Question

我正在尝试在数据库中插入记录（使用Java EE 6，EJB 3.1，JPA 2.0）。我收到一个错误，accountTypeId字段为空，但我已将其设置为autogenerate。任何人都可以建议我做错了什么？

以下是创建表查询：

create table example.account_type(
    account_type_id INT NOT null PRIMARY KEY GENERATED ALWAYS AS IDENTITY (START WITH 1, INCREMENT BY 1),
    account_type_desc varchar(20)
);

以下是实体类：

编辑：更新了由NetBeans生成的实体类，该实体类不起作用。我还添加了@GeneratedValue注释，但它仍无效。

@Entity
@Table(name = "ACCOUNT_TYPE")
@NamedQueries({
    @NamedQuery(name = "AccountType.findAll", query = "SELECT a FROM AccountType a"),
    @NamedQuery(name = "AccountType.findByAccountTypeId", query = "SELECT a FROM AccountType a WHERE a.accountTypeId = :accountTypeId"),
    @NamedQuery(name = "AccountType.findByAccountTypeDesc", query = "SELECT a FROM AccountType a WHERE a.accountTypeDesc = :accountTypeDesc")})
public class AccountType implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)  // ADDED THIS LINE
    @Basic(optional = false)
    @NotNull
    @Column(name = "ACCOUNT_TYPE_ID")
    private Integer accountTypeId;
    @Size(max = 50)
    @Column(name = "ACCOUNT_TYPE_DESC")
    private String accountTypeDesc;

    public AccountType() {
    }

    public AccountType(Integer accountTypeId) {
        this.accountTypeId = accountTypeId;
    }

    public Integer getAccountTypeId() {
        return accountTypeId;
    }

    public void setAccountTypeId(Integer accountTypeId) {
        this.accountTypeId = accountTypeId;
    }

    public String getAccountTypeDesc() {
        return accountTypeDesc;
    }

    public void setAccountTypeDesc(String accountTypeDesc) {
        this.accountTypeDesc = accountTypeDesc;
    }

    @Override
    public int hashCode() {
        int hash = 0;
        hash += (accountTypeId != null ? accountTypeId.hashCode() : 0);
        return hash;
    }

    @Override
    public boolean equals(Object object) {
        // TODO: Warning - this method won't work in the case the id fields are not set
        if (!(object instanceof AccountType)) {
            return false;
        }
        AccountType other = (AccountType) object;
        if ((this.accountTypeId == null && other.accountTypeId != null) || (this.accountTypeId != null && !this.accountTypeId.equals(other.accountTypeId))) {
            return false;
        }
        return true;
    }

    @Override
    public String toString() {
        return "Entities.AccountType[ accountTypeId=" + accountTypeId + " ]";
    }

}

以下是会话bean接口：

@Remote
public interface AccountTypeSessionBeanRemote {
    public void createAccountType();
    public void createAccountType(String accDesc);
}

以下是会话bean实现类：

@Stateless
public class AccountTypeSessionBean implements AccountTypeSessionBeanRemote {
    @PersistenceContext(unitName="ExamplePU")
    private EntityManager em;

    @Override
    public void createAccountType(String accDesc) {
        AccountType emp = new AccountType(accDsc);
       try {
           this.em.persist(emp);
           System.out.println("after persist");
       } catch(Exception ex) {
           System.out.println("ex: " + ex.getMessage());
       }
    }
}

以下是Main类：

public class Main {
    @EJB
    private static AccountTypeSessionBeanRemote accountTypeSessionBean;

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        accountTypeSessionBean.createAccountType("test");
    }
}

以下是错误：

INFO: ex: Object: Entities.AccountType[ accountTypeId=null ] is not a known entity type.

Answer 1

由于“accountTypeId字段为空”，您没有收到错误。如错误消息所示，发生错误是因为“Entities.AccountType [accountTypeId = null]不是已知的实体类型”。

最可能的原因是AccountType未注明@Entity。添加它可能会解决此问题。另外，使用

是有意义的

@GeneratedValue(strategy = GenerationType.IDENTITY)

而不是AUTO。 Auto表示提供程序根据目标数据库的功能选择策略。根据表格定义，似乎很清楚首选策略是IDENTITY。

Answer 2

我更改了create table query，如下所示：

create table example.account_type(
    account_type_id INT NOT null PRIMARY KEY,
    account_type_desc varchar(20)
);

然后必须将以下行添加到实体类（Netbeans不添加）：

@GeneratedValue(strategy = GenerationType.AUTO)

Java EE 6，EJB 3.1，JPA 2.0 - 在数据库中插入记录时出错

2 个答案: