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無論輸入什么帳號密碼都沒有通過action成功轉跳

提交表單后控制臺輸出語句

WARN: [DEPRECATION] Encountered positional parameter near line 1, column 34 in HQL: [from entity.Users where username=? and password=? ].? Positional parameter are considered deprecated; use named parameters or JPA-style positional parameters instead.

三月 01, 2018 6:34:23 下午 org.hibernate.hql.internal.ast.HqlSqlWalker generatePositionalParameter

WARN: [DEPRECATION] Encountered positional parameter near line 1, column 49 in HQL: [from entity.Users where username=? and password=? ].? Positional parameter are considered deprecated; use named parameters or JPA-style positional parameters instead.

Hibernate:?

? ? select

? ? ? ? users0_.uid as uid1_1_,

? ? ? ? users0_.username as username2_1_,

? ? ? ? users0_.password as password3_1_?

? ? from

? ? ? ? USERS users0_?

? ? where

? ? ? ? users0_.username=??

? ? ? ? and users0_.password=?

public?class?UserDAOImpl?implements?UsersDAO{
?
????@Override
????public?boolean?usersLogin(Users?u)?{
????????//?TODO?Auto-generated?method?stub
????????//創建事物對象
????????//事務(Transaction)是訪問并可能更新數據庫中各種數據項的一個程序執行單元(unit)
????????Transaction?ts?=?null;
????????String?hql?=?"";
?????????
????????try?{
????????????Session?session?=?MyHibernateSessionFactory.getSessionFactory().getCurrentSession();
????????????//開啟事物
????????????ts?=session.beginTransaction();
????????????//hql查詢語句
????????????hql?=?"from?Users?where?username=??and?password=??";
????????????//Query是Hibernate的查詢接口，用于從數據存儲源查詢對象及控制執行查詢的過程，Query包裝了一個HQL查詢語句
????????????org.hibernate.Query?query?=session.createQuery(hql);
????????????//傳遞參數
????????????query.setParameter(0,?u.getUsername());
????????????query.setParameter(1,?u.getPassword());
????????????//調用其中的方法得到結果??
????????????List?list?=query.list();
????????????ts.commit();
????????????if?(list.size()>0)?{
????????????????return?true;
????????????}?else?{
????????????????return?false;
????????????}
????????}
????????catch?(Exception?ex)?{
????????????//?TODO:?handle?exception???
????????????//打印異常對象
????????????ex.printStackTrace();
????????????return?false;
????????}
????????finally{
????????????//釋放事物對象資源
????????????if?(ts!=null)?{
????????????????ts=null;
????????????}
????????}
?????
????}
?????????
}
?
?
public?class?UsersAction?extends?superAction?implements?ModelDriven<Users>{
?????
????/**
?????*?
?????*/
????private?static?final?long?serialVersionUID?=?1L;
????private?Users?user?=?new?Users();
?????
????//用戶登錄動作
????public?String?login()?{
????????UsersDAO?udao?=new?UserDAOImpl();
????????if?(udao.usersLogin(user))?{
????????????return?"login_success";
????????}
????????else?{
????????????return?"login_failure";
????????}
?????????
????}
????@Override
????public?Users?getModel()?{
????????//?TODO?Auto-generated?method?stub
????????return?this.user;
????}
?
}

foggentle3137565

2018-03-05

源自：使用Struts2+Hibernate開發學生信息管理功能 2-6

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1 回答

2841liuhai
2018-03-09

public?boolean?usersLogin(Users?users)?{
????//創建事務對象
????Transaction?tx?=?null;
????try?{
????????Session?session?=?MyHibernateSessionFactory.getSessionObject();
????????tx?=?session.beginTransaction();

????????Criteria?criteria?=?session.createCriteria(Users.class);
????????criteria.add(Restrictions.eq("username",users.getUsername()));
????????criteria.add(Restrictions.eq("password",users.getPassword()));

????????List<Users>?usersList?=?criteria.list();
????????tx.commit();

????????if(usersList.size()>0){
????????????return?true;
????????}else?{
????????????return?false;
????????}
????}catch?(Exception?e){
????????e.printStackTrace();
????????return?false;
????}
}

換一種查詢方式就對了