我正在尝试使用hibernate criteriabuilder加入4个表.
下面分别是表格..
`
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,我如何使用Hibernate CriteriaBuilder实现这一目标？请帮忙…
提前致谢…….
解决方法
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,TestDetails> testJoin = buildDetailsTable.join(TestDetails_.build,JoinType.INNER); List<Predicate> predicates = new ArrayList<>(); predicates.add(cb.equal(BuildDetails_.buildNumber,"1.0.0.1")); predicates.add(cb.equal(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();
我假设你为你的类构建了JPA元模型.如果您没有元模型或者您根本不想使用它,只需将BuildDetails_.buildNumber替换为其余部分,并将该列的实际名称替换为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它),但它应该至少包含构建查询所需要知道的所有内容.
如何构建元模型？看看hibernate tooling(或其他替代方案请咨询How to generate JPA 2.0 metamodel?).如果您正在使用maven,那么就像将hibernate-jpamodelgen-dependency添加到构建类路径一样简单.因为我现在没有任何这样的项目,所以我对以下内容不太确定(所以请稍等一下).将以下内容添加为依赖项可能就足够了：
<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>