使用查询生成器从另一个 table 中查询 table 中的所有行 link

Question

我有一个 table，其中只有几行变量。这一行 link 到另一行 table 有更多行。

说白了，1table叫做Connection。变量是：

name, groupname, etc.

组名应该 link 到第二个 table 称为 ConnectionGroup。变量是：

name, id.

我的想法是通过名称查询 ConnectionGroup table。 Connection和ConnectionGroup的抽象class是这样的：

public abstract class Connection_ {

    public static volatile SingularAttribute<Connection, String>name;
    public static volatile SingularAttribute<Connection, String>host;
    public static volatile SetAttribute<Connection, ConnectionGroup>connectionGroups;
}

public abstract class ConnectionGroup_ {

    public static volatile SingularAttribute<ConnectionGroup, String> name;
    public static volatile SingularAttribute<ConnectionGroup, Long> id;

}

要查询这个，我假设我必须加入这 2 个 table，然后只查询它们。这是我试过的代码：

@PersistenceContext
private EntityManager em;

public List<Connection> retrieveAll( String groupFilter, int start, int length) {
  ServiceUtil.requireAdmin();
  CriteriaBuilder cb = em.getCriteriaBuilder();
  CriteriaQuery<Connection> q = cb.createQuery(Connection.class);
  Root<Connection> c = q.from(Connection.class);
  Join<Connection, ConnectionGroup> join = c.join(Connection_.connectionGroups);
  q.select(c);

  c.fetch(Connection_.connectionGroups).fetch(ConnectionGroup_.id);
  Predicate groupPredicate = cb.equal(
  c.get(Connection_.connectionGroups), "%" + groupFilter + "%");
  q.where(groupPredicate);

  List<Connection> results = em.createQuery(q).setFirstResult(start)
            .setMaxResults(length).getResultList();
  for (Connection conn : results) {
      logger.info( "getconnectionGroups =["+ conn.getConnectionGroups() + "]");
      for (ConnectionGroup conngroup : conn.getConnectionGroups()) {
            logger.info("connectiongroups = [" + conngroup.getName() + "]");
      }
  }
}

我尝试过的东西，改变这个：

Predicate groupPredicate = cb.equal(
  c.get(Connection_.connectionGroups), "%" + groupFilter + "%");

对此：

Predicate groupPredicate = cb.equal(join.get(ConnectionGroup_.name),
            "%" + groupFilter + "%");

改变这个：

Join<Connection, ConnectionGroup> join = c
            .join(Connection_.connectionGroups);

为此：

Join<Connection, ConnectionGroup> join = c.join("connectionGroups");

当我尝试这些方法时，我总是收到异常 Cannot join to attribute of basic type

我也试过把代码改成这样：

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Connection> q = cb.createQuery(Connection.class);
Root<Connection> c = q.from(Connection.class);
Join<Connection, ConnectionGroup> join = (Join<Connection, ConnectionGroup>) c.fetch(Connection_.connectionGroups);
q.select(c);

Predicate groupPredicate = cb.equal(join.get(ConnectionGroup_.name), "%" + groupFilter + "%");
q.where(groupPredicate);
List<Connection> results = em.createQuery(q).setFirstResult(start)
            .setMaxResults(length).getResultList();

其中return一个例外：

query specified join fetching, but the owner of the fetched association was not present in the select list

我使用这些站点作为编写代码的参考：

JPA CriteriaBuilder using fetch joins and result objects

JPA 2 Criteria Fetch Path Navigation

wiki.eclipse.org

developer.com

如何查询 ConnectionGroup table 中的名称？我的方法错了吗？

Answer 1

如果您尝试通过字符串匹配使用名称过滤，CriteriaBuilder.like() 比 CriteriaBuilder.equal() 更合适。

要实现您要查询的内容，您可以使用 ff.日联QL查询：

SELECT DISTINCT conn FROM Connection conn JOIN conn.connectionGroups connGrp
WHERE connGrp.name LIKE :groupFilter

将 JP QL 转换为 CriteriaQuery，您将拥有：

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Connection> q = cb.createQuery(Connection.class);
Root<Connection> conn = q.from(Connection.class);
Join<Connection, ConnectionGroup> connGrp = conn.join(Connection_.connectionGroups);
q.select(conn).distinct(true);
ParameterExpression<String> param = cb.parameter(String.class, "%"+ groupFilter + "%");
q.where(cb.like(connGrp.get(ConnectionGroup_name), param));

使用查询生成器从另一个 table 中查询 table 中的所有行 link

query all row in table that is link from another table using query builder

java

jpa

criteria-api

criteriaquery