如何在jpa查询上传递参数?

Yub*_*raj 2 java sql-server jpa-2.0

我有两个实体,如SmsOutSmsIn.两个实体之间的关系包含OneToMany,其中smsIn.id主键,smsOut.sms_in_id外键.

现在我想在查询上传递smsIn.mobileNumber,smsIn.smsText等参数

SELECT so FROM SmsOut so order by id desc
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以下是我的数据库图表:

在此输入图像描述

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();
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例外是:

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)
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请帮我.谢谢

chr*_*ke- 9

您还没有解释的JPA关系SmsInSmsOut,所以我会假设SmsOut有一个getSmsIn()与在关系id领域.

当你有一个EntityManager em,你可以调用em.createQuery,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();
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有关可以指定参数的不同方法,请参阅JavadocQuery.