Question

在我通过以下方式添加字节码检测之前，我没有在任何一个类上进行延迟加载：

  <plugin>
    <artifactId>maven-antrun-plugin</artifactId>
    <executions>
      <execution>
        <phase>process-classes</phase>
        <goals>
          <goal>run</goal>
        </goals>
      </execution>
    </executions>
    <configuration>
      <tasks>
        <taskdef name="instrument" classname="org.hibernate.tool.instrument.javassist.InstrumentTask">
          <classpath>
            <path refid="maven.runtime.classpath" />
            <path refid="maven.plugin.classpath" />
          </classpath>
        </taskdef>
        <instrument verbose="false">
          <fileset dir="${project.build.outputDirectory}">
            <include name="**/db/**/*.class" />
          </fileset>
        </instrument>
      </tasks>
    </configuration>
  </plugin>

这是我的两个实体类，大大减少了：

此表“来自Ininvgrmtr”按我的意愿工作（没有n + 1期）：

@Entity
@Table(name = "ININVGRMTR", catalog = "CO05IN", schema = "")
@XmlRootElement
public class Ininvgrmtr implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 8)
    @Column(name = "IGMGRUP", nullable = false, length = 8)
    private String igmgrup;

    //other attributes

    @JoinColumn(name = "IGMGRUP", referencedColumnName = "IGGRUP", nullable = false, insertable = false, updatable = false)
    @OneToOne(optional = false, fetch=FetchType.LAZY)
    private Ininvgrp ininvgrp;
}

此表“来自Ininvgrp”不会：

@Entity
@Table(name = "ININVGRP", catalog = "CO05IN", schema = "")
@XmlRootElement
public class Ininvgrp implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 8)
    @Column(name = "IGGRUP", nullable = false, length = 8)
    private String iggrup;

    //other attributes

    @OneToOne(cascade = CascadeType.ALL, mappedBy = "ininvgrp", fetch=FetchType.LAZY)
    private Ininvgrmtr ininvgrmtr;

    //getters setters
}

说明问题：

entityManagerFactory.createEntityManager().createQuery("from Ininvgrmtr").getResultList();

将以下内容打印到日志中（这很好）：

INFO: Hibernate: select ininvgrmtr0_.IGMGRUP as IGMGRUP96_, ininvgrmtr0_.IGMTRACE as IGMTRACE96_ from CO05IN.ININVGRMTR ininvgrmtr0_

虽然

entityManagerFactory.createEntityManager().createQuery("from Ininvgrp").getResultList();

打印打印以下内容：

INFO: Hibernate: select ininvgrp0_.IGGRUP as IGGRUP97_, ininvgrp0_.Added as Added97_, ininvgrp0_.IGABCF as IGABCF97_, ininvgrp0_.IGADCN as IGADCN97_, ininvgrp0_.IGADDT as IGADDT97_, ininvgrp0_.IGADUS as IGADUS97_, ininvgrp0_.IGCAT as IGCAT97_, ininvgrp0_.IGDESC as IGDESC97_, ininvgrp0_.IGMDCN as IGMDCN97_, ininvgrp0_.IGMDDT as IGMDDT97_, ininvgrp0_.IGMDUS as IGMDUS97_, ininvgrp0_.IGRETH as IGRETH97_, ininvgrp0_.IGSTA as IGSTA97_, ininvgrp0_.IGTYPE as IGTYPE97_, ininvgrp0_.IGUBAS as IGUBAS97_, ininvgrp0_.IGUSEL as IGUSEL97_, ininvgrp0_.IGUWGT as IGUWGT97_, ininvgrp0_.Modified as Modified97_ from CO05IN.ININVGRP ininvgrp0_
INFO: Hibernate: select ininvgrmtr0_.IGMGRUP as IGMGRUP96_0_, ininvgrmtr0_.IGMTRACE as IGMTRACE96_0_ from CO05IN.ININVGRMTR ininvgrmtr0_ where ininvgrmtr0_.IGMGRUP=?
INFO: Hibernate: select ininvgrmtr0_.IGMGRUP as IGMGRUP96_0_, ininvgrmtr0_.IGMTRACE as IGMTRACE96_0_ from CO05IN.ININVGRMTR ininvgrmtr0_ where ininvgrmtr0_.IGMGRUP=?
INFO: Hibernate: select ininvgrmtr0_.IGMGRUP as IGMGRUP96_0_, ininvgrmtr0_.IGMTRACE as IGMTRACE96_0_ from CO05IN.ININVGRMTR ininvgrmtr0_ where ininvgrmtr0_.IGMGRUP=?
INFO: Hibernate: select ininvgrmtr0_.IGMGRUP as IGMGRUP96_0_, ininvgrmtr0_.IGMTRACE as IGMTRACE96_0_ from CO05IN.ININVGRMTR ininvgrmtr0_ where ininvgrmtr0_.IGMGRUP=?
INFO: Hibernate: select ininvgrmtr0_.IGMGRUP as IGMGRUP96_0_, ininvgrmtr0_.IGMTRACE as IGMTRACE96_0_ from CO05IN.ININVGRMTR ininvgrmtr0_ where ininvgrmtr0_.IGMGRUP=?
INFO: Hibernate: select ininvgrmtr0_.IGMGRUP as IGMGRUP96_0_, ininvgrmtr0_.IGMTRACE as IGMTRACE96_0_ from CO05IN.ININVGRMTR ininvgrmtr0_ where ininvgrmtr0_.IGMGRUP=?
INFO: Hibernate: select ininvgrmtr0_.IGMGRUP as IGMGRUP96_0_, ininvgrmtr0_.IGMTRACE as IGMTRACE96_0_ from CO05IN.ININVGRMTR ininvgrmtr0_ where ininvgrmtr0_.IGMGRUP=?
INFO: Hibernate: select ininvgrmtr0_.IGMGRUP as IGMGRUP96_0_, ininvgrmtr0_.IGMTRACE as IGMTRACE96_0_ from CO05IN.ININVGRMTR ininvgrmtr0_ where ininvgrmtr0_.IGMGRUP=?
INFO: Hibernate: select ininvgrmtr0_.IGMGRUP as IGMGRUP96_0_, ininvgrmtr0_.IGMTRACE as IGMTRACE96_0_ from CO05IN.ININVGRMTR ininvgrmtr0_ where ininvgrmtr0_.IGMGRUP=?
...

造成这种情况的原因是什么？

Answer 1

Ininvgrmtr上的属性被标记为不可为空且不是可选的，因此hibernate知道必须有一个具有给定id的实体。然后，Hibernate可以创建动态代理并将其设置为属性值。然后，只有在访问其属性时才会初始化此代理。

另一方面，从Ininvgrp开始，默认情况下该属性被标记为可选。在这种情况下，Hibernate不能使用动态代理，因为如果没有匹配的实体，它必须返回null。

使用字节码编织可以修改对字段本身的每次访问以查询数据库。如果您可以在构建过程中使用字节码检测，那么这将是首选解决方案。

另一种解决方法是将属性声明为OneToMany关系，并将空/一个元素列表转换为null或其在getter和setter中的第一个元素，如下所示：

@OneToMany(cascade = CascadeType.ALL, mappedBy = "ininvgrp", fetch=FetchType.LAZY)
private List<Ininvgrmtr> ininvgrmtr;

public void setIninvgrmtr(Ininvgrmtr ininvgrmtr) {
    if (this.ininvgrmtr == null || this.ininvgrmtr.isEmpty()) {
        this.ininvgrmtr = Collections.singletonList(ininvgrmtr);
    } else {
        this.ininvgrmtr.set(0, ininvgrmtr);
    }
}

public Ininvgrmtr getIninvgrmtr() {
    return ininvgrmtr == null || ininvgrmtr.isEmpty() ? null : ininvgrmtr.get(0);
}

修改：this blog post可以找到有关某些参考文献问题的更详细说明。

Hibernate JPA2 - n + 1选择问题仅影响OneToOne关系的一侧

1 个答案: