20

我有一张正在使用的旧表,如下所示:

+------------------+--------------+------+-----+---------+-------+
| Field            | Type         | Null | Key | Default | Extra |
+------------------+--------------+------+-----+---------+-------+
| BINARY_DATA_ID   | varchar(255) | NO   | PRI |         |       |
| BINARY_DATA      | longblob     | YES  |     | NULL    |       |
| BINARY_DATA_NAME | varchar(255) | YES  |     | NULL    |       |
+------------------+--------------+------+-----+---------+-------+

这样做的主要问题是BinaryDataJava 类会加载BINARY_DATA列,即使我只需要BINARY_DATA_NAME. 我知道构建此架构的最佳方法是将数据从元数据(如文件名)中拆分出来,以便它们存在于单独的表中。从那里开始,使数据延迟加载是微不足道的。这就是它应该首先完成的方式。

不幸的是,由于组织限制,我可能无法执行上述操作。作为一种解决方法,是否可以使用一些注释使该列延迟加载,而不是将内容拆分到单独的表中?我已经修改了这个BinaryData类,使它有一个内部静态BinaryDataData类,它@Embedded的属性是@Basic(fetch=FetchType.LAZY)

@Entity
@Table
@Proxy(lazy=false)
@Inheritance(strategy=InheritanceType.JOINED)
public class BinaryData implements Serializable, Persistable<BinaryData>, Cloneable {

    private static final long serialVersionUID = /** blah */;

    @Id @Column @GeneratedValue(generator="uuid") @GenericGenerator(name="uuid", strategy="uuid")
    private String id;

    @Column
    private String binaryDataName;

    @Embedded
    @Basic(fetch = FetchType.LAZY)
    private BinaryDataData binaryData;

    @Transient
    private String cacheId;

    /**
     * Hibernate constructor
     */
    public BinaryData() { /* Creates a new instance of Attachment. */}

    public BinaryData(byte[] binaryData, String binaryDataName) {
        this.binaryData = new BinaryDataData(ArrayUtils.clone(binaryData));
        this.binaryDataName = binaryDataName;
    }

    /**
     * Returns the BinaryData byte stream.
     *
     * @return binaryData byte stream
     */
    @Embedded
    @Basic(fetch = FetchType.LAZY)
    public byte[] getBinaryData() {
        if (this.binaryData == null) {
            return new byte[0];
        }
        return binaryData.getActualData();
    }

    @Embeddable
    public static class BinaryDataData implements Serializable {
        @Column(length=32*1024*1024, columnDefinition="longblob", name="BINARY_DATA") @Lob
        private byte[] actualData;

        public BinaryDataData() { }

        public BinaryDataData(byte[] data) {
            this.actualData = data;
        }

        public byte[] getActualData() {
            if (this.actualData == null) {
                return new byte[0];
            }
            return this.actualData;
        }

        public void setBinaryData(byte[] newData) {
            this.actualData = newData;
        }

        @Override public boolean equals(Object obj) {
            if (this == obj) {
                return true;
            }
            if (obj == null) {
                return false;
            }
            if (!(obj instanceof BinaryDataData)) {
                return false;
            }
            final BinaryDataData other = (BinaryDataData) obj;
            if (!Arrays.equals(actualData, other.actualData)) {
                return false;
            }
            return true;
        }
    }

    /** onwards... */

不幸的是,这不起作用。即使未请求二进制数据,我看到的 SQL 仍然显示对象的完整获取:

select ideaattach0_.BINARY_DATA_ID as BINARY1_9_, ideaattach0_1_.BINARY_DATA as BINARY2_9_, ideaattach0_1_.BINARY_DATA_NAME as BINARY3_9_, ideaattach0_.IDEA_BUCKET_ID as IDEA2_136_ from IDEA_ATTACHMENT ideaattach0_ inner join BINARY_DATA ideaattach0_1_ on ideaattach0_.BINARY_DATA_ID=ideaattach0_1_.BINARY_DATA_ID where ideaattach0_.BINARY_DATA_ID=?

有任何想法吗?谢谢你。

4

3 回答 3

14

来自Hibernate,第 19 章。提高性能

惰性属性获取:访问实例变量时获取属性或单值关联。这种方法需要构建时字节码检测,并且很少需要。

于 2012-07-17T16:00:29.370 回答
4

对于 maven 项目,需要在 pom.xml 中添加以下插件依赖项:

<plugin>
    <groupId>org.hibernate.orm.tooling</groupId>
    <artifactId>hibernate-enhance-maven-plugin</artifactId>
    <version>${hibernate.version}</version>
    <executions>
        <execution>
            <configuration>
                <failOnError>true</failOnError>
                <enableLazyInitialization>true</enableLazyInitialization>
            </configuration>
            <goals>
                <goal>enhance</goal>
            </goals>
        </execution>
    </executions>
</plugin>

我已经在我的项目中检查了它并且它有效,实体示例:

@Entity(name = "processing_record")
public class ProcessingRecord {

/**
 * Why uuid: https://www.clever-cloud.com/blog/engineering/2015/05/20/why-auto-increment-is-a-terrible-idea/
 */
@Id
@Column(name = "record_id")
@org.hibernate.annotations.Type(type = "pg-uuid")
private UUID id;

...

/**
 * Processing result.
 */
@Column(name = "result")
@Basic(fetch = FetchType.LAZY)
private String result;
...

有关更多详细信息,请查看以下文章:LINK

于 2017-10-30T12:05:58.063 回答
0

我知道此查询的日期,但是,我也会尝试使用映射为列子集的投影值类,并将该投影与指定的命名查询一起使用,该查询实例化该投影值对象,而不是此处引用的基础对象。

我正在研究使用此方法的解决方案,因此我目前没有完整的示例。但是,基本思想是您将创建一个使用“select NEW Projection_Object_Target”语法的 JPA 查询,其中字段在“Projection_Object_Target”的构造函数中直接引用。

IE 使用构造函数表达式如下:

SELECT NEW fully.qualified.package.name.ProjectionObject(baseObject.column_target_0,baseObject.column_target_1,...,baseObject.column_target_n) FROM BaseObjectMappedInDBTable AS baseObject

通用示例用例:

String queryStr =
  "SELECT NEW fully.qualified.package.name.ProjectionObject(baseObject.column_target_0) " +
  "FROM BaseObjectMappedInTable AS baseObject";
TypedQuery<ProjectionObject> query =
  em.createQuery(queryStr, ProjectionObject.class);
List<ProjectionObject> results = query.getResultList();
于 2016-03-13T00:56:11.217 回答