本文介绍了关键字'table'附近的语法不正确,无法提取ResultSet的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我已经用SQL Server创建了一个项目,其中包含以下文件:
UserDAO.java *
public class UserDAO
{
private static SessionFactory sessionFactory;
static
{
sessionFactory = HibernateUtility.getSessionFactory();
}
@SuppressWarnings(unchecked)
public static List< User> findAll()
{
Session session = sessionFactory.openSession();
Criteria crit = session.createCriteria(User.class);
列表<用户> userList = crit.list();
返回userList;
}
}
UserService.java
public class UserService
{
public static void main(String [] args)
{
List< User> ; listUsers = UserDAO.findAll();
for(User u:listUsers)
{
System.out.println(User is =+ u.getUserName());
$ b
hibernate.cfg.xml
<?xml version =1.0encoding =utf-8?>
<!DOCTYPE hibernate-configuration PUBLIC
- // Hibernate / Hibernate Configuration DTD // EN
http://www.hibernate.org/dtd/hibernate-configuration- 3.0.dtd>
< hibernate-configuration>
< session-factory>
< property name =hibernate.dialect> org.hibernate.dialect.SQLServer2008Dialect< / property>
< property name =hibernate.connection.driver_class> com.microsoft.sqlserver.jdbc.SQLServerDriver< / property>
< property name =hibernate.connection.url> jdbc:sqlserver:// localhost:1433; database = happy< / property>
< property name =hibernate.connection.username> lm< / property>
< property name =hibernate.connection.password> pp< / property>
< property name =hibernate.hbm2ddl.auto>建立< / property>
< property name =show_sql> true< / property>
< property name =hibernate.current_session_context_class>线程< / property>
< property name =hibernate.hbm2ddl.auto>验证< / property>
< mapping class =com.annotation.day1.entity.User/>
< / session-factory>
< / hibernate-configuration>
运行该项目后,会显示以下异常:
Hibernate:
选择
this_.id作为id1_0_0_,
this_.password作为password2_0_0_,
this_.userName as userName3_0_0_
from
用户this_
Aug 07,2016 9:29:00 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
WARN:SQL错误:156,SQLState:S0001
Aug 07,2016 9:29:00 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
错误:关键字'User'附近的语法不正确。在线程 主要 org.hibernate.exception.SQLGrammarException
例外:在org.hibernate.exception.internal.SQLStateConversionDelegate.convert不能提取的ResultSet
(SQLStateConversionDelegate.java:123)
。在org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:49)
在org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:126)
。在组织。 hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:112)
在org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:89)
。在组织。 hibernate.loader.Loader.getResultSet(Loader.java:2065)
在org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1862)
在org.hibernate.loader.Loader.executeQueryStatement( Loader.java:1838)
在org.hibernate.loader.Loader.doQuery(Loader.java:909)
在org.hibernate.loader.Loader.doQueryAndInitialize NonLazyCollections(Loader.java:354)
在org.hibernate.loader.Loader.doList(Loader.java:2553)
在org.hibernate.loader.Loader.doList(Loader.java:2539)
在org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2369)
在org.hibernate.loader.Loader.list(Loader.java:2364)
在org.hibernate .loader.criteria.CriteriaLoader.list(CriteriaLoader.java:126)
在org.hibernate.internal.SessionImpl.list(SessionImpl.java:1682)
在org.hibernate.internal.CriteriaImpl.list (CriteriaImpl.java:380)
at com.annotation.day1.dao.UserDAO.findAll(UserDAO.java:74)
at com.annotation.day1.service.UserService.main(UserService.java :24)
导致:com.microsoft.sqlserver.jdbc.SQLServerException:关键字'User'附近的语法不正确。
at com.microsoft.sqlserver.jdbc.SQLServerException.makeFromDatabaseError(SQLServerException.java:197)
任何帮助表示赞赏。 是,需要在查询时使用方括号[]
进行查询转义。
如果您使用注释,请使用单引号。 ''
,如下
@Table(name =`user `)
另请参阅类似的问题。
I have created a project with SQL Server which the files:
UserDAO.java*
public class UserDAO
{
private static SessionFactory sessionFactory;
static
{
sessionFactory = HibernateUtility.getSessionFactory();
}
@SuppressWarnings("unchecked")
public static List<User> findAll()
{
Session session = sessionFactory.openSession();
Criteria crit = session.createCriteria(User.class);
List<User> userList = crit.list();
return userList;
}
}
UserService.java
public class UserService
{
public static void main(String[] args)
{
List<User> listUsers = UserDAO.findAll();
for(User u : listUsers)
{
System.out.println("User is = " + u.getUserName());
}
}
}
hibernate.cfg.xml
<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD//EN"
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
<property name="hibernate.dialect">org.hibernate.dialect.SQLServer2008Dialect</property>
<property name="hibernate.connection.driver_class">com.microsoft.sqlserver.jdbc.SQLServerDriver</property>
<property name="hibernate.connection.url">jdbc:sqlserver://localhost:1433;database=happy</property>
<property name="hibernate.connection.username">lm</property>
<property name="hibernate.connection.password">pp</property>
<property name="hibernate.hbm2ddl.auto">create</property>
<property name="show_sql">true</property>
<property name="hibernate.current_session_context_class">thread</property>
<property name="hibernate.hbm2ddl.auto">validate</property>
<mapping class="com.annotation.day1.entity.User"/>
</session-factory>
</hibernate-configuration>
After running the project, the exception bellow displayed:
Hibernate:
select
this_.id as id1_0_0_,
this_.password as password2_0_0_,
this_.userName as userName3_0_0_
from
User this_
Aug 07, 2016 9:29:00 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
WARN: SQL Error: 156, SQLState: S0001
Aug 07, 2016 9:29:00 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
ERROR: Incorrect syntax near the keyword 'User'.
Exception in thread "main" org.hibernate.exception.SQLGrammarException: could not extract ResultSet
at org.hibernate.exception.internal.SQLStateConversionDelegate.convert(SQLStateConversionDelegate.java:123)
at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:49)
at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:126)
at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:112)
at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:89)
at org.hibernate.loader.Loader.getResultSet(Loader.java:2065)
at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1862)
at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1838)
at org.hibernate.loader.Loader.doQuery(Loader.java:909)
at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:354)
at org.hibernate.loader.Loader.doList(Loader.java:2553)
at org.hibernate.loader.Loader.doList(Loader.java:2539)
at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2369)
at org.hibernate.loader.Loader.list(Loader.java:2364)
at org.hibernate.loader.criteria.CriteriaLoader.list(CriteriaLoader.java:126)
at org.hibernate.internal.SessionImpl.list(SessionImpl.java:1682)
at org.hibernate.internal.CriteriaImpl.list(CriteriaImpl.java:380)
at com.annotation.day1.dao.UserDAO.findAll(UserDAO.java:74)
at com.annotation.day1.service.UserService.main(UserService.java:24)
Caused by: com.microsoft.sqlserver.jdbc.SQLServerException: Incorrect syntax near the keyword 'User'.
at com.microsoft.sqlserver.jdbc.SQLServerException.makeFromDatabaseError(SQLServerException.java:197)
Any help is appreciated. Thanks in advance.
解决方案
USER
is a Reserve Word and needs to be escaped in query using square bracket []
while querying.
If you are using annotations, escape through single quotes. ''
, like below
@Table(name="`user`")
Also refer here for similar issue.
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