Question

实体：帖子，空间和个人资料。总结：

class Post { Space space; String text; }

class Space { List<Profile> members; }

class Profile { String username; List<Space> spaces; }

如何在@AdditionalCriteria上设置Post仅返回属于当前用户所属空间的帖子。

到目前为止我所尝试的内容如下。

＃1 - ：space.members中的currentUserProfile

@AdditionalCriteria(":currentUserProfile in (this.space.members)")

Profile profile = new Profile();
em.setProperty("currentUserProfile", profile);

结果：

Exception [EclipseLink-6015] (Eclipse Persistence Services - 2.6.2.v20151217-774c696): org.eclipse.persistence.exceptions.QueryException
Exception Description: Invalid query key [space] in expression.
Query: ReadObjectQuery(name="readPost" referenceClass=Post )

＃2 - ：space.members的currentUserProfile成员

@AdditionalCriteria(":currentUserProfile member of this.space.members")

Profile profile = new Profile();
em.setProperty("currentUserProfile", profile);

结果：

Caused by: org.postgresql.util.PSQLException: Não pode inferir um tipo SQL a ser usado para uma instância de br.com.senior.social.model.profile.Profile. Use setObject() com um valor de Types explícito para especificar o tipo a ser usado.
    at org.postgresql.jdbc.PgPreparedStatement.setObject(PgPreparedStatement.java:1039)

英文：“无法推断用于_的实例的SQL类型。使用带有显式值的类型的setObject（）来设置要使用的类型。”

＃3 - profile.spaces中的this.space

@AdditionalCriteria("this.space in (select p.spaces from Profile p where p.username = :currentUsername)")

String username = "foo";
em.setProperty("currentUsername", username);

结果：

Exception [EclipseLink-0] (Eclipse Persistence Services - 2.6.2.v20151217-774c696): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Problem compiling [this.space in (select p.spaces from Profile p where p.username = :currentUsername)]. 
[131, 139] The state field path 'p.spaces' cannot be resolved to a collection type.

＃4 - ：currentUsername =（subselect）

@AdditionalCriteria(":currentUsername = (select p.username from Profile p where p.username = :currentUsername and this.space in (p.spaces))")

String currentUsername = "foo";
em.setProperty("currentUsername", currentUsername);

结果：

org.postgresql.util.PSQLException: ERROR: relation "post" does not exist
  Posição: 423

EclipseLink在不使用描述符（前缀）的情况下为表发出SQL。

Answer 1

使用不那么优雅的解决方案来解决它：

首先，我在Post-Space关系中添加了一个原始ID属性：

class Post {
    @Column(name = "SPACE_ID", updatable = false, insertable = false)
    private String spaceId;
}

然后在附加标准中引用它：

@AdditionalCriteria("this.spaceId in :currentUserSpaceIds")

并将:currentUserSpaceIds初始化为当前用户所属的实际空格。

EntityManager em = ...;
Profile profile = ...; // current user profile
List<Space> spaces = profile.getSpaces();
List<String> spaceIds = new ArrayList<>();
for (Space space : spaces) { spaceIds.add(space.getId()); }
em.setProperty("currentUserSpaceIds", spaceIds);

在AdditionalCriteria

1 个答案: