单行不同列中的Oracle树结构层次结构数据显示
我在oracle中有一个名为 Employee_Hierarchy 的分层表，列名为 entity_code ， parent_entity_code ，entity_name和entity_role，没有循环。 在名为客户端的其他表中具有最低子项的子项与层次表的最低子项与entity_code相关联。 我必须在单行层次结构中显示数据，其中列名将由角色附加。
示例示例：
树形结构：
分层表：
大多数孩子的表：
预期结果：
有没有办法通过oracle查询获得预期的结果？ 并且预期结果取决于输入，这意味着它不总是从根元素开始，它可以从任何节点开始，例如team-lead（Shail）到最下面的子节点。
（注意：如果缺少上层层次结构，则当前节点的parent_code将是其上层次结构的parent_code，并且缺少的层次结构元素在预期结果中将为空白。）< / p>
先谢谢。
4 个答案:
答案 0 :(得分：4)
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 |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
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 |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
答案 1 :(得分：4)
在查询结果中，您希望某些角色在层次结构中具有特定位置。经理是第一级，团队领导者是第二级，开发者是第三级。因此，您可以将您的分层表视为不是。这使得查询非常易读：
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;
如果您想将结果限制在团队负责人Shail的部门，只需添加相应的WHERE条款：
where t.entity_name = 'Shail'
答案 2 :(得分：2)
这是使用连接执行此操作的一种方法。另一种方法是组合这两个表并进行分层查询 - 但这实际上是相同的（分层查询只不过是一个递归的自连接）。
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
;
<强>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.
答案 3 :(得分：2)
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 |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
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 |
+---------------------+---------------------+-----------------------+-----------------------+-----------------------+-----------------------+-------------+
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