我将show_sql设置为true并运行此查询：
Query q = session.createQuery("select inc FROM Incident inc "
+ "inner join fetch inc.workType "
+ " WHERE inc.incidentId =:id");
q.setLong("id", id);
控制台：
Hibernate:
select
incident0_.incident_id as incident1_3_0_,
worktype1_.work_id as work1_28_1_,
incident0_.ccs_incident_id as ccs2_3_0_,
incident0_.process_status as process3_3_0_,
incident0_.time_created as time4_3_0_,
incident0_.is_finish as is5_3_0_,
incident0_.type as type3_0_,
incident0_.subra_customer_file_id as subra7_3_0_,
incident0_.description as descript8_3_0_,
incident0_.group_id as group9_3_0_,
incident0_.package_size as package10_3_0_,
incident0_.work_id as work11_3_0_,
worktype1_.name as name28_1_,
worktype1_.state_id as state3_28_1_
from
ra.incident incident0_
inner join
ra.work_type worktype1_
on incident0_.work_id=worktype1_.work_id
where
incident0_.incident_id=?
Hibernate:
select
group0_.group_id as group1_6_1_,
group0_.name as name6_1_,
group0_.code as code6_1_,
group0_.type as type6_1_,
group0_.is_active as is5_6_1_,
group0_.group_remote as group6_6_1_,
group0_.send_data as send7_6_1_,
employees1_.group_id as group1_3_,
employee2_.employee_id as employee2_3_,
employee2_.employee_id as employee1_20_0_,
employee2_.username as username20_0_,
employee2_.password as password20_0_,
employee2_.full_name as full4_20_0_,
employee2_.ip_address as ip5_20_0_,
employee2_.last_login as last6_20_0_,
employee2_.is_locked as is7_20_0_,
employee2_.ccs_username as ccs8_20_0_,
employee2_.base64_certificate as base9_20_0_,
employee2_.is_active as is10_20_0_,
employee2_.time_created as time11_20_0_,
employee2_.is_absent as is12_20_0_,
employee2_.is_system_account as is13_20_0_
from
ra.group group0_
left outer join
group_employee employees1_
on group0_.group_id=employees1_.group_id
left outer join
ra.employee employee2_
on employees1_.employee_id=employee2_.employee_id
where
group0_.group_id=?
Hibernate:
select
workconfig0_.employee_id as employee1_1_,
workconfig0_.work_config_id as work2_1_,
workconfig1_.work_config_id as work1_14_0_,
workconfig1_.work_percent as work2_14_0_,
workconfig1_.total_work as total3_14_0_,
workconfig1_.group_id as group4_14_0_,
workconfig1_.is_template as is5_14_0_,
workconfig1_.work_type_id as work6_14_0_
from
mn_config workconfig0_
left outer join
ra.work_config workconfig1_
on workconfig0_.work_config_id=workconfig1_.work_config_id
where
workconfig0_.employee_id=?
Hibernate:
select
group0_.employee_id as employee2_1_,
group0_.group_id as group1_1_,
group1_.group_id as group1_6_0_,
group1_.name as name6_0_,
group1_.code as code6_0_,
group1_.type as type6_0_,
group1_.is_active as is5_6_0_,
group1_.group_remote as group6_6_0_,
group1_.send_data as send7_6_0_
from
group_employee group0_
left outer join
ra.group group1_
on group0_.group_id=group1_.group_id
where
group0_.employee_id=?
最后两次查询被执行了很多次。我只需要第一个被执行。这是xml文件中的映射关系：
<many-to-one name="group" class="Group"
column="group_id" fetch="join" unique="true" lazy="proxy"
not-found="ignore" />
<set name="certificateRequestHistory" table="certificate_request_history"
fetch="select" lazy="extra" cascade="delete" >
<key>
<column name="incident_id" not-null="true" />
</key>
<one-to-many class="CertificateRequestHistory"
not-found="ignore" />
</set>
<set name="businessLogs" table="business_log" inverse="true"
fetch="select" lazy="true" order-by="time_created desc">
<key>
<column name="incident_id" not-null="true" />
</key>
<one-to-many class="BusinessLog"
not-found="ignore" />
</set>
<set name="revocationRequestHistory" table="revocation_request_history"
inverse="true" fetch="select" lazy="true">
<key>
<column name="incident_id" not-null="true" />
</key>
<one-to-many class="RevocationRequestHistory"
not-found="ignore" />
</set>
<set name="mnProcess" table="mn_process" inverse="true" fetch="select"
lazy="true">
<key column="incident_id" not-null="true" />
<one-to-many class="MnProcess" />
</set>
<set name="certificate" table="certificate" inverse="true" fetch="select"
lazy="extra">
<key column="incident_id" not-null="true" />
<one-to-many class="Certificate" />
</set>
<set name="unlockRequestHistory" table="unlock_request_history"
inverse="true" fetch="select" lazy="extra">
<key column="incident_id" not-null="true" />
<one-to-many class="UnlockRequestHistory" />
</set>
<set name="ccsRequests" table="ccs_request" inverse="true" fetch="select"
lazy="extra">
<key column="incident_id" not-null="true" />
<one-to-many class="CcsRequest" />
</set>
<many-to-one name="workType" class="WorkType"
column="work_id" not-null="false" fetch="join" update="true" insert="true"
unique="true" lazy="no-proxy" />
我一直在寻找：懒人取，代理等，但没有希望。任何使Hibernate运行这两次的建议？甚至没有？非常感谢！