Question

我有两个表（JPA实体）

JOB
申请者

JOBID是这两个表之间的公共列。我正在尝试选择用户已应用的作业实体（申请人表具有userId字段）

SELECT j from JOB j, APPLICANTS a WHERE j.jobId=:a.jobId AND a.userId=:userId
query.setParameter("userId", userId);

这会引发以下错误

java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Syntax error parsing [SELECT j from JOB j, APPLICANTS a WHERE j.jobId=:a.jobId AND a.uuId=:uuId]. 
[50, 58] The named input parameter ''{0}'' is not following the rules for a Java identifier.
    org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1605)

Answer 1

您不必在WHERE中使用:为字段值添加前缀，它仅用于参数。因此，如果您想加入Job上的Applicant和jobId，请使用此

SELECT j from JOB j, APPLICANTS a WHERE j.jobId = a.jobId AND a.userId = :userId
query.setParameter("userId", userId);

如果您想将jobId作为参数传递，请使用此

SELECT j from JOB j, APPLICANTS a WHERE j.jobId = :jobId AND a.userId = :userId
query.setParameter("userId", userId);
query.setParameter("jobId", jobId);

Answer 2

您需要使用Annotation指定Job和Applicants实体之间的关系申请人实体

@manyToOne(fetch=FetchType.LAZY)
@joinColumns(name="job_id", insertable=false , updatable=false)
private Job job ;

你的查询应该是这样的

select  NEW YourClass( Your Column List  ) from Applicants a left join a.job where j.jobId=?1 and a.userId=?2

query.setParameter(1, userId);
query.setParameter(2, jobId);

这个JPQL的正确语法是什么

2 个答案: