Hibernate CriteriaBuilder 加入多个表

Question

我正在尝试使用休眠条件生成器加入 4 个表..
下面分别是表格.. `

@Entity
public class BuildDetails {
    @Id
    private long id;
    @Column
    private String buildNumber; 
    @Column
    private String buildDuration;
    @Column
    private String projectName;

}   

@Entity
public class CodeQualityDetails{
    @Id
    private long id;
    @Column
    private String codeHealth;
    @ManyToOne
    private BuildDetails build; //columnName=buildNum
}

@Entity
public class DeploymentDetails{
    @Id
    private Long id;
    @Column
    private String deployedEnv;
    @ManyToOne
    private BuildDetails build; //columnName=buildNum
}

@Entity
public class TestDetails{
    @Id
    private Long id;
    @Column
    private String testStatus;
    @ManyToOne
    private BuildDetails build; //columnName=buildNum
}

---

在这 4 个表中，我想为 MySQL 执行以下 sql 脚本：

SELECT b.buildNumber, b.buildDuration,
       c.codeHealth, d.deployedEnv, t.testStatus
FROM BuildDetails b
INNER JOIN CodeQualityDetails c ON b.buildNumber=c.buildNum
INNER JOIN DeploymentDetails d ON b.buildNumber=d.buildNum
INNER JOIN TestDetails t ON b.buildNumber=t.buildNum
WHERE b.buildNumber='1.0.0.1' AND
      b.projectName='Tera'

---

那么，如何使用 Hibernate CriteriaBuilder 来实现这一点？请帮忙...

提前谢谢.......

Answer 1

CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery query = cb.createQuery(/* Your combined target type, e.g. MyQueriedBuildDetails.class, containing buildNumber, duration, code health, etc.*/);

Root<BuildDetails> buildDetailsTable = query.from(BuildDetails.class);
Join<BuildDetails, CopyQualityDetails> qualityJoin = buildDetailsTable.join(CopyQualityDetails_.build, JoinType.INNER);
Join<BuildDetails, DeploymentDetails> deploymentJoin = buildDetailsTable.join(DeploymentDetails_.build, JoinType.INNER);
Join<BuildDetails, TestDetails> testJoin = buildDetailsTable.join(TestDetails_.build, JoinType.INNER);

List<Predicate> predicates = new ArrayList<>();
predicates.add(cb.equal(buildDetailsTable.get(BuildDetails_.buildNumber), "1.0.0.1"));
predicates.add(cb.equal(buildDetailsTable.get(BuildDetails_.projectName), "Tera"));

query.multiselect(buildDetails.get(BuildDetails_.buildNumber),
                  buildDetails.get(BuildDetails_.buildDuration),
                  qualityJoin.get(CodeQualityDetails_.codeHealth),
                  deploymentJoin.get(DeploymentDetails_.deployedEnv),
                  testJoin.get(TestDetails_.testStatus));
query.where(predicates.stream().toArray(Predicate[]::new));

TypedQuery<MyQueriedBuildDetails> typedQuery = entityManager.createQuery(query);

List<MyQueriedBuildDetails> resultList = typedQuery.getResultList();

我假设您为自己的类构建了 JPA 元模型。如果您没有元模型或者您只是不想使用它，只需将BuildDetails_.buildNumber 替换为列的实际名称String，例如"buildNumber".

请注意，我无法测试答案（也是在没有编辑器支持的情况下编写的），但它至少应该包含构建查询所需的所有内容。

如何构建您的元模型？看看hibernate tooling （或咨询How to generate JPA 2.0 metamodel? 了解其他替代方案）。如果您使用的是 maven，那么只需将 hibernate-jpamodelgen-dependency 添加到您的构建类路径中即可。由于我现在没有任何此类项目可用，因此我对以下内容不太确定（因此请谨慎对待）。只需添加以下作为依赖项就足够了：

<dependency>
  <groupId>org.hibernate</groupId>
  <artifactId>hibernate-jpamodelgen</artifactId>
  <version>5.3.7.Final</version>
  <scope>provided</scope> <!-- this might ensure that you do not package it, but that it is otherwise available; untested now, but I think I used it that way in the past -->
</dependency>