JPA +如何选择使用第三个表作为一对多关系中的连接记录？

Question

我想要获取具有特定Tag的所有Query对象。

---

实体类新：

public abstract class IdentifiableEntity implements Serializable{ 
     private long id; 
     //set/get 
     } 

    public class Query extends IdentifiableEntity { 
     private String query; 
     private Set<Tag> tags; 
     //set/get 
    } 

    public class Tag extends IdentifiableEntity { 
     private String tagName; 
     //set/get  
    } 

---

表：

CREATE TABLE IF NOT EXISTS `mydb`.`tbl_map_query_tag` (
     `map_query_tag_id` INT NOT NULL AUTO_INCREMENT, 
     `query_id` INT NOT NULL, 
     `tag_id` INT NOT NULL, 
     `comments` VARCHAR(100) NULL, 
     `map_query_tag_fut1` VARCHAR(45) NULL, 
     `map_query_tag_fut2` VARCHAR(45) NULL, 
     PRIMARY KEY (`map_query_tag_id`), 
     INDEX `fk_tbl_map_query_tag_tbl_query_idx` (`query_id` ASC), 
     INDEX `fk_tbl_map_query_tag_tbl_tag1_idx` (`tag_id` ASC), 
     CONSTRAINT `fk_tbl_map_query_tag_tbl_query` 
     FOREIGN KEY (`query_id`) 
     REFERENCES `mydb`.`tbl_query` (`query_id`) 
     ON DELETE CASCADE 
     ON UPDATE CASCADE, 
     CONSTRAINT `fk_tbl_map_query_tag_tbl_tag1` 
     FOREIGN KEY (`tag_id`) 
     REFERENCES `mydb`.`tbl_tag` (`tag_id`) 
     ON DELETE CASCADE 
     ON UPDATE CASCADE) 
    ENGINE = InnoDB; 


CREATE TABLE IF NOT EXISTS `mydb`.`tbl_query` (
    `query_id` INT NOT NULL AUTO_INCREMENT, 
    `query_desc` VARCHAR(2000) NOT NULL, 
    PRIMARY KEY (`query_id`)) 
ENGINE = InnoDB; 


CREATE TABLE IF NOT EXISTS `mydb`.`tbl_tag` (
    `tag_id` INT NOT NULL AUTO_INCREMENT, 
    `tag_name` VARCHAR(45) NOT NULL, 
    PRIMARY KEY (`tag_id`)) 
ENGINE = InnoDB; 

---

orm.xml中

<?xml version="1.0" encoding="utf-8"?> 
<entity-mappings xmlns="http://java.sun.com/xml/ns/persistence/orm" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xsi:schemaLocation="http://java.sun.com/xml/ns/persistence/orm 
     http://java.sun.com/xml/ns/persistence/orm_2_0.xsd" 
    version="2.0"> 
    <mapped-superclass class="com.qa.models.IdentifiableEntity"> 
      <attributes> 
       <id name="id"> 
        <column name="id" nullable="false" column-definition="INT" /> 
        <generated-value strategy="AUTO" /> 
       </id> 
      </attributes> 
     </mapped-superclass> 
    <entity name="Tag" class="com.qa.models.Tag" cacheable="true"> 
      <table name="tbl_tag" /> 

      <attribute-override name="id"> 
       <column name="tag_id" nullable="false" column-definition="INT" /> 
      </attribute-override> 
      <attributes> 
       <basic name="tagName"> 
        <column name="tag_name" nullable="false" /> 
       </basic> 
      </attributes> 
    </entity> 

    <entity name="Query" class="com.qa.models.Query" cacheable="true"> 
      <table name="tbl_query" /> 
      <attribute-override name="id"> 
       <column name="query_id" nullable="false" column-definition="INT" /> 
      </attribute-override> 
      <attributes> 
       <basic name="query"> 
        <column name="query_desc" nullable="false" /> 
       </basic> 
       <one-to-many name="tags" target-entity="com.qa.models.Tag" fetch="LAZY"> 
        <join-table name="tbl_map_query_tag"> 
         <join-column name="query_id" referenced-column-name="query_id" /> 
         <inverse-join-column name="tag_id" referenced-column-name="tag_id" unique="true" /> 
        </join-table> 
        <cascade> 
         <cascade-all /> 
        </cascade> 
       </one-to-many> 
      </attributes> 
     </entity> 
</entity-mappings>  

等效SQL查询

select q.* from tbl_query q left outer join tbl_map_query_tag tm on q.query_id=tm.query_id left outer join tbl_tag t on t.tag_id=tm.tag_id where t.tag_id=:tagId 

吾道方法是

public List<Query> searchQueryByTag(Tag tag) throws QADAOException { 
     javax.persistence.Query query = getEntityManager().createQuery(SQL_SEL_QUERY_TAG_ID); 
     query.setParameter("tag", tag); 
     List<Query> queries = query.getResultList(); 
     return queries; 

    } 

当我转换此查询到JPQL这是行不通的。我的问题是我如何使用JPQL实现等效于SQL查询结果的结果？

Answer 1

的JPQL是

select q from Query q join q.tags ts where ts in(select t from Tag t where t.id=:tagId) 

而在DAO方法稍加修改

public List<Query> searchQueryByTag(Tag tag) throws QADAOException { 
     javax.persistence.Query query = getEntityManager().createQuery(SQL_SEL_QUERY_TAG_ID); 
     query.setParameter("tag", tag.getId()); 
     List<Query> queries = query.getResultList(); 
     return queries; 

    }