java - hibernate :org.hibernate.hql.ast.QuerySyntaxException:意外的 token

Question

我正在使用 Hibernate 并且我有这个查询:

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

通过这条语句，我从数据库中获取了所有人。但是现在，我只想要一些人。

我的数据库方案:

项目 <- Project_Person -> Person

所以我只想要作为项目成员的人员。

通过数据库中的 SQL 语句，我得到了想要的结果:

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

所以我想，我可以用 Hibernate 写这个:

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

但是这里我得到一个错误:

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)
 ...

有人知道这里出了什么问题吗？

最好的问候。

新错误:

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关系:

@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>();

完全错误

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 查询是针对对象模型而不是数据库模式编写的。

因此，您的查询取决于您如何映射人与项目之间的关系。例如，在 Person 中，通过 project 属性与 Project 存在多对一关系，查询将如下所示:

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

编辑:在多对多关系的情况下，您需要

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

也不是说通过字符串连接传递参数是一种不好的做法，而是使用 setParameter()。