【发布时间】:2016-02-01 19:54:16
【问题描述】:
我有一个名为查询的插入 JPA,如下所示。当我尝试使用实体管理器从我的 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。
在字符 1 处遇到“INSERT”,但应为:[“DELETE”、“SELECT”、“UPDATE”]。
@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();
【问题讨论】:
标签: java oracle hibernate jpa openjpa