JPA fetchType.Lazy 不起作用

Question

我正在尝试< code>fetchType的示例。懒惰，但是在调试代码时，似乎< code>fetchType。懒惰不起作用。

实体bean：地址

添加了属性为fetch=fetchType的注释@Basic。懒散地在地区赛场上。

我的实体bean由以下代码定义：

    package model;

    import java.io.Serializable;
    import javax.persistence.*;
    import java.util.List;


    /**
     * The persistent class for the address database table.
     */
    @Entity
    @Table(name="address", schema="home")
    public class Address implements Serializable {
        private static final long serialVersionUID = 1L;

        @TableGenerator(name = "addr_gen", table = "table_generator", pkColumnName = "gen_name", valueColumnName = "gen_val", allocationSize=1)
        @Id
        @GeneratedValue(strategy=GenerationType.TABLE, generator="addr_gen")
        private String addressId;

        private String city;

        @Basic(fetch=FetchType.LAZY)
        @Column(name="district")
        private String district;

        private String houseNumber;

        private String pincode;

        private String state;

        private String street;

        //bi-directional many-to-one association to Employee
        @OneToMany(mappedBy="address")
        private List<Employee> employees;

        public Address() {
        }


        public String getAddressId() {
            return this.addressId;
        }

        public void setAddressId(String addressId) {
            this.addressId = addressId;
        }

        public String getCity() {
            return this.city;
        }

        public void setCity(String city) {
            this.city = city;
        }

        public String getDistrict() {
            return this.district;
        }

        public void setDistrict(String district) {
            this.district = district;
        }

        public String getHouseNumber() {
            return this.houseNumber;
        }

        public void setHouseNumber(String houseNumber) {
            this.houseNumber = houseNumber;
        }

        public String getPincode() {
            return this.pincode;
        }

        public void setPincode(String pincode) {
            this.pincode = pincode;
        }

        public String getState() {
            return this.state;
        }

        public void setState(String state) {
            this.state = state;
        }

        public String getStreet() {
            return this.street;
        }

        public void setStreet(String street) {
            this.street = street;
        }

        public List<Employee> getEmployees() {
            return this.employees;
        }

        public void setEmployees(List<Employee> employees) {
            this.employees = employees;
        }


        public Employee addEmployees(Employee employees) {
            getEmployees().add(employees);
            employees.setAddress(this);

            return employees;
        }

        public Employee removeEmployees(Employee employees) {
            getEmployees().remove(employees);
            employees.setAddress(null);

            return employees;
        }

        /*@Override
        public String toString() {
            return "Address [addressId=" + addressId + ", city=" + city
                    + ", district=" + district + ", houseNumber=" + houseNumber
                    + ", pincode=" + pincode + ", state=" + state + ", street="
                    + street + ", employees=" + employees + "]";
        }*/
    }

使用上述实体 Bean 的方法：

public Address findAddress(EntityManagerFactory emf, UserTransaction tx) {
    EntityManager em = emf.createEntityManager();
    Address addr = null;
    try {
        tx.begin();
        addr = em.find(Address.class, new String("154"));
        tx.commit();
    } catch (Exception e) {
        e.printStackTrace();
    }

    return addr;
}

因此，在查找方法调用之后，当我检查地址对象时，它已经填充了区域字段。

请让我知道我是否缺少某些配置，或者代码本身是否存在一些问题。

Answer 1

要在JPA中初始化laze，您需要调用一个jar库并启动它，如果它是使用maven或手动por示例，如果您需要laze，请在maven jar de.empulse.eclipselink中使用jar 更多信息 http://wiki.eclipse.org/EclipseLink/UserGuide/JPA/Advanced_JPA_Development/Performance/Weaving/Static_Weaving#Use_the_Maven_plugin

如果你不使用maven，你可以让手动启用jar

Answer 2

确保您已启用编织。

http://wiki . eclipse . org/eclipse link/user guide/JPA/Advanced _ JPA _ Development/Performance/Weaving

Answer 3

问题是，当在< code>@Basic注释中使用时，应用服务器可以/可能自己决定何时获取数据更好。如此处记录的

EAGER策略是持久性提供程序运行时的一项要求，即必须急切地获取值。懒惰策略是对持久性提供程序运行时的提示。

@OneTo很多类注释根据留档也有同样的问题，但在这种情况下，JPA提供程序更有可能会考虑FetchType提示。

另一方面，你可以尝试使用大数据字段，比如

@Lob
@Basic(fetch=FetchType.LAZY)
private byte[] document;

查看是否有更改（尽管应用程序服务器可以决定始终获取字段）。

为了检查是否加载了收集字段，我将检查数据库查询日志或执行以下操作:

@PersistenceContext private EntityManager em;
.....
Person person = em.find(Person.class, 1L);//Person entity has a OneToMany relationship to entity Address
em.detach(person);
person.getAddresses().size();//if the Address are now not fetched, it the call should  throw an error