我在MySQL数据库中有两个表:
1-广告系列
2-内容:
我有一个叫做CampaignData的java bean(不是hibernate实体)
public class CampaignData {
private long contentId;
private long contentSubTypeId;
private Long distributionGroupId;
}
这是我如何进行查询:
CampaignData campaignData = (CampaignData) session
.createSQLQuery(
"select camp.fk_Content as contentId,camp.tk_DistributionGroup as distributionGroupId,cont.tk_contentSubtype as contentSubTypeId "
+ "from campaign camp,content cont"
+ " where camp.pkid=:campaignId and camp.fk_Content=cont.pkid")
.setLong("campaignId", campaignId)
.setResultTransformer(
Transformers.aliasToBean(CampaignData.class))
.uniqueResult();
生成hibernate查询:
select
camp.fk_Content as contentId,
camp.tk_DistributionGroup as distributionGroupId,
cont.tk_contentSubtype as contentSubTypeId
from
campaign camp,
content cont
where
camp.pkid=?
and camp.fk_Content=cont.pkid
当我在数据库中尝试生成的SQL查询时,它运行正常并且数据被成功检索,但是在运行应用程序时我得到了异常:
Exception in thread "main" org.hibernate.exception.SQLGrammarException: could not execute query
at org.hibernate.exception.SQLStateConverter.convert(SQLStateConverter.java:92)
at org.hibernate.exception.JDBCExceptionHelper.convert(JDBCExceptionHelper.java:66)
at org.hibernate.loader.Loader.doList(Loader.java:2297)
at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2172)
at org.hibernate.loader.Loader.list(Loader.java:2167)
at org.hibernate.loader.custom.CustomLoader.list(CustomLoader.java:316)
at org.hibernate.impl.SessionImpl.listCustomQuery(SessionImpl.java:1832)
at org.hibernate.impl.AbstractSessionImpl.list(AbstractSessionImpl.java:165)
at org.hibernate.impl.SQLQueryImpl.list(SQLQueryImpl.java:179)
at org.hibernate.impl.AbstractQueryImpl.uniqueResult(AbstractQueryImpl.java:859)
at com.xeno.xecamp.desktopManagement.Main.getCampaignSMSs(Main.java:43)
at com.xeno.xecamp.desktopManagement.Main.main(Main.java:18)
Caused by: java.sql.SQLException: Column 'fk_Content' not found.
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1073)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:987)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:982)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:927)
at com.mysql.jdbc.ResultSetImpl.findColumn(ResultSetImpl.java:1144)
at com.mysql.jdbc.ResultSetImpl.getBigDecimal(ResultSetImpl.java:1414)
at org.hibernate.type.BigIntegerType.get(BigIntegerType.java:57)
at org.hibernate.type.NullableType.nullSafeGet(NullableType.java:184)
at org.hibernate.type.NullableType.nullSafeGet(NullableType.java:210)
at org.hibernate.loader.custom.CustomLoader$ScalarResultColumnProcessor.extract(CustomLoader.java:501)
at org.hibernate.loader.custom.CustomLoader$ResultRowProcessor.buildResultRow(CustomLoader.java:447)
at org.hibernate.loader.custom.CustomLoader.getResultColumnOrRow(CustomLoader.java:344)
at org.hibernate.loader.Loader.getRowFromResultSet(Loader.java:647)
at org.hibernate.loader.Loader.doQuery(Loader.java:745)
at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:270)
at org.hibernate.loader.Loader.doList(Loader.java:2294)
... 9 more
请告知我为什么会收到例外。
更新:这是第三方应用程序,它在数据库上连接另一个应用程序。
hibernate.hbm2ddl.auto=create-drop
答案 0 :(得分:5)
为了从数据库中正确地获取您请求的内容,hibernate使用的实体类应该匹配100%的数据库表。
您有fk_Content
列,但私有字段为contentId
。仅使用as
将无法获得所需的结果。如果您想使用不同的名称(就像您一样),则需要使用@Column(name = "")
为休眠提供正确的列名。此外,不建议使用基本数据类型。您的CampaignData类看起来像:
@Entity
@Table(name = "campaign")
public class CampaignData {
private Long contentId;
private Long contentSubTypeId;
private Long distributionGroupId;
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name = "pkid", unique = true, nullable = false)
public Long getContentId() {
return this.contentId;
}
public void setContentId(Long contentId){
this.contentId = contentId;
}
@Column(name = "fk_Content")
public Long getContentSubTypeId() {
return this.contentSubTypeId;
}
public void setContentSubTypeId(Long contentSubTypeId){
this.contentSubTypeId= contentSubTypeId;
}
@Column(name = "tk_DistributionGroup")
public Long getDistributionGroupId() {
return this.distributionGroupId;
}
public void setDistributionGroupId(Long distributionGroupId){
this.distributionGroupId= distributionGroupId;
}
}
这应该这样做。另外,尝试学习使用Hibernate's Criteria。这是一种比硬编码的SQL语句更好的做法。
答案 1 :(得分:5)
要使其工作,我需要使用addScalar,如下所示:
.createSQLQuery(
"select camp.fk_Content as contentId,camp.tk_DistributionGroup as distributionGroupId,cont.tk_contentSubtype as contentSubTypeId "
+ "from campaign camp,content cont"
+ " where camp.pkid=:campaignId and camp.fk_Content=cont.pkid")
.addScalar("contentId")
.addScalar("distributionGroupId")
.addScalar("contentSubTypeId")