休眠：org.hibernate.hql.ast.QuerySyntaxException：意外的標記

Question

我使用Hibernate和我有這個疑問：

List<Person> list = sess.createQuery("from Person").list(); 

有了這個說法，我從數據庫中獲取的所有人員。 但現在，我只想要一些人。

我的數據庫方案：

項目< - Project_Person - >人員

所以我只希望這是一個項目的成員人數。

隨着數據庫中，我得到了想要的結果的SQL語句：

select * from Person inner join Project_Person 
    on person_id = id 
    where project_id = 1; 

所以我想，我可以用Hibernate這樣寫：

List<Person> list = 
    sess.createQuery(
     "from Person inner join Project_Person 
      on person_id = id 
      where project_id = "+projectId).list(); 

但在這裏我得到一個錯誤：

SERVE: Servlet.service() for servlet myproject3 threw exception 
org.hibernate.hql.ast.QuerySyntaxException: unexpected token: on near line 1, column 65 [from com.mydomain.myproject.domain.Person inner join Project_Person on person_id = id where project_id = 1] 
at org.hibernate.hql.ast.QuerySyntaxException.convert(QuerySyntaxException.java:54) 
at org.hibernate.hql.ast.QuerySyntaxException.convert(QuerySyntaxException.java:47) 
at org.hibernate.hql.ast.ErrorCounter.throwQueryException(ErrorCounter.java:82) 
at org.hibernate.hql.ast.QueryTranslatorImpl.parse(QueryTranslatorImpl.java:284) 
at org.hibernate.hql.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:182) 
at org.hibernate.hql.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:136) 
at org.hibernate.engine.query.HQLQueryPlan.<init>(HQLQueryPlan.java:101) 
at org.hibernate.engine.query.HQLQueryPlan.<init>(HQLQueryPlan.java:80) 
at org.hibernate.engine.query.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:124) 
at org.hibernate.impl.AbstractSessionImpl.getHQLQueryPlan(AbstractSessionImpl.java:156) 
at org.hibernate.impl.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:135) 
at org.hibernate.impl.SessionImpl.createQuery(SessionImpl.java:1770) 
at sun.reflect.GeneratedMethodAccessor33.invoke(Unknown Source) 
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25) 
at java.lang.reflect.Method.invoke(Method.java:597) 
at org.hibernate.context.ThreadLocalSessionContext$TransactionProtectionWrapper.invoke(ThreadLocalSessionContext.java:344) 
at $Proxy26.createQuery(Unknown Source) 
... 

有沒有人有一個想法這裏有什麼問題？

最好的問候。

新的錯誤：

SERVE: Servlet.service() for servlet myproject3 threw exception 
org.hibernate.QueryException: could not resolve property: project of: com.mydomain.myproject.domain.Person [from com.mydomain.myproject.domain.Person p where p.project.id = :id] 

N：M關係：

@ManyToMany(cascade = CascadeType.ALL) 
@JoinTable(name = "Project_Person", 
    joinColumns = {@JoinColumn(name="project_id", referencedColumnName="id")}, 
    inverseJoinColumns = {@JoinColumn(name="person_id", referencedColumnName="id")} 
) 
private Set<Person> persons = new HashSet<Person>(); 


@ManyToMany(mappedBy="persons") 
private Set<Project> projects = new HashSet<Project>(); 

完全錯誤

Hibernate: select project0_.id as id1_, project0_.createDate as create2_1_, project0_.description as descript3_1_, project0_.name as name1_ from Project project0_ where project0_.id=1 
Hibernate: select person0_.id as id0_0_, project2_.id as id1_1_, person0_.email as email0_0_, person0_.firstName as firstName0_0_, person0_.lastName as lastName0_0_, project2_.createDate as create2_1_1_, project2_.description as descript3_1_1_, project2_.name as name1_1_ from Person person0_ inner join Project_Person projects1_ on person0_.id=projects1_.person_id inner join Project project2_ on projects1_.project_id=project2_.id where project2_.id=? 
15.12.2010 16:42:26 org.apache.catalina.core.ApplicationDispatcher invoke 
SERVE: Servlet.service() for servlet myproject3 threw exception 
java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to com.mydomain.myproject.domain.Person 

Answer 1

HQL查詢被寫入侵害的對象模型，而不是針對數據庫架構。

因此，您的查詢取決於您如何映射人與項目之間的關係。例如，在Person具有通過project財產多到一個關係到Project，查詢會是這樣的：

List<Person> list = sess.createQuery(
    "from Person p where p.project.id = :id") 
    .setParameter("id", projectId) 
    .list(); 

編輯：在很多一對多關係的情況下，你需要

select p from Person p join p.projects proj where proj.id = :id 

而不是通過字符串連接傳遞參數是一種不好的做法，請使用setParameter()來代替。