Question

我有一个插入JPA命名查询，如下所示。当我尝试使用enity manger从我的DAO执行它时，我收到以下错误。我正在使用Apache OpenJPA，我无法确定此问题的根本原因。由于我的域对象的复杂性，我无法使用persist方法。

遇到＆＃34; INSERT＆＃34;在角色1，但预期：[＆＃34; DELETE＆＃34;，＆＃34; SELECT＆＃34;，＆＃34; UPDATE＆＃34;]。

@NamedQuery(
 name=IuaPersistenceConstants.QUERY_INSERT_AGREEMENT_ACKNOWLEDGEMENT,
 query="INSERT INTO AgreementAcknowledgement agrackn VALUES agrackn.id.agrmntCntntUrl = :agrmntCntntUrl, agrackn.id.prvsndUserSqn = :prvsndUserSqn, agrackn.id.prvsnUserEffD = :userEffDate, " +
 "agrackn.id.agrmntacknEffD = :agrmntacknEffD, agrackn.acknExpD = :acknExpD, agrackn.crtTs = :crtTs, agrackn.crtUidC = :crtUidC, agrackn.lstUpdtTs = :lstUpdtTs, agrackn.lstUpdtUidC = :lstUpdtUidC")


Query ackQuery =getEntityManager().createNamedQuery(IuaPersistenceConstants.QUERY_INSERT_AGREEMENT_ACKNOWLEDGEMENT);
 ackQuery.setParameter("agrmntCntntUrl", agreementDO.getAgreementURL());
 ackQuery.setParameter("agrmntacknEffD", agreementDO.getAgreementEffDate());
 ackQuery.setParameter("acknExpD", agreementDO.getAgreementExpDate());
 ackQuery.setParameter("prvsndUserSqn", userDO.getUserSequence());
 ackQuery.setParameter("userEffDate", userDO.getUserEffDate());
 ackQuery.setParameter("crtTs", new Date());
 ackQuery.setParameter("crtUidC", "WS");
 ackQuery.setParameter("lstUpdtTs", new Date());
 ackQuery.setParameter("lstUpdtUidC", "WS");
 ackQuery.executeUpdate();

Answer 1

OpenJPA doesn't seem to support JPQL "INSERT" queries。但是，这并不奇怪，因为JPA没有定义INSERT查询。它定义了SELECT，UPDATE，DELETE查询，这就是全部。参见JPA规范。这不是SQL。

遭遇＆＃34; INSERT＆＃34;在角色1，但预期：[＆＃34; DELETE＆＃34;，＆＃34; SELECT＆＃34;，＆＃34; UPDATE＆＃34;]

1 个答案: