You Can Retrive all data in Many-To-Many Relationship Using the following join Query
Hbm File is Like these
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
<joined-subclass entity-name="test"
table="TEST" dynamic-update="true" dynamic-insert="true">
<key column="ID"/>
<property name="name" type="string" column="NAME" />
<set name="assistantName"
table="TEST_1"
inverse="false"
lazy = "false"
cascade="none"
sort="unsorted">
<key column="ID"/>
<element column="ASSISTANTNAME" type="string"/>
</set>
</joined-subclass>
</hibernate-mapping>
Use Of Following Hql join Query Retrive Both Table data TEST and TEST_1
select
fetchInfo.id as id,
fetchInfo.name as name,
Test_1.id as assistantName
from
test as fetchInfo join fetchInfo.test_1 as Test_1
where
fetchInfo.id =:someValue
Hbm File is Like these
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
<joined-subclass entity-name="test"
table="TEST" dynamic-update="true" dynamic-insert="true">
<key column="ID"/>
<property name="name" type="string" column="NAME" />
<set name="assistantName"
table="TEST_1"
inverse="false"
lazy = "false"
cascade="none"
sort="unsorted">
<key column="ID"/>
<element column="ASSISTANTNAME" type="string"/>
</set>
</joined-subclass>
</hibernate-mapping>
Use Of Following Hql join Query Retrive Both Table data TEST and TEST_1
select
fetchInfo.id as id,
fetchInfo.name as name,
Test_1.id as assistantName
from
test as fetchInfo join fetchInfo.test_1 as Test_1
where
fetchInfo.id =:someValue
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