hibernate中的org.hibernate.loader.custom.CustomLoader.java中是否有個(gè)BUG ？ 如果有，麻煩在hibernate搞個(gè)BUG上去，不會(huì)英文描述。protected void autoDiscoverTypes(ResultSet rs) { try { Metadata metadata = new Metadata( getFactory(), rs ); List aliases = new ArrayList(); List types = new ArrayList(); rowProcessor.prepareForAutoDiscovery( metadata ); for ( int i = 0; i < rowProcessor.columnProcessors.length; i++ ) { rowProcessor.columnProcessors[i].performDiscovery( metadata, types, aliases ); } //======================bug處============================= // resultTypes的設(shè)值是有問(wèn)題的，應(yīng)該是將types中 // 的數(shù)據(jù)放入resultTypes，而不是直接賦值。 resultTypes = ArrayHelper.toTypeArray( types ); transformerAliases = ArrayHelper.toStringArray( aliases ); } catch ( SQLException e ) { throw new HibernateException( "Exception while trying to autodiscover types.", e ); } } 我用的Hibernate是3.2.7.ga，我下過(guò)幾個(gè)版本的源碼，這段都是一樣的。 <dependency> <groupId>org.hibernate</groupId> <artifactId>hibernate</artifactId> <version>3.2.7.ga</version> </dependency>
2 回答
慕蓋茨4494581
TA貢獻(xiàn)1850條經(jīng)驗(yàn) 獲得超11個(gè)贊
給你一段我的代碼
package com.s2sh.dao.impl;
import java.util.List;
import org.springframework.orm.hibernate3.support.HibernateDaoSupport;
import com.s2sh.dao.UserDao;
import com.s2sh.mobel.User;
public class UserDaoImpl extends HibernateDaoSupport implements UserDao{
//用戶登錄
public User findUser(User user) {
try{
String hql="from User where name=? and password=?";
List list = (List) this.getHibernateTemplate().find(hql,new String[]{user.getName(),user.getPassword()});
if(list.size()>=1){
return (User)list.get(0);
}else {
return null;
}
}catch (Exception e) {
e.printStackTrace();
}
return null;
}
}
