Question

我需要具有客户ID的客户列表，该ID的客户（客户）和客户名称。问题是该表来自没有PK的dblink。

@Entity

@Table(name = "mytable", schema = "myschema")

public class Client {

    @Column(name = "clientid")
    @Id
    private Integer clientid;

    @Column(name = "client")
    private Integer client;

    @Column(name = "name")
    private String name;
}

使用此代码，我得到的名字一遍又一遍地重复到列表的末尾，因为没有PK，没有PK就无法启动应用程序。当我将client列设置为PK时，也会发生类似的情况（我正确地获得了所有名称，但是clientid字段显示了错误的值。是否有解决方法？

样本数据：

错误的数据（将clientid设置为@Id时）：

[
  {
    "clientid": 99,
    "client": 81,
    "name": "Organization 1"
  },
  {
    "clientid": 99,
    "client": 81,
    "name": "Organization 1"
  },
  {
    "clientid": 99,
    "client": 81,
    "name": "Organization 1"
  }
]

错误的数据（将客户端设置为@Id时）：

[
  {
    "clientid": 99,
    "client": 81,
    "name": "Organization 1"
  },
  {
    "clientid": 3,
    "client": 99,
    "name": "Organization 2"
  },
  {
    "clientid": 3,
    "client": 127,
    "name": "Organization 3"
  }
]

我应该得到的：（在所有情况下，clientid都是正确的）

[
  {
    "clientid": 99,
    "client": 81,
    "name": "Organization 1"
  },
  {
    "clientid": 3,
    "client": 99,
    "name": "Organization 2"
  },
  {
    "clientid": 1,
    "client": 127,
    "name": "Organization 3"
  }
]

Answer 1

您应该创建一个复合键来正确处理您的数据。首先，创建另一个代表键的类：

public class ClientPK implements Serializable{

    private static final long serialVersionUID = 1L;

    private Integer clientid;

    private Integer client;

    /*Constructor, getters and setters here*/
}

然后按如下所示更新类Client：

@Entity
@Table(name = "mytable", schema = "myschema")
@IdClass(ClientPK.class)
public class Client {

    @Column(name = "clientid")
    @Id
    private Integer clientid;

    @Column(name = "client")
    private Integer client;

    @Column(name = "name")
    private String name;
}

最好在hashCode类中实现equals和ClientPK方法。

带有JPA的组合键，用于没有PK的表

1 个答案: