QueryDSL中的查询谓词

Question

环境是Java、Spring-boot、Hibernat、QueryDSL、MySQL。

我有表结构

艾碧索

+----+-------------+--------
| id | address_id  | eventno
+----+-------------+--------
|  5 |         27  | F123
|  6 |         30  | F456
|  7 |         45  | F789
+----+-------------+--------

@Entity
public class Episode {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@NotEmpty
private String eventno;
@ManyToOne(cascade = CascadeType.ALL)
private Address address;

+----+--------------+--------------+------------+-----------+
| id | episode_role | primary_flag | episode_id | person_id |
+----+--------------+--------------+------------+-----------+
| 19 | Buyer        |              |          5 |         1 |
| 20 | Subject      |              |          5 |         2 |
| 23 | Witness      |              |          6 |         3 |
| 24 | Child        |              |          6 |         4 |
| 27 | Buyer        |              |          5 |         3 |
| 63 | Investor     |              |          5 |         4 |
| 64 | Subject      |              |          7 |         1 |
| 65 | Subject      |              |          7 |         3 |

@Entity
public class EpisodePerson {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;

@ManyToOne
@Valid
private Person person;

@ManyToOne
private Episode episode;

+----+-----------+----------+
| id | firstname | surname  |
+----+-----------+----------+
|  1 | Clint     | eastwood |
|  2 | Angelina  | joilee   |
|  3 | Brad      | pitt     |
|  4 | Jennifer  | aniston  |

@Entity
@Table(uniqueConstraints = @UniqueConstraint(columnNames = {"nia"}))
public class Person {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String surname;
private String firstname;
private String gender;

            BooleanBuilder where = new BooleanBuilder();
        if (pagination.getFilterBy().getMapOfFilters().get("eventno")!=null) {
            where.and(qEpisode.eventno.containsIgnoreCase(pagination.getFilterBy().getMapOfFilters().get("eventno")));
        }
        if (pagination.getFilterBy().getMapOfFilters().get("address")!=null) {
            where.and(qEpisode.address.formattedAddress.containsIgnoreCase(pagination.getFilterBy().getMapOfFilters().get("address")));
        }
        where.and(qEpisode.creatingUser.eq(user));
        List<Episode> e = episodeRepository.findAll(where);

更新

为了澄清起见，支持UI视图的DTO包含“casename”属性。它是在将域对象转换为DTO时在服务层创建的：

episodeDashboard.setNames(episodePersonList.get(0).getPerson().getSurname().toUpperCase() +" & " +episodePersonList.get(1).getPerson().getSurname().toUpperCase());

Answer 1

不容易，除非您将一些处理委托给数据库。

如果我们可以在数据库层填充case_name属性，而不是作为应用程序逻辑中的派生属性，那么前端代码就变得很简单。

我们可以通过一个视图来做到这一点。它的确切定义将取决于您的数据库，但是输出如下所示：

+------------+-------------------------+
| epsiode_id | case_name               |
+------------+-------------------------+
|  5         |        Eastwood & Joilee| 
|  6         |           Pitt & Aniston| 
|  7         |           Aniston & Pitt| 
+------------+-------------------------+

CREATE VIEW episode_summary_vw as
SELECT ep.episode_id, GROUP_CONCAT(p.surname SEPARATOR ' & ')
FROM episode_person ep
INNER JOIN person p on p.id = ep.person_id
GROUP BY ep.episode_id;
-- todo: needs limit to first 2 records

@Entity
@Table(name = "episodes")
@SecondaryTable(name = "episode_summary_vw", primaryKeyJoinColumna = @PrimaryKeyJoinColumn(name="episode_id", reference_column_name="id"))
public class Episode {

    @Column(name ="case_name", table = "episode_summary_vw")
    private String caseName;
}

然后对该属性进行筛选和排序，就像对任何其他字段一样：

if (pagination.getFilterBy().getMapOfFilters().get("caseName")!=null) {

    where.and(qEpisode.caseName.containsIgnoreCase(pagination.getFilterBy().
       getMapOfFilters().get("caseName")));
}