hibernate - 这可能吗：JPA/Hibernate query with list property in result？

Question

在休眠中，我想运行这个 JPQL / HQL 查询：

select new org.test.userDTO( u.id, u.name, u.securityRoles)
FROM User u
WHERE u.name = :name

userDTO 类：

public class UserDTO {
   private Integer id;
   private String name;
   private List<SecurityRole> securityRoles;

   public UserDTO(Integer id, String name, List<SecurityRole> securityRoles) {
     this.id = id;
     this.name = name;
     this.securityRoles = securityRoles;
   }

   ...getters and setters...
}

用户实体：

@Entity
public class User {

  @id
  private Integer id;

  private String name;

  @ManyToMany
  @JoinTable(name = "user_has_role",
      joinColumns = { @JoinColumn(name = "user_id") },
      inverseJoinColumns = {@JoinColumn(name = "security_role_id") }
  )
  private List<SecurityRole> securityRoles;

  ...getters and setters...
}

但是当 Hibernate 3.5 (JPA 2) 启动时，我得到这个错误：

org.hibernate.hql.ast.QuerySyntaxException: Unable to locate appropriate 
constructor on class [org.test.UserDTO] [SELECT NEW org.test.UserDTO (u.id,
u.name, u.securityRoles) FROM nl.test.User u WHERE u.name = :name ]

是否无法选择包含列表 (u.securityRoles) 的结果？我应该只创建 2 个单独的查询吗？

Answer 1

没有NEW（选择标量值和集合值路径表达式）的查询无效，因此我认为添加 aNEW不会使事情正常进行。

作为记录，这是 JPA 2.0 规范在第4.8 节 SELECT 子句中所说的：

SELECT 子句具有以下语法：

select_clause ::= SELECT [DISTINCT] select_item {, select_item}*
select_item ::= select_expression [ [AS] result_variable]
select_expression ::=
         single_valued_path_expression |
         scalar_expression |
         aggregate_expression |
         identification_variable |
         OBJECT(identification_variable) |
         constructor_expression
constructor_expression ::=
         NEW constructor_name ( constructor_item {, constructor_item}* )
constructor_item ::=
         single_valued_path_expression |
         scalar_expression |
         aggregate_expression |
         identification_variable
aggregate_expression ::=
         { AVG | MAX | MIN | SUM } ([DISTINCT] state_field_path_expression) |
         COUNT ([DISTINCT] identification_variable | state_field_path_expression |
                  single_valued_object_path_expression)

Answer 2

我相信你需要在你的UserDTO类中声明一个 0-arg 构造函数。

编辑：或者一个构造函数，Integer而不是int作为第一个参数。当使用反射查找构造函数时，Hibernate 可能不会将它们视为“兼容”类型。

基本上，我会专注于Unable to locate appropriate constructor on class [...UserDTO]信息的一部分。

Answer 3

我认为您应该尝试以下方法：

select new org.test.userDTO( u.id, u.name, u.securityRoles) AS uDTO,
  uDTO.setRoles(u.securityRoles)
 FROM User u
 WHERE u.name = :name