Is this possible: JPA/Hibernate query with list property in result?

问题

In hibernate I want to run this JPQL / HQL query:

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

userDTO class:

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...
}

User Entity:

@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...
}

But when Hibernate 3.5 (JPA 2) starts I get this error:

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 ]

Is a select that includes a list (u.securityRoles) as a result not possible? Should I just create 2 seperate queries?

回答1:

The query without the NEW (selecting a scalar value and a collection-valued path expression) isn't valid so I don't think that adding a NEW will make things work.

For the record, this is what the JPA 2.0 specification says in the section 4.8 SELECT Clause:

The SELECT clause has the following syntax:

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)

回答2:

I believe you need to declare a 0-arg constructor in your UserDTO class.

EDIT: Or a constructor which takes Integer instead of int as the first argument. When looking up constructors using reflection Hibernate might not treat them as 'compatible' types.

Basically, I would focus on the Unable to locate appropriate constructor on class [...UserDTO] part of the message.

回答3:

I think that you should try something like:

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

来源：https://stackoverflow.com/questions/2678634/is-this-possible-jpa-hibernate-query-with-list-property-in-result

标签

hibernate

jpa

hql

jpql

jpa-2.0