这是可能的:JPA / Hibernate查询结果中的列表属性? [英] Is this possible: JPA/Hibernate query with list property in result?
问题描述
select new org。 test.userDTO(u.id,u.name,u.securityRoles)
FROM用户u
其中u.name =:name
userDTO class:
public class UserDTO {
private Integer ID;
私人字符串名称;
私人列表< SecurityRole> securityRoles;
$ b $ public UserDTO(Integer id,String name,List< SecurityRole> securityRoles){
this.id = id;
this.name = name;
this.securityRoles = securityRoles;
}
...获得者和设置者...
}
用户实体:
@Entity
公共类用户{
@id
私人整数ID;
私人字符串名称;
@ManyToMany
@JoinTable(name =user_has_role,
joinColumns = {@JoinColumn(name =user_id)},
inverseJoinColumns = {@JoinColumn (name =security_role_id)}
)
private List< SecurityRole> securityRoles;
...获得者和设置者...
}
但是当Hibernate 3.5(JPA 2)启动时,我得到这个错误:
org.hibernate.hql.ast.QuerySyntaxException:Unable在类[org.test.UserDTO]上找到合适的
构造函数[SELECT new org.test.UserDTO(u.id,
u.name,u.securityRoles)FROM nl.test.User u WHERE u.name =:name]
是一个包含列表的选择(u.securityRoles )因此不可能?我应该创建2个单独的查询吗?
code>(选择标量值和集合值路径表达式)无效,所以我不认为添加 NEW
会使事情发挥作用。 为了记录,这就是JPA 2.0规范在 4.8 SELECT子句部分所说的内容:
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)
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?
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)
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