单行异列Oracle树状结构层次数据展示
Oracle Tree Structure Hierarchy Data Display in Single Row Different Column
我在 oracle 中有一个名为 Employee_Hierarchy 的层次结构 table，列名为 entity_code , parent_entity_code, entity_name 和 entity_role 没有循环。
在其他 table 名为 Client 的最底层 child 与层级 table 的最底层 child 和 entity_code 相连.
我必须在单行层次结构中显示数据，其中列名将按角色附加。
示例：
树结构：
分层Table:
最低 Child 的 Table:
预期结果：
有没有办法通过oracle查询得到预期的结果？
预期的结果取决于输入，这意味着它不会总是从根元素开始，它可能从任何节点开始，例如 team-lead (Shail) 到最低的 child。
(注意:如果缺少上层那么当前节点的parent_code将是其上层的parent_code以及缺失的层级元素在预期结果中将是空白。）
提前致谢。
这是执行此操作的一种方法，使用联接。另一种方法是合并两个表并进行分层查询——但这实际上是一回事（分层查询只不过是递归自连接）。
with
hierarchical_table ( entity_code, entity_name, entity_role, parent_entity_code ) as (
select 100, 'Mack' , 'Manager', cast (null as number) from dual union all
select 200, 'Shail', 'Team-Lead', 100 from dual union all
select 300, 'Jack' , 'Team-Lead', 100 from dual union all
select 400, 'Teju' , 'Developer', 200 from dual union all
select 500, 'Neha' , 'Developer', 200 from dual union all
select 600, 'Rocky', 'Developer', 300 from dual
),
client_table ( entity_code, client_name, address ) as (
select 600, 'Voda' , 'Pune' from dual union all
select 600, 'Rel' , 'Mumbai' from dual union all
select 600, 'Airtel', 'Pune' from dual union all
select 500, 'Tata' , 'Mumbai' from dual
)
-- end of test data (not part of the solution)
-- SQL query begins BELOW THIS LINE; use your actual table names
select h1.entity_code as manager_code, h1.entity_name as manager_name,
h2.entity_code as teamlead_code, h2.entity_name as teamlead_name,
h3.entity_code as developer_code, h3.entity_name as developer_name,
c.client_name
from hierarchical_table h1 left join hierarchical_table h2
on h2.parent_entity_code = h1.entity_code
left join hierarchical_table h3
on h3.parent_entity_code = h2.entity_code
left join client_table c
on c.entity_code = h3.entity_code
where h1.parent_entity_code is null
order by manager_code, teamlead_code, developer_code, client_name
;
输出:
MANAGER_CODE MANAGER_NAME TEAMLEAD_CODE TEAMLEAD_NAME DEVELOPER_CODE DEVELOPER_NAME CLIENT
------------ ------------ ------------- ------------- -------------- -------------- ------
100 Mack 200 Shail 400 Teju
100 Mack 200 Shail 500 Neha Tata
100 Mack 300 Jack 600 Rocky Airtel
100 Mack 300 Jack 600 Rocky Rel
100 Mack 300 Jack 600 Rocky Voda
5 rows selected.
select h.Manager_entity_code
,h.Manager_entity_name
,h.Team_Lead_entity_code
,h.Team_Lead_entity_name
,h.Developer_entity_code
,h.Developer_entity_name
,c.client_name
from (select trim (both ',' from sys_connect_by_path (case when entity_role = 'Manager' then entity_code end,',')) as Manager_entity_code
,trim (both ',' from sys_connect_by_path (case when entity_role = 'Manager' then entity_name end,',')) as Manager_entity_name
,trim (both ',' from sys_connect_by_path (case when entity_role = 'Team-Lead' then entity_code end,',')) as Team_Lead_entity_code
,trim (both ',' from sys_connect_by_path (case when entity_role = 'Team-Lead' then entity_name end,',')) as Team_Lead_entity_name
,trim (both ',' from sys_connect_by_path (case when entity_role = 'Developer' then entity_code end,',')) as Developer_entity_code
,trim (both ',' from sys_connect_by_path (case when entity_role = 'Developer' then entity_name end,',')) as Developer_entity_name
from hierarchical_table
where connect_by_isleaf = 1
connect by parent_entity_code = prior entity_code
start with entity_code = 100
) h
join client_table c
on c.entity_code =
h.Developer_entity_code
order by h.Manager_entity_code
,h.Team_Lead_entity_code
,h.Developer_entity_code
;
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| MANAGER_ENTITY_CODE | MANAGER_ENTITY_NAME | TEAM_LEAD_ENTITY_CODE | TEAM_LEAD_ENTITY_NAME | DEVELOPER_ENTITY_CODE | DEVELOPER_ENTITY_NAME | CLIENT_NAME |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| 100 | Mack | 200 | Shail | 500 | Neha | Tata |
| 100 | Mack | 300 | Jack | 600 | Rocky | Rel |
| 100 | Mack | 300 | Jack | 600 | Rocky | Voda |
| 100 | Mack | 300 | Jack | 600 | Rocky | Airtel |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
对于start with entity_code = 300
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| MANAGER_ENTITY_CODE | MANAGER_ENTITY_NAME | TEAM_LEAD_ENTITY_CODE | TEAM_LEAD_ENTITY_NAME | DEVELOPER_ENTITY_CODE | DEVELOPER_ENTITY_NAME | CLIENT_NAME |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| (null) | (null) | 300 | Jack | 600 | Rocky | Airtel |
| (null) | (null) | 300 | Jack | 600 | Rocky | Voda |
| (null) | (null) | 300 | Jack | 600 | Rocky | Rel |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
select regexp_substr (h.entity,'Manager~([^~]*)~' ,1,1,'',1) as Manager_entity_code
,regexp_substr (h.entity,'Manager~([^~]*)~([^,]*)' ,1,1,'',2) as Manager_entity_name
,regexp_substr (h.entity,'Team-Lead~([^~]*)~' ,1,1,'',1) as Team_Lead_entity_code
,regexp_substr (h.entity,'Team-Lead~([^~]*)~([^,]*)',1,1,'',2) as Team_Lead_entity_name
,regexp_substr (h.entity,'Developer~([^~]*)~' ,1,1,'',1) as Developer_entity_code
,regexp_substr (h.entity,'Developer~([^~]*)~([^,]*)',1,1,'',2) as Developer_entity_name
,c.client_name
from (select sys_connect_by_path (entity_role || '~' || entity_code || '~' || entity_name,',') as entity
,entity_code
from hierarchical_table
where connect_by_isleaf = 1
connect by parent_entity_code = prior entity_code
start with entity_code = 200
) h
join client_table c
on c.entity_code =
h.entity_code
order by Manager_entity_code
,Team_Lead_entity_code
,Developer_entity_code
;
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| MANAGER_ENTITY_CODE | MANAGER_ENTITY_NAME | TEAM_LEAD_ENTITY_CODE | TEAM_LEAD_ENTITY_NAME | DEVELOPER_ENTITY_CODE | DEVELOPER_ENTITY_NAME | CLIENT_NAME |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| 100 | Mack | 200 | Shail | 500 | Neha | Tata |
| 100 | Mack | 300 | Jack | 600 | Rocky | Rel |
| 100 | Mack | 300 | Jack | 600 | Rocky | Voda |
| 100 | Mack | 300 | Jack | 600 | Rocky | Airtel |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
对于start with entity_code = 600
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| MANAGER_ENTITY_CODE | MANAGER_ENTITY_NAME | TEAM_LEAD_ENTITY_CODE | TEAM_LEAD_ENTITY_NAME | DEVELOPER_ENTITY_CODE | DEVELOPER_ENTITY_NAME | CLIENT_NAME |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| (null) | (null) | (null) | (null) | 600 | Rocky | Airtel |
| (null) | (null) | (null) | (null) | 600 | Rocky | Voda |
| (null) | (null) | (null) | (null) | 600 | Rocky | Rel |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
在您的查询结果中，您希望某些角色在层次结构中具有特定的位置。经理是第一层，团队领导是第二层，开发人员是第三层。所以你可以把你的分层 table 当作没有。这使得查询非常可读：
with manager as (select * from employee_hierarchy where entity_role = 'Manager')
, teamleader as (select * from employee_hierarchy where entity_role = 'Team-Lead')
, developer as (select * from employee_hierarchy where entity_role = 'Developer')
select
m.entity_code as manager_entity_code,
m.entity_name as manager_entity_name,
t.entity_code as team_lead_entity_code,
t.entity_name as team_lead__entity_name,
d.entity_code as developer_entity_code,
d.entity_name as developer_entity_name,
c.client_name
from manager m
join teamleader t on t.parent_entity_code = m.entity_code
join developer d on d.parent_entity_code = t.entity_code
left join client_table c on c.entity_code = d.entity_code;
如果你想将结果限制在团队负责人 Shail 的部门，只需添加相应的 WHERE 子句：
where t.entity_name = 'Shail'
我在 oracle 中有一个名为 Employee_Hierarchy 的层次结构 table，列名为 entity_code , parent_entity_code, entity_name 和 entity_role 没有循环。 在其他 table 名为 Client 的最底层 child 与层级 table 的最底层 child 和 entity_code 相连. 我必须在单行层次结构中显示数据，其中列名将按角色附加。
示例：
树结构：
分层Table:
最低 Child 的 Table:
预期结果：
有没有办法通过oracle查询得到预期的结果？ 预期的结果取决于输入，这意味着它不会总是从根元素开始，它可能从任何节点开始，例如 team-lead (Shail) 到最低的 child。
(注意:如果缺少上层那么当前节点的parent_code将是其上层的parent_code以及缺失的层级元素在预期结果中将是空白。）
提前致谢。
这是执行此操作的一种方法，使用联接。另一种方法是合并两个表并进行分层查询——但这实际上是一回事（分层查询只不过是递归自连接）。
with
hierarchical_table ( entity_code, entity_name, entity_role, parent_entity_code ) as (
select 100, 'Mack' , 'Manager', cast (null as number) from dual union all
select 200, 'Shail', 'Team-Lead', 100 from dual union all
select 300, 'Jack' , 'Team-Lead', 100 from dual union all
select 400, 'Teju' , 'Developer', 200 from dual union all
select 500, 'Neha' , 'Developer', 200 from dual union all
select 600, 'Rocky', 'Developer', 300 from dual
),
client_table ( entity_code, client_name, address ) as (
select 600, 'Voda' , 'Pune' from dual union all
select 600, 'Rel' , 'Mumbai' from dual union all
select 600, 'Airtel', 'Pune' from dual union all
select 500, 'Tata' , 'Mumbai' from dual
)
-- end of test data (not part of the solution)
-- SQL query begins BELOW THIS LINE; use your actual table names
select h1.entity_code as manager_code, h1.entity_name as manager_name,
h2.entity_code as teamlead_code, h2.entity_name as teamlead_name,
h3.entity_code as developer_code, h3.entity_name as developer_name,
c.client_name
from hierarchical_table h1 left join hierarchical_table h2
on h2.parent_entity_code = h1.entity_code
left join hierarchical_table h3
on h3.parent_entity_code = h2.entity_code
left join client_table c
on c.entity_code = h3.entity_code
where h1.parent_entity_code is null
order by manager_code, teamlead_code, developer_code, client_name
;
输出:
MANAGER_CODE MANAGER_NAME TEAMLEAD_CODE TEAMLEAD_NAME DEVELOPER_CODE DEVELOPER_NAME CLIENT
------------ ------------ ------------- ------------- -------------- -------------- ------
100 Mack 200 Shail 400 Teju
100 Mack 200 Shail 500 Neha Tata
100 Mack 300 Jack 600 Rocky Airtel
100 Mack 300 Jack 600 Rocky Rel
100 Mack 300 Jack 600 Rocky Voda
5 rows selected.
select h.Manager_entity_code
,h.Manager_entity_name
,h.Team_Lead_entity_code
,h.Team_Lead_entity_name
,h.Developer_entity_code
,h.Developer_entity_name
,c.client_name
from (select trim (both ',' from sys_connect_by_path (case when entity_role = 'Manager' then entity_code end,',')) as Manager_entity_code
,trim (both ',' from sys_connect_by_path (case when entity_role = 'Manager' then entity_name end,',')) as Manager_entity_name
,trim (both ',' from sys_connect_by_path (case when entity_role = 'Team-Lead' then entity_code end,',')) as Team_Lead_entity_code
,trim (both ',' from sys_connect_by_path (case when entity_role = 'Team-Lead' then entity_name end,',')) as Team_Lead_entity_name
,trim (both ',' from sys_connect_by_path (case when entity_role = 'Developer' then entity_code end,',')) as Developer_entity_code
,trim (both ',' from sys_connect_by_path (case when entity_role = 'Developer' then entity_name end,',')) as Developer_entity_name
from hierarchical_table
where connect_by_isleaf = 1
connect by parent_entity_code = prior entity_code
start with entity_code = 100
) h
join client_table c
on c.entity_code =
h.Developer_entity_code
order by h.Manager_entity_code
,h.Team_Lead_entity_code
,h.Developer_entity_code
;
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| MANAGER_ENTITY_CODE | MANAGER_ENTITY_NAME | TEAM_LEAD_ENTITY_CODE | TEAM_LEAD_ENTITY_NAME | DEVELOPER_ENTITY_CODE | DEVELOPER_ENTITY_NAME | CLIENT_NAME |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| 100 | Mack | 200 | Shail | 500 | Neha | Tata |
| 100 | Mack | 300 | Jack | 600 | Rocky | Rel |
| 100 | Mack | 300 | Jack | 600 | Rocky | Voda |
| 100 | Mack | 300 | Jack | 600 | Rocky | Airtel |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
对于start with entity_code = 300
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| MANAGER_ENTITY_CODE | MANAGER_ENTITY_NAME | TEAM_LEAD_ENTITY_CODE | TEAM_LEAD_ENTITY_NAME | DEVELOPER_ENTITY_CODE | DEVELOPER_ENTITY_NAME | CLIENT_NAME |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| (null) | (null) | 300 | Jack | 600 | Rocky | Airtel |
| (null) | (null) | 300 | Jack | 600 | Rocky | Voda |
| (null) | (null) | 300 | Jack | 600 | Rocky | Rel |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
select regexp_substr (h.entity,'Manager~([^~]*)~' ,1,1,'',1) as Manager_entity_code
,regexp_substr (h.entity,'Manager~([^~]*)~([^,]*)' ,1,1,'',2) as Manager_entity_name
,regexp_substr (h.entity,'Team-Lead~([^~]*)~' ,1,1,'',1) as Team_Lead_entity_code
,regexp_substr (h.entity,'Team-Lead~([^~]*)~([^,]*)',1,1,'',2) as Team_Lead_entity_name
,regexp_substr (h.entity,'Developer~([^~]*)~' ,1,1,'',1) as Developer_entity_code
,regexp_substr (h.entity,'Developer~([^~]*)~([^,]*)',1,1,'',2) as Developer_entity_name
,c.client_name
from (select sys_connect_by_path (entity_role || '~' || entity_code || '~' || entity_name,',') as entity
,entity_code
from hierarchical_table
where connect_by_isleaf = 1
connect by parent_entity_code = prior entity_code
start with entity_code = 200
) h
join client_table c
on c.entity_code =
h.entity_code
order by Manager_entity_code
,Team_Lead_entity_code
,Developer_entity_code
;
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| MANAGER_ENTITY_CODE | MANAGER_ENTITY_NAME | TEAM_LEAD_ENTITY_CODE | TEAM_LEAD_ENTITY_NAME | DEVELOPER_ENTITY_CODE | DEVELOPER_ENTITY_NAME | CLIENT_NAME |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| 100 | Mack | 200 | Shail | 500 | Neha | Tata |
| 100 | Mack | 300 | Jack | 600 | Rocky | Rel |
| 100 | Mack | 300 | Jack | 600 | Rocky | Voda |
| 100 | Mack | 300 | Jack | 600 | Rocky | Airtel |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
对于start with entity_code = 600
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| MANAGER_ENTITY_CODE | MANAGER_ENTITY_NAME | TEAM_LEAD_ENTITY_CODE | TEAM_LEAD_ENTITY_NAME | DEVELOPER_ENTITY_CODE | DEVELOPER_ENTITY_NAME | CLIENT_NAME |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
| (null) | (null) | (null) | (null) | 600 | Rocky | Airtel |
| (null) | (null) | (null) | (null) | 600 | Rocky | Voda |
| (null) | (null) | (null) | (null) | 600 | Rocky | Rel |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
在您的查询结果中，您希望某些角色在层次结构中具有特定的位置。经理是第一层，团队领导是第二层，开发人员是第三层。所以你可以把你的分层 table 当作没有。这使得查询非常可读：
with manager as (select * from employee_hierarchy where entity_role = 'Manager')
, teamleader as (select * from employee_hierarchy where entity_role = 'Team-Lead')
, developer as (select * from employee_hierarchy where entity_role = 'Developer')
select
m.entity_code as manager_entity_code,
m.entity_name as manager_entity_name,
t.entity_code as team_lead_entity_code,
t.entity_name as team_lead__entity_name,
d.entity_code as developer_entity_code,
d.entity_name as developer_entity_name,
c.client_name
from manager m
join teamleader t on t.parent_entity_code = m.entity_code
join developer d on d.parent_entity_code = t.entity_code
left join client_table c on c.entity_code = d.entity_code;
如果你想将结果限制在团队负责人 Shail 的部门，只需添加相应的 WHERE 子句：
where t.entity_name = 'Shail'