Java hibernate-减少查询的数量

一个产品可以有一个或多个子产品形成树状结构。子产品也将引用父产品。在我的测试代码中，我创建了一个家长（

mortgage

）和两个孩子（

ml和me

）。为什么我尝试加载父级

Product entity=（Product）hibernateTemplate.get（Product.class，mortgage.getId（））它正在正确使用连接

然而，hibernate也在尝试加载
ml和me
的子项，即使它们都没有任何子项。有没有办法避免这种情况，这样我就可以将查询的数量从3个减少到1个

生成的SQL使用动态生成的别名，如
children0\u0
。有没有办法让hibernate使用我自己的别名说“p”

这是我的hbm和测试代码

hbm

<class name="Product" table="PRODUCT">
        <id name="id" type="java.lang.Long" column="ID">
            <generator class="native">
                <param name="sequence">PRODUCT_SN</param>
            </generator>
        </id>

        <many-to-one name="parent" class="Product" lazy="false" column="PARENT" />

        <set name="children" lazy="false" fetch="join"  table="PRODUCT" cascade="all">
            <key>
                <column name="PARENT" />
            </key>
            <one-to-many class="Product" />
        </set>
        <property name="code" type="java.lang.String" column="CODE" not-null="true" />
        <property name="name" type="java.lang.String" column="NAME" />
        <property name="startDate" type="java.util.Date" column="STARTDATE" />
        <property name="endDate" type="java.util.Date" column="ENDDATE" />
        <property name="decisionable" type="boolean" column="ISDECISIONABLE" />
        <property name="selectable" type="boolean" column="ISSELECTABLE" />
    </class>

日志

Hibernate: select product0_.ID as ID0_1_, product0_.PARENT as PARENT0_1_, product0_.CODE as CODE0_1_, product0_.NAME as NAME0_1_, product0_.STARTDATE as STARTDATE0_1_, product0_.ENDDATE as ENDDATE0_1_, product0_.ISDECISIONABLE as ISDECISI7_0_1_, product0_.ISSELECTABLE as ISSELECT8_0_1_, children1_.PARENT as PARENT3_, children1_.ID as ID3_, children1_.ID as ID0_0_, children1_.PARENT as PARENT0_0_, children1_.CODE as CODE0_0_, children1_.NAME as NAME0_0_, children1_.STARTDATE as STARTDATE0_0_, children1_.ENDDATE as ENDDATE0_0_, children1_.ISDECISIONABLE as ISDECISI7_0_0_, children1_.ISSELECTABLE as ISSELECT8_0_0_ from PRODUCT product0_ left outer join PRODUCT children1_ on product0_.ID=children1_.PARENT where product0_.ID=?
DEBUG [org.hibernate.type.LongType] binding '1' to parameter: 1

DEBUG [org.hibernate.type.LongType] returning '2' as column: ID0_0_
DEBUG [org.hibernate.type.LongType] returning '1' as column: PARENT0_0_
DEBUG [org.hibernate.type.StringType] returning 'Mortgage Loan' as column: CODE0_0_
DEBUG [org.hibernate.type.StringType] returning null as column: NAME0_0_
DEBUG [org.hibernate.type.TimestampType] returning null as column: STARTDATE0_0_
DEBUG [org.hibernate.type.TimestampType] returning null as column: ENDDATE0_0_
DEBUG [org.hibernate.type.BooleanType] returning 'false' as column: ISDECISI7_0_0_
DEBUG [org.hibernate.type.BooleanType] returning 'false' as column: ISSELECT8_0_0_

DEBUG [org.hibernate.type.LongType] returning null as column: PARENT0_1_
DEBUG [org.hibernate.type.StringType] returning 'Mortgage' as column: CODE0_1_
DEBUG [org.hibernate.type.StringType] returning null as column: NAME0_1_
DEBUG [org.hibernate.type.TimestampType] returning null as column: STARTDATE0_1_
DEBUG [org.hibernate.type.TimestampType] returning null as column: ENDDATE0_1_
DEBUG [org.hibernate.type.BooleanType] returning 'false' as column: ISDECISI7_0_1_
DEBUG [org.hibernate.type.BooleanType] returning 'false' as column: ISSELECT8_0_1_

DEBUG [org.hibernate.type.LongType] returning '1' as column: PARENT3_
DEBUG [org.hibernate.type.LongType] returning '2' as column: ID3_
DEBUG [org.hibernate.type.LongType] returning '3' as column: ID0_0_
DEBUG [org.hibernate.type.LongType] returning '1' as column: PARENT0_0_
DEBUG [org.hibernate.type.StringType] returning 'Home Equity LOC' as column: CODE0_0_
DEBUG [org.hibernate.type.StringType] returning null as column: NAME0_0_
DEBUG [org.hibernate.type.TimestampType] returning null as column: STARTDATE0_0_
DEBUG [org.hibernate.type.TimestampType] returning null as column: ENDDATE0_0_
DEBUG [org.hibernate.type.BooleanType] returning 'false' as column: ISDECISI7_0_0_
DEBUG [org.hibernate.type.BooleanType] returning 'false' as column: ISSELECT8_0_0_
DEBUG [org.hibernate.type.LongType] returning '1' as column: PARENT3_
DEBUG [org.hibernate.type.LongType] returning '3' as column: ID3_

Hibernate: select children0_.PARENT as PARENT1_, children0_.ID as ID1_, children0_.ID as ID0_0_, children0_.PARENT as PARENT0_0_, children0_.CODE as CODE0_0_, children0_.NAME as NAME0_0_, children0_.STARTDATE as STARTDATE0_0_, children0_.ENDDATE as ENDDATE0_0_, children0_.ISDECISIONABLE as ISDECISI7_0_0_, children0_.ISSELECTABLE as ISSELECT8_0_0_ from PRODUCT children0_ where children0_.PARENT=?
DEBUG [org.hibernate.type.LongType] binding '3' to parameter: 1

Hibernate: select children0_.PARENT as PARENT1_, children0_.ID as ID1_, children0_.ID as ID0_0_, children0_.PARENT as PARENT0_0_, children0_.CODE as CODE0_0_, children0_.NAME as NAME0_0_, children0_.STARTDATE as STARTDATE0_0_, children0_.ENDDATE as ENDDATE0_0_, children0_.ISDECISIONABLE as ISDECISI7_0_0_, children0_.ISSELECTABLE as ISSELECT8_0_0_ from PRODUCT children0_ where children0_.PARENT=?
DEBUG [org.hibernate.type.LongType] binding '2' to parameter: 1

您可能应该将关系标记为lazy，以便仅在必要时加载子对象。
如果要加载节点及其子节点，请使用专用查询：

select p from Product left join fetch p.children where p.id = :id

我认为你看到所有
孩子的原因是你在
中定义了
fetch=“join”
。如果你把它去掉，我想Hibernate只会拉
产品
，我不希望他们懒惰。我只是想知道是否有办法告诉hibernate抑制最后两个sql。不，如果要抑制它们，应该使用延迟抓取。您可以在模型中配置它，也可以将其配置为此特定查询的获取配置文件（以便它在应用程序中的任何其他地方都可以使用）。
select p from Product left join fetch p.children where p.id = :id