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我有一个 Hibernate Web 应用程序,目前正在运行 tomcat/CloudFoundry,但是当我尝试在 JBoss 上运行它时,我遇到了一些错误。

在我的 DAO 中,我正在创建一个查询以按用户名加载用户,如下所示:

Query query = getEntityManager().createQuery("select u from com.tmm.enterprise.socialcv.security.Account u where u.userName = ?1");
query.setParameter(1, userName);

当我在 Tomcat 上运行应用程序时,上述方法有效,但是当我在 JBoss 上运行它时,出现以下错误:

16:31:47,639 DEBUG [org.springframework.web.servlet.DispatcherServlet] (http-localhost-127.0.0.1-8080-1) Handler execution resulted in exception - forwarding to resolved error view: ModelAndView: reference to view with name 'dataAccessFailure'; model is {exception=org.springframework.dao.InvalidDataAccessApiUsageException: org.hibernate.QueryParameterException: Position beyond number of declared ordinal parameters. Remember that ordinal parameters are 1-based! Position: 1; nested exception is java.lang.IllegalArgumentException: org.hibernate.QueryParameterException: Position beyond number of declared ordinal parameters. Remember that ordinal parameters are 1-based! Position: 1}: org.springframework.dao.InvalidDataAccessApiUsageException: org.hibernate.QueryParameterException: Position beyond number of declared ordinal parameters. Remember that ordinal parameters are 1-based! Position: 1; nested exception is java.lang.IllegalArgumentException: org.hibernate.QueryParameterException: Position beyond number of declared ordinal parameters. Remember that ordinal parameters are 1-based! Position: 1
    at org.springframework.orm.jpa.EntityManagerFactoryUtils.convertJpaAccessExceptionIfPossible(EntityManagerFactoryUtils.java:301) [org.springframework.orm-3.1.1.RELEASE.jar:3.1.1.RELEASE]
    at org.springframework.orm.jpa.aspectj.JpaExceptionTranslatorAspect.ajc$afterThrowing$org_springframework_orm_jpa_aspectj_JpaExceptionTranslatorAspect$1$18a1ac9(JpaExceptionTranslatorAspect.aj:15) [spring-aspects-3.1.1.RELEASE.jar:3.1.1.RELEASE]
    at com.tmm.enterprise.socialcv.security.dao.AccountDAO.loadAccountByUserName(AccountDAO.java:28) [classes:]
    at com.tmm.enterprise.socialcv.service.AccountService.loadAccountByUserName(AccountService.java:48) [classes:]
    at com.tmm.enterprise.socialcv.service.AccountService.setCredentials(AccountService.java:241) [classes:]
    at com.tmm.enterprise.socialcv.controller.HomeController.signup(HomeController.java:59) [classes:]
    ...
Caused by: java.lang.IllegalArgumentException: org.hibernate.QueryParameterException: Position beyond number of declared ordinal parameters. Remember that ordinal parameters are 1-based! Position: 1
    at org.hibernate.ejb.QueryImpl.setParameter(QueryImpl.java:446) [hibernate-entitymanager-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.ejb.QueryImpl.setParameter(QueryImpl.java:67) [hibernate-entitymanager-4.0.1.Final.jar:4.0.1.Final]
    ... 80 more
Caused by: org.hibernate.QueryParameterException: ***Position beyond number of declared ordinal parameters. Remember that ordinal parameters are 1-based! Position: 1***
    at org.hibernate.engine.query.spi.ParameterMetadata.getOrdinalParameterDescriptor(ParameterMetadata.java:80) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.engine.query.spi.ParameterMetadata.getOrdinalParameterExpectedType(ParameterMetadata.java:86) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.internal.AbstractQueryImpl.determineType(AbstractQueryImpl.java:444) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.internal.AbstractQueryImpl.setParameter(AbstractQueryImpl.java:416) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.ejb.QueryImpl.setParameter(QueryImpl.java:440) [hibernate-entitymanager-4.0.1.Final.jar:4.0.1.Final]
    ... 81 more

我尝试更改以下两个查询,但仍然没有运气:

Query query = getEntityManager().createQuery("select u from com.tmm.enterprise.socialcv.security.Account u where u.userName = ?");
        query.setParameter(0, userName);

以上给了我同样的错误。就像这样:

Query query = getEntityManager().createQuery("select u from com.tmm.enterprise.socialcv.security.Account u where u.userName = ?");
        query.setParameter(1, userName);

切换到命名参数会给我一个关于无法找到命名参数的错误。

有任何想法吗?(顺便说一句,我还必须在 JBoss 上的查询中将 DAO 更新为完全合格的 Account - 在 tomcat 上它只是在查询 Account)

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4 回答 4

1

我以这种方式修复了这样的错误

schema.table = ' ?' ->error

schema.table = '?' ->no error

也许它可以帮助某人。

于 2013-04-24T06:08:04.667 回答
0

基于 Hibernate HQL参考,如果你使用 JDBC 位置样式,你需要使用?. 如果您使用命名参数,那么您应该使用:paramName.

于 2012-04-09T16:03:56.557 回答
0
query = getEntityManager().createQuery(
"select u from com.tmm.enterprise.socialcv.security.Account u where u.userName = :userName");
query.setParameter("userName", userName);
于 2013-04-26T13:41:20.063 回答
0

即使在我的情况下也没有任何效果,我已经尝试过

  1. 索引为 0 的位置参数
  2. 索引为 1 的位置参数
  3. 命名参数

最后我发现结果查询对象的参数列表每次都是空的,导致调用时出错setParameter()

解决方法:文件中缺少实体映射persistence.xml,添加后就开始工作了!!

于 2015-09-24T08:20:58.557 回答