java - 如何在jpa查询中传递参数？

Question

我有两个实体，例如 SmsOut 和 SmsIn。两个实体之间的关系包含 OneToMany，其中 smsIn.id 是主键，smsOut.sms_in_id 是 外键。

现在我想在查询中传递 smsIn.mobileNumber、smsIn.smsText 等参数

SELECT so FROM SmsOut so order by id desc

以下是我的数据库图表:

已编辑

以下是我的代码:

String sql = "SELECT so FROM SmsOut so WHERE so.smsInId.mobileNumber =:mobileNumber AND so.smsInId.smsText =:smsText AND so.smsInId.shortCode =:shortCode between so.smsOutDate =:startDate and so.smsOutDate =:endDate order by id desc";
    Query query = em.createQuery(sql);
    query.setParameter("mobileNumber", mobileNumber);
    query.setParameter("smsText", smsText);
    query.setParameter("shortCode", shortCode);
    query.setParameter("smsOutDate", startDate);
    query.setParameter("smsOutDate", endDate);
    smsOutList = query.getResultList();

异常(exception)是:

SEVERE: line 1:188: expecting "and", found '='
java.lang.IllegalArgumentException: org.hibernate.hql.ast.QuerySyntaxException: expecting "and", found '=' near line 1, column 188 [SELECT so FROM     com.f1soft.SMSC.entities.SmsOut so WHERE so.smsInId.mobileNumber =:mobileNumber AND so.smsInId.smsText =:smsText AND so.smsInId.shortCode =:shortCode between so.smsOutDate =:startDate and so.smsOutDate =:endDate order by id desc]
at org.hibernate.ejb.AbstractEntityManagerImpl.throwPersistenceException(AbstractEntityManagerImpl.java:624)
at org.hibernate.ejb.AbstractEntityManagerImpl.createQuery(AbstractEntityManagerImpl.java:96)

请帮助我。谢谢

Answer 1

您还没有解释 SmsIn 和 SmsOut 之间的 JPA 关系，因此我假设 SmsOut 有一个 getSmsIn( ) 与 id 字段上的关系。

当你有一个EntityManager em时，你可以调用em.createQuery，这就像SQL prepare，然后setParameter :

TypedQuery<SmsOut> q = em.createQuery("SELECT so FROM SmsOut so WHERE so.smsIn.mobileNumber = :number ORDER BY id DESC");
q.setParameter("number", "12345678");
List<SmsOut> results = q.getResultList();

请参阅 the Javadoc for Query 了解指定参数的不同方式。