Question

我目前正尝试通过JPQL中的构造函数进行查询。例如：

em这里是EntityManager

em.createQuery("SELECT NEW {packagePath}.UserDTO(name, status) FROM User", UserDTO.class);

但我想在status中使用Case When，所以我试图查询：

em.createQuery("SELECT NEW {packagePath}.UserDTO(name, case when status = 'A' then '1' else '0' end as status) FROM User", UserDTO.class);

我试图在SQL Developer中查询：

SELECT name, CASE WHEN status = 'A' THEN '1' ELSE '0' END AS status FROM User

它完美无缺。

样本DTO：

UserDTO.class

public class UserDTO {
    private String name;
    private Character status;

    // Suggested by Viacheslav Shalamov but not working.
    public UserDTO(){}

    // EDIT
    public UserDTO(String name, Character status){
         this.name = name;
         this.status = status;
    }
}

编辑：User.class

@Entity
@Table(name = "User")
public class User {

    @Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    @Column(name = "name", length = 100, columnDefinition = "VARCHAR2(100)")
    private String name;
    @Column(name = "pc_no", length = 3, columnDefinition = "VARCHAR2(3)")
    private String pc_no;
    @Column(name = "status")
    private Character status;

    // constructor / get and set
}

错误：

Unable to locate appropriate constructor on class. Expected arguments are: java.lang.String, java.lang.String.

还有其他方法可以包含Case When方法吗？如果没有，请给我一些关于如何以不同的方法做到这一点的建议。谢谢。

Answer 1

您应该提供一种方法，在失败的查询中将VARCHAR2 status从String转换为Character。

试试这个：

public class UserDTO {
    private String name;
    private Character status;

    public UserDTO(String name, String status){
        this.name = name;
        this.status = status.charAt(0);
    }
}

在构造函数JPQL中使用Case

1 个答案: