我需要以下面的方式提取数据
Compare_Result Association_Type_name Association_Role
Type in Dev,but not Prod ProcessingSite Null
Role in Dev,but not Prod ProcessingSite MoodedActivity
Role in Dev,but not Prod Specimen ProductRelationship
要求：我需要比较开发和生产。如果生产中不存在任何association_type或association_role，那么我需要返回一条消息，如“在开发中找到的类型对象，但不是生产”
任何人都可以给我一个想法。
示例数据
create table nodes
(
node_id number(5),
parent_node_id number(5),
object_id number(5)
);
begin
insert into nodes values(4001,7001,2001);
insert into nodes values(4002,4001,2005);
insert into nodes values(4003,4002,2002);
insert into nodes values(4004,4003,2004);
insert into nodes values(4005,4004,2003);
insert into nodes values(4006,4004,2006);
insert into nodes values(4007,4004,2007);
insert into nodes values(4008,4003,2008);
insert into nodes values(4009,4008,2011);
insert into nodes values(4010,4008,2013);
insert into nodes values(4011,4008,2014);
insert into nodes values(4012,4003,2009);
insert into nodes values(4013,4012,2015);
insert into nodes values(4014,4012,2017);
insert into nodes values(4015,4001,2020);
insert into nodes values(4016,4015,2012);
insert into nodes values(4017,4016,2016);
insert into nodes values(4018,4017,2024);
insert into nodes values(4019,4017,2023);
insert into nodes values(4020,4016,2029);
insert into nodes values(4021,4020,2022);
insert into nodes values(4022,4020,2021);
insert into nodes values(4023,4020,2019);
insert into nodes values(4024,4016,2010);
insert into nodes values(4025,4024,2018);
end;
/
create table objects
(
object_id number(5),
object_type_id number(5),
name varchar2(40)
);
begin
insert into objects values ( 2001,5001,'iad_Dictionary_area');
insert into objects values ( 2002,5003,'iad_Association_Dictionary');
insert into objects values ( 2003,5005,'TreatingSite');
insert into objects values ( 2004,5004,'ProcessingSite');
insert into objects values ( 2005,5002,'Development');
insert into objects values ( 2006,5005,'MoodedActivity');
insert into objects values ( 2007,5005,'MaterialName');
insert into objects values ( 2008,5004,'PerformedClass');
insert into objects values ( 2009,5004,'Specimen');
insert into objects values ( 2010,5004,'Specimen');
insert into objects values ( 2011,5005,'RegulatoryAssessment');
insert into objects values ( 2012,5003,'iad_Association_Dictionary');
insert into objects values ( 2013,5005,'Submission');
insert into objects values ( 2014,5005,'Mooded');
insert into objects values ( 2015,5005,'class13th');
insert into objects values ( 2016,5004,'ProcessingSite');
insert into objects values ( 2017,5005,'ProductRelationship');
insert into objects values ( 2018,5005,'class13th');
insert into objects values ( 2019,5005,'Distributor');
insert into objects values ( 2020,5006,'Production');
insert into objects values ( 2021,5005,'Assessor');
insert into objects values ( 2022,5005,'StudyInvestigator');
insert into objects values ( 2023,5005,'MaterialName');
insert into objects values ( 2024,5005,'TreatingSite');
insert into objects values ( 2029,5004,'BiologicEntityGroup');
end;
/
create table object_types
(
object_type_id number(5),
name varchar2(35)
);
begin
insert into object_types values (5001,'Dictionary_area');
insert into object_types values (5002,'Development');
insert into object_types values (5003,'Association_Dictionary');
insert into object_types values (5004,'Association_Type');
insert into object_types values (5005,'Association_Role');
insert into object_types values (5006,'Production');
end;
/
让我们假设我有以下结构
Dictionary_area
development
Association_Dictionary
ProcessingSite(type1)
TreatingSite(role1)
MoodedActivity(role2)
MaterialName(role3)
PerformedClass(type2)
RegulatoryAssessment(role1)
Submission(role2)
Mooded(role3)
Specimen(type3)
class13th(role1)
ProductRelationship(role2)
production
Association_Dictionary
ProcessingSite(type1)
TreatingSite(role1)
MaterialName(role2)
BiologicEntityGroup(type2)
StudyInvestigator(role1)
Assessor(role2)
Distributor(role3)
Specimen (type3)
class13th(role1)
ProcessingSite是一个association_type，它有3个association_roles。我们可以在生产中观察到相同的association_type，但是MoodedActivity（association_role）不可用。在这种情况下，我们需要返回“在开发中找到的类型对象，而不是生产”。
节点：此表将具有节点和父节点信息。例如，如果开发对象的某个节点为100，其父节点为99。
对象：对象信息。
object_types：object_type信息。例如，ProcessingSite object_type是“Association_Type”。
我尝试了以下方式
WITH childs AS
(SELECT n.object_id, n.node_id, n.parent_node_id, ot.NAME, "path",
rootnode
FROM (SELECT object_id, node_id, parent_node_id,
SYS_CONNECT_BY_PATH (object_id, '/') "path",
CONNECT_BY_ROOT (object_id) rootnode
FROM nodes
START WITH object_id = 2001
CONNECT BY PRIOR node_id = parent_node_id) n
JOIN
objects o ON n.object_id = o.object_id
JOIN object_types ot ON o.object_type_id = ot.object_type_id
),
devobj AS
(SELECT n.object_id, n.NAME
FROM (SELECT *
FROM childs) n
WHERE n.NAME = 'Development'),
prodobj AS
(SELECT n.object_id, n.NAME
FROM (SELECT *
FROM childs) n
WHERE n.NAME = 'Production'),
dev_associ_type_obj AS
(SELECT *
FROM (SELECT object_id, NAME
FROM (SELECT *
FROM childs)
START WITH object_id = (SELECT object_id
FROM devobj)
CONNECT BY PRIOR node_id = parent_node_id) t
WHERE t.NAME = 'Association_Type'),
dev_associ_roles_obj AS
(SELECT object_id, NAME, CONNECT_BY_ROOT (object_id) rootnode
FROM (SELECT *
FROM childs)
START WITH object_id IN (SELECT object_id
FROM dev_associ_type_obj)
CONNECT BY PRIOR node_id = parent_node_id),
prod_associ_type_obj AS
(SELECT *
FROM (SELECT object_id, NAME
FROM (SELECT *
FROM childs)
START WITH object_id = (SELECT object_id
FROM devobj)
CONNECT BY PRIOR node_id = parent_node_id) t
WHERE t.NAME = 'Association_Type'),
prod_associ_roles_obj AS
(SELECT object_id, NAME, CONNECT_BY_ROOT (object_id) rootnode
FROM (SELECT *
FROM childs)
START WITH object_id IN (SELECT object_id
FROM prod_associ_type_obj)
CONNECT BY PRIOR node_id = parent_node_id)
SELECT *
FROM prod_associ_roles_obj
MINUS
SELECT *
FROM dev_associ_roles_obj