Question

我正在使用spring 4.1.4.RELEASE + hibernate 4.3.6.Final，这是我的实体代码：

public class BaseEntity implements Serializable {
}


public class MarketInfo extends BaseEntity {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id")
    private int id;

    @Column(name = "market_id", unique = true, length = 15)
    private String marketId;

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "market")
    private List<MarketChannelGroup> channelGroups;

    public List<MarketChannelGroup> getChannelGroups() {
        return channelGroups;
    }

    public void setChannelGroups(List<MarketChannelGroup> channelGroups) {
        this.channelGroups = channelGroups;
    }

...
}

public class MarketChannelGroup extends BaseEntity {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id")
    private int id;


    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "market_id", referencedColumnName = "market_id")
    private MarketInfo market;
...
}

从我的测试中我可以看到MarketInfo中的channelGroup正常工作（如果我不调用getChannelGroups（），那么channelGroups为null），但是如果我调用getChannelGroups（），则会获取每个MarketChannelGroup中的MarketInfo虽然这不应该发生，因为市场的提取模式是FetchType.LAZY。

从控制台我可以看到以下休眠日志，当我调用它的getter：

Hibernate: select channelgro0_.market_id as market_i5_12_1_, channelgro0_.id as id1_9_1_, channelgro0_.id as id1_9_0_, channelgro0_.channel_group_id as channel_2_9_0_, channelgro0_.channel_group_name as channel_3_9_0_, channelgro0_.channel_group_type as channel_4_9_0_, channelgro0_.market_id as market_i5_9_0_ from market_channel_group channelgro0_ where channelgro0_.market_id=?
Hibernate: select marketinfo0_.id as id1_12_0_, marketinfo0_.enable_flag as enable_f2_12_0_, marketinfo0_.enable_time as enable_t3_12_0_, marketinfo0_.market_id as market_i4_12_0_, marketinfo0_.market_name as market_n5_12_0_, marketinfo0_.stb_count as stb_coun6_12_0_ from market_info marketinfo0_ where marketinfo0_.market_id=?
Hibernate: select marketinfo0_.id as id1_12_0_, marketinfo0_.enable_flag as enable_f2_12_0_, marketinfo0_.enable_time as enable_t3_12_0_, marketinfo0_.market_id as market_i4_12_0_, marketinfo0_.market_name as market_n5_12_0_, marketinfo0_.stb_count as stb_coun6_12_0_ from market_info marketinfo0_ where marketinfo0_.market_id=?
Hibernate: select marketinfo0_.id as id1_12_0_, marketinfo0_.enable_flag as enable_f2_12_0_, marketinfo0_.enable_time as enable_t3_12_0_, marketinfo0_.market_id as market_i4_12_0_, marketinfo0_.market_name as market_n5_12_0_, marketinfo0_.stb_count as stb_coun6_12_0_ from market_info marketinfo0_ where marketinfo0_.market_id=?

有人可以帮忙吗？

更新

没有针对ManyToOne注释的可选方法，因此OneToOne中的解决方案对我的案例不起作用。

Answer 1

将@OneToMany批注从声明移到getter方法，如下所示：

private List<MarketChannelGroup> channelGroups;

@OneToMany(fetch = FetchType.LAZY, mappedBy = "market")
public List<MarketChannelGroup> getChannelGroups() {
    return channelGroups;
}

Answer 2

您的配置存档如何？看看您是否使用过滤器 OpenEntityManagerInViewFilter 。如果您正在使用它，那么总是调用方法Lazy getChannelGroups（），hibernate将在每个元素中获取。

Answer 3

如果@id在getter上，则必须使用oneToMany注释来获取。惰性工作，但是如果您或某个框架在该转换中获得了惰性属性，则get方法将触发选择。

Hibernate ManyToOne FetchType.LAZY无法正常工作？

3 个答案: