'CLIENT-APPROVER2')
客户 - 批准人 3 : (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.
Table 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（正在运行）：
但它显示信息而不显示列名（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 |
我很迷茫，你能帮帮我吗？
编辑：
The maximum amount of approvers is: 4
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 window 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
该解决方案使用您的 SQL 查询并为 table 格式添加必要的信息：(1) row_id，以及 (2) col_id。然后将这些值用于条件聚合以创建 table.
我有不同的 table，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 table 称为实体
(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，实体内部存在不同的批准人 table。
当我说客户 "could have" 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...（希望您明白）
table 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');
table 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 - 批准人 1 : (459,'CLIENT-APPROVER1')
客户 - 批准人 2 : (471,'CLIENT-APPROVER2')
客户 - 批准人 3 : (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.
Table 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（正在运行）：
但它显示信息而不显示列名（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 |
我很迷茫，你能帮帮我吗？
编辑：
The maximum amount of approvers is: 4
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 window 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
该解决方案使用您的 SQL 查询并为 table 格式添加必要的信息：(1) row_id，以及 (2) col_id。然后将这些值用于条件聚合以创建 table.