Kde*_*per 13 hibernate jpa hql jpql jpa-2.0
在hibernate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
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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...
}
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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...
}
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但是当Hibernate 3.5(JPA 2)启动时,我收到此错误:
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 ]
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是否包含列表(u.securityRoles)的select不可能?我应该创建2个单独的查询吗?
Pas*_*ent 10
没有NEW(选择标量值和集合值路径表达式)的查询无效,所以我不认为添加一个NEW会使事情有效.
为了记录,这是JPA 2.0规范在4.8 SELECT子句中所述的内容:
SELECT子句具有以下语法:
Run Code Online (Sandbox Code Playgroud)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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