Hibernate 搜索多个标记的文章-休眠多对多
我存储具有多个标签的文章,如下所示:Hibernate 搜索多个标记的文章-休眠多对多,hibernate,many-to-many,Hibernate,Many To Many,我存储具有多个标签的文章,如下所示: import java.util.ArrayList; import java.util.Arrays; import java.util.List; import javax.persistence.Entity; import javax.persistence.GeneratedValue; import javax.persistence.Id; import javax.persistence.ManyToMany; @Entity publi
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.ManyToMany;
@Entity
public class Article {
@Id
@GeneratedValue
private Integer id;
@ManyToMany
private List<Tag> tags;
private String subject;
}
我想搜索包含一些标签的文章,例如,像这个问题hibernate
和many-to-many
。所以我试着:
import static org.hibernate.criterion.Restrictions.*;
// ...
Criteria criteria = session.createCriteria(Article.class);
Criteria tagCriteria = criteria.createCriteria("tags");
tagCriteria.add(and(eq("name", "hibernate"), eq("name", "many-to-many")));
@SuppressWarnings("unchecked")
List<Article> list = criteria.list();
当replace和
到或
返回过多的项目时,因为生成的SQL是。。。其中(tag1_u2;.name=?或tag1_2;.name=?)
我想要的正是梅的样子:
select
this_.id as id1_1_,
this_.subject as subject1_1_,
tags3_.Article_id as Article1_,
tag1_.name as tags4_,
tag1_.name as name0_0_,
tag1_.description as descript2_0_0_
from
Article this_
inner join Article_Tag tags3_ on this_.id=tags3_.Article_id
inner join Tag tag1_ on tags3_.tags_name=tag1_.name
// following 2 lines are added
inner join Article_Tag tags4_ on this_.id=tags4_.Article_id
inner join Tag tag5_ on tags4_.tags_name=tag5_.name
where
tag1_.name=? // assign 'hibernate'
and
tag5_.name=? // assign 'many-to-many'
当我只使用HQL时是否可能?您应该尝试类似的方法:
Criteria criteria = getSession().createCriteria(Article.class);
int i=0;
for ( String tagName : tagNames ) {
String aliasName = "alias_" + i;
criteria.createAlias("tags", aliasName, Criteria.INNER_JOIN);
criteria.add( Restrictions.eq(aliasName+".name",tagName) );
i++;
}
实际上,问题是当你做一个简单的内部连接时,你已经用标记连接了你的结果,但问题是数据实际上在两行上。。。因此,如果您加入两次,就可以在一行中获取数据
我认为它不是很优雅,但应该可以工作…谢谢你的回答,但它抛出了
org.hibernate.QueryException:replicate association path:tags
的异常。标准字符串是criteriampl(com.cremoi.model.article.article:this[Subcriteria(tags:alias\u 0),Subcriteria(tags:alias\u 1)][alias\u 0.name=hibernate,alias\u 1.name=many to many])
,看起来效果不错,但是没有。我找到了另一个解决方案,使用了连接词和分离标准:这很有效,但也不优雅。坦率地说,这太难看了(
select
this_.id as id1_1_,
this_.subject as subject1_1_,
tags3_.Article_id as Article1_,
tag1_.name as tags4_,
tag1_.name as name0_0_,
tag1_.description as descript2_0_0_
from
Article this_
inner join Article_Tag tags3_ on this_.id=tags3_.Article_id
inner join Tag tag1_ on tags3_.tags_name=tag1_.name
// following 2 lines are added
inner join Article_Tag tags4_ on this_.id=tags4_.Article_id
inner join Tag tag5_ on tags4_.tags_name=tag5_.name
where
tag1_.name=? // assign 'hibernate'
and
tag5_.name=? // assign 'many-to-many'
Criteria criteria = getSession().createCriteria(Article.class);
int i=0;
for ( String tagName : tagNames ) {
String aliasName = "alias_" + i;
criteria.createAlias("tags", aliasName, Criteria.INNER_JOIN);
criteria.add( Restrictions.eq(aliasName+".name",tagName) );
i++;
}