如何在基于休眠的 dao 实现中减少样板代码

Question

我为我的一个 dao 接口编写了一个基于休眠的实现，但现在我意识到有很多重复的代码，比如打开 sesison、开始运行、提交和回滚，每个方法都有相同的代码，我想知道如何将它保存在一个地方，我听说过一个叫做 AbstractDao 的东西，但是这个概念如何适合这里，它真的有效吗，我想减少样板代码。

这是我的实现类的sn-p

package com.javarnd.cip.dao;

import java.util.ArrayList;
import java.util.List;

import org.hibernate.Session;
import org.hibernate.Transaction;

import com.javarnd.cip.db.HibernateUtil;
import com.javarnd.cip.model.Sports;

public class SportsDaoImpl implements SportsDao {

    public void save(Sports sports) {
        Session session = HibernateUtil.openSession();
        Transaction tx = null;
        try {
            tx = session.beginTransaction();
            session.save(sports);
            session.getTransaction().commit();
        } catch (Exception e) {
            if (tx != null) {
                tx.rollback();
            }
            e.printStackTrace();
        } finally {
            session.close();
        }

    }

    public List<Sports> findAll() {
        Session session = HibernateUtil.openSession();
        Transaction tx = null;
        List<Sports> sportsList = new ArrayList<>();
        try {
            tx = session.beginTransaction();
            sportsList = session.createQuery("FROM Sports", Sports.class).list();
            session.getTransaction().commit();

        } catch (Exception e) {
            if (tx != null) {
                tx.rollback();
                return null;
            }
            e.printStackTrace();
        } finally {
            session.close();
        }
        return sportsList;

    }

    public void update(Sports sports) {
        Session session = HibernateUtil.openSession();
        Transaction tx = null;
        try {
            tx = session.beginTransaction();

            Sports sportsData = session.get(Sports.class, sports.getSportId());
            sportsData.setName(sports.getName());
            session.getTransaction().commit();

        } catch (Exception e) {
            if (tx != null) {
                tx.rollback();
            }
            e.printStackTrace();
        } finally {
            session.close();
        }
    }

    public void delete(Sports sports) {
        Session session = HibernateUtil.openSession();
        Transaction tx = null;
        try {
            tx = session.beginTransaction();

            Sports sportsData = session.get(Sports.class, sports.getSportId());
            session.delete(sportsData);

            session.getTransaction().commit();

        } catch (Exception e) {
            if (tx != null) {
                tx.rollback();
            }
            e.printStackTrace();
        } finally {
            session.close();
        }
    }

    @Override
    public Sports findById(long sportsId) {
        Session session = HibernateUtil.openSession();
        Transaction tx = null;
        Sports sportsData = null;
        try {
            tx = session.beginTransaction();
            sportsData = session.get(Sports.class, sportsId);
            session.getTransaction().commit();
        } catch (Exception e) {
            if (tx != null) {
                tx.rollback();
            }
            e.printStackTrace();
        } finally {
            session.close();
        }
        return sportsData;
    }

}

Answer 1

如果你可以使用 java 8，它可能是这样的：

private static <OUTPUT> OUTPUT runInTransaction(Function<Session, OUTPUT> function) {
    Transaction tx = null;
    try (Session session = HibernateUtil.openSession()) {
        tx = session.beginTransaction();
        final OUTPUT result = function.apply(session);
        session.getTransaction().commit();
        return result;
    } catch (Exception e) {
        if (tx != null) {
            tx.rollback();
        }
        e.printStackTrace();
        return null;
    }
}

然后：

runInTransaction(session -> {
    return session.get(Sports.class, sportsId);
});