Question

我对Hibernate不是很有经验，而且我现在很快就遇到了Hibernate的无效列名问题。我只是尝试使用Criteria API从数据库中选择数据，如下所示：

public Subject lookupByID(SubjectValue subjectValue) {
    List list;
    Iterator it;
    Criteria criteria = getBaseCriteria();
    criteria = criteria.add(Restrictions.eq("subjectId", subjectValue.getSubjectId()));
    list = criteria.list();
    it = list.iterator();
    if(it.hasNext()){
        return (Subject)it.next();
    }
    else{
        return null;
    }
}

我启用了Hibernate调试模式，现在我可以看到它生成的选择：

Hibernate: select this_.Subject_id as Subject1_11_0_, this_.Site_id as Site2_11_0_, this_.Notification_id as Notifica3_11_0_, this_.Name as Name11_0_, this_.Description as Descript5_11_0_, this_.UserName as UserName11_0_, this_.DisplayOrder as DisplayO7_11_0_, this_.InsertDate as InsertDate11_0_, this_.insertedBy_id as insertedBy9_11_0_, this_.UpdateDate as UpdateDate11_0_, this_.updatedBy_id as updatedBy11_11_0_, this_.StatusCode as StatusCode11_0_, this_.Template_id as Template13_11_0_, this_.ForwardMail as Forward14_11_0_, this_.AutoReply as AutoReply11_0_, this_.FromName as FromName11_0_, this_.FromMail as FromMail11_0_, this_.ResponseRequired as Respons18_11_0_ from MYSCHEMA.Subject this_ where this_.Subject_id = ?

直接在数据库上运行此查询它工作正常，但在休眠时它显示：

org.hibernate.exception.SQLGrammarException: could not execute query
Caused by: java.sql.SQLException: Invalid column name 'Template_id'.

这是我的映射XML：

<hibernate-mapping>
<class name="MYPACKAGE.model.Subject" table="Subject" dynamic-update="true" >
    <id name="subjectId" type="integer" column="Subject_id">
        <generator class="identity" />
    </id>
    <property name="siteId" column="Site_id" type="integer" />
    <property name="notificationId" column="Notification_id" type="integer" not-null="false" />
    <property name="name" column="Name" type="string" />
    <property name="description" column="Description" type="string" />
    <property name="userName" column="UserName" type="string" />
    <property name="displayOrder" column="DisplayOrder" type="integer" />
    <property name="insertDate" column="InsertDate" type="timestamp" />
    <property name="insertedBy_id" column="insertedBy_id" type="integer" />
    <property name="updateDate" column="UpdateDate" type="timestamp" />
    <property name="updatedBy_id" column="updatedBy_id" type="integer" />
    <property name="statusCode" column="StatusCode" type="string" />
    <property name="templateId" column="Template_id" type="integer" />
    <property name="forwardMail" column="ForwardMail" type="string" />
    <property name="autoReply" column="AutoReply" type="boolean" />
    <property name="fromName" column="FromName" type="string" />
    <property name="fromMail" column="FromMail" type="string" />
    <property name="responseRequired" column="ResponseRequired" type="boolean" />

</class>

我正在运行Java 1.5.022，Hibernate 3.26GA和JBOSS 6.0 EAP（非常类似于JBOSS AS7）。有人可以帮忙吗？如果您需要任何进一步的细节，请告诉我，我很乐意提供。

度过美好的一天！

编辑：表结构：

 CREATE TABLE [ITrInnoFramework].[Subject](
    [Subject_id] [int] IDENTITY(1,1) NOT NULL,
    [Site_id] [int] NOT NULL,
    [Notification_id] [int] NULL,
    [Name] [nvarchar](50) NOT NULL,
    [Description] [nvarchar](1000) NOT NULL,
    [DisplayOrder] [int] NOT NULL,
    [InsertDate] [datetime] NOT NULL,
    [InsertedBy_id] [int] NOT NULL,
    [UpdateDate] [datetime] NOT NULL,
    [UpdatedBy_id] [int] NOT NULL,
    [StatusCode] [char](10) NULL,
    [UserName] [varchar](50) NULL,
    [Template_id] [int] NULL,
    [ForwardMail] [varchar](255) NULL,
    [ResponseRequired] [bit] NULL,
    [AutoReply] [bit] NULL,
    [FromName] [varchar](255) NULL,
    [FromMail] [varchar](255) NULL,
 CONSTRAINT [PK_Subject] PRIMARY KEY CLUSTERED 
(
    [Site_id] ASC,
    [Name] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY],
 CONSTRAINT [IX_Subject_1] UNIQUE NONCLUSTERED 
(
    [Subject_id] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]

GO

SET ANSI_PADDING OFF
GO

ALTER TABLE [ITrInnoFramework].[Subject]  WITH CHECK ADD  CONSTRAINT [FK_Subject_Notification] FOREIGN KEY([Notification_id])
REFERENCES [ITrInnoFramework].[Notification] ([Notification_id])
GO

ALTER TABLE [ITrInnoFramework].[Subject] CHECK CONSTRAINT [FK_Subject_Notification]
GO

ALTER TABLE [ITrInnoFramework].[Subject]  WITH CHECK ADD  CONSTRAINT [FK_Subject_SiteInfo] FOREIGN KEY([Site_id])
REFERENCES [ITrInnoFramework].[SiteInfo] ([Site_id])
GO

ALTER TABLE [ITrInnoFramework].[Subject] CHECK CONSTRAINT [FK_Subject_SiteInfo]
GO

ALTER TABLE [ITrInnoFramework].[Subject]  WITH CHECK ADD  CONSTRAINT [FK_Subject_StatusLkp] FOREIGN KEY([StatusCode])
REFERENCES [ITrInnoFramework].[StatusLkp] ([StatusCode])
GO

ALTER TABLE [ITrInnoFramework].[Subject] CHECK CONSTRAINT [FK_Subject_StatusLkp]
GO

ALTER TABLE [ITrInnoFramework].[Subject]  WITH CHECK ADD  CONSTRAINT [FK_Subject_Template] FOREIGN KEY([Template_id])
REFERENCES [ITrInnoFramework].[Template] ([Template_id])
GO

ALTER TABLE [ITrInnoFramework].[Subject] CHECK CONSTRAINT [FK_Subject_Template]
GO

ALTER TABLE [ITrInnoFramework].[Subject] ADD  CONSTRAINT [DF_Subject_InsertDate]  DEFAULT (getdate()) FOR [InsertDate]
GO

ALTER TABLE [ITrInnoFramework].[Subject] ADD  CONSTRAINT [DF_Subject_InsertedBy_id]  DEFAULT ((-1)) FOR [InsertedBy_id]
GO

ALTER TABLE [ITrInnoFramework].[Subject] ADD  CONSTRAINT [DF_Subject_UpdateDate]  DEFAULT (getdate()) FOR [UpdateDate]
GO

ALTER TABLE [ITrInnoFramework].[Subject] ADD  CONSTRAINT [DF_Subject_UpdatedBy_id]  DEFAULT ((-1)) FOR [UpdatedBy_id]
GO

Answer 1

感谢您的帮助，但我结束了放弃并将hibernate版本更改为4.0.1。这解决了我的问题！

Answer 2

这可能是你的问题吗？

 this_.Template_id as Template13_11_0_

Template_id被截断为Template，Template是SQL关键字。尝试将列Template_id更改为templateId。

Hibernate java.sql.SQLException：列名无效

2 个答案: