Hibernate:org.hibernate.hql.ast。QuerySyntaxException:意想不到的令牌

I am using Hibernate and I have this query:

我在使用Hibernate，我有这个查询:

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

With this statement, I get all persons from the database. But now, I only want some persons.

通过这个语句，我从数据库中获得所有人。但是现在，我只想要一些人。

My database scheme:

我的数据库方案:

Project <- Project_Person -> Person

项目<- Project ect_person -> Person

So I only want Persons which are a member of a project.

所以我只需要项目成员。

With the SQL statement on the database I get the desired result:

通过数据库上的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;

So I thought, I can write this with Hibernate:

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();

But here I get an error:

但是这里我有一个错误

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)
 ...

Does anyone has an idea what's wrong here?

有人知道这里有什么问题吗?

Best Regards.

致以最亲切的问候。

New Error:

新的错误:

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 relation:

护士: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>();

Full Error

完整的错误

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

1 个解决方案

#1

HQL queries are written against the object model, not against the database schema.

HQL查询是针对对象模型而不是针对数据库模式编写的。

Therefore your query depends on how you mapped the relationship between persons and projects. For example, in Person has a many-to-one relationship to Project via project property, the query will look like this:

因此，查询取决于您如何映射人员和项目之间的关系。例如，通过项目属性与项目具有多对一关系，查询将如下所示:

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

EDIT: In the case of many-to-many relationship you need

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

Also not that passing parameters via string concatenation is a bad practice, use setParameter() instead.

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

#1