如何使用 JPA 和 Hibernate 加入两个不相关的实体 [英] How to join two unrelated entities using JPA and Hibernate

问题描述

我有两张表 - 一张包含地址，另一张包含照片.它们之间唯一的公共字段是 PersonID.这些被映射到两个 POJO 类地址和照片.我能够通过创建条件和对字段添加限制来获取这些表中的详细信息.我们应该如何在两个表上编写连接.是否可以将结果作为两个对象 - 地址和照片.

I have two tables - one containing Address and another containing Photographs. The only common field between them is the PersonID. These were mapped to two POJO Classes Address and Photo. I was able to fetch details in these tables by creating criteria and adding restrictions on the fields . How should we write a join on the two tables. Is it possible to get the result as two objects -Address and Photo.

我想做一个左连接，这样我也可以得到没有照片的人的记录.我已经读到这只能使用 hql 但也可以使用标准来完成吗?

I want to do a left join so that i can get records of persons without photos as well. I have read that this is possible only using hql but Can this be done using criterias as well?

推荐答案

您可以轻松编写 HQL 查询，使用 Theta Join 将结果作为两个对象返回(如 Adrian 所述).下面是一个例子:

You can easily write HQL query that will return result as two objects using Theta Join (as Adrian noted). Here is an example:

String queryText = "select address, photo from Address address, Photo photo " 
                 + " where address.personID=photo.personId";
List<Object[]> rows = session.createQuery(queryText).list();

for (Object[] row: rows) {
    System.out.println(" ------- ");
    System.out.println("Address object: " + row[0]);
    System.out.println("Photo object: " + row[1]);
}

如您所见，查询返回代表每个提取行的 Object[] 数组列表.此数组的第一个元素将包含一个对象和第二个元素 - 另一个.

As you can see query returns list of Object[] arrays that represents each fetched row. First element of this array will contain one obejct and second element - another.

在左连接的情况下，我认为您需要使用本机 SQL 查询(而不是 HQL 查询).您可以在此处执行此操作:

In case of left join I think you need to use native SQL query (not HQL query). Here how you can do this:

String queryText = "select address.*, photo.* from ADDRESS address 
                    left join PHOTO photo on (address.person_id=photo.person_id)";

List<Object[]> rows = sess.createSQLQuery(queryText)
                          .addEntity("address", Address.class)
                          .addEntity("photo", Photo.class)
                          .list();

这应该适用于您的情况.

This should work for your case.

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