Question

当我使用Hibernate 5时，它显示错误：

2018年4月28日下午12:24:45 org.hibernate.internal.SessionImpl createCriteria警告：HHH90000022：Hibernate的遗产 org.hibernate.Criteria API已弃用;使用JPA 而是javax.persistence.criteria.CriteriaQuery

@Transactional
public class AccountDAOImpl implements AccountDAO {

    @Autowired
    private SessionFactory sessionFactory;


    @Override
    public ACCOUNTS findAccount(String userName) {
        // TODO Auto-generated method stub//

        Session session=sessionFactory.getCurrentSession();
        Criteria crit=session.createCriteria(ACCOUNTS.class).add(Restrictions.eq("user_name", userName));
        //crit.add(Restrictions.eq("user_name", userName));
        return (ACCOUNTS)crit.uniqueResult();
    }

}

interaface accountDAO

public interface AccountDAO {

     public ACCOUNTS findAccount(String userName );

}

和模型类帐户

package org.vikas.shoppingCart.entity;

import java.io.Serializable;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name="ACCOUNTS")
public class ACCOUNTS implements Serializable
{
    private static final long serialVersionUID = 1L;
    public static final String ROLE_MANAGER = "MANAGER";
    public static final String ROLE_EMPLOYEE = "EMPLOYEE";

    private String user_name;
    private String user_password;
    private boolean active;
    private String user_role;

    @Id
    @Column(name="user_name",length=50,nullable=false)
    public String getUser_name() {
        return user_name;
    }
    public void setUser_name(String user_name) {
        this.user_name = user_name;
    }

    @Column(name="user_password",length=20,nullable=false)
    public String getUser_password() {
        return user_password;
    }
    public void setUser_password(String user_password) {
        this.user_password = user_password;
    }

    @Column(name="active",length=1,nullable=false)
    public boolean isActive() {
        return active;
    }
    public void setActive(boolean active) {
        this.active = active;
    }

    @Column(name="user_role",length=20,nullable=false)
    public String getUser_role() {
        return user_role;
    }
    public void setUser_role(String user_role) {
        this.user_role = user_role;
    }
    @Override
    public String toString() {
        return "ACCOUNTS [user_name=" + user_name + ", user_password=" + user_password + ", active=" + active
                + ", user_role=" + user_role + "]";
    }
}

当我搜索它并说它使用CreateBuilder来避免弃用（因为我的示例仅显示List中的所有数据），但我在这里使用添加条件有限制，请帮助我如何在使用时使用createBuilder像我在代码中使用 add（Restrictions.eq（＆＃34; user_name＆＃34;，userName））; 这样的条件，还是有其他解决方案可以避免弃用？

Answer 1

此日志是警告而非错误：

2018年4月28日下午12:24:45 org.hibernate.internal.SessionImpl createCriteria警告：HHH90000022：Hibernate的遗产 org.hibernate.Criteria API已弃用;使用JPA 而是javax.persistence.criteria.CriteriaQuery

它鼓励您在使用Hibernate时通过JPA API。

问题的根源是，您操纵Session（休眠）而不是EntityManager（JPA）。
反过来会引起实际代码的许多变化，因为大多数导入和处理方式都不同你不能改变一件事。

这是JPA方式（包括导入）：

import javax.persistence.EntityManager;
import javax.persistence.criteria.CriteriaBuilder;
import javax.persistence.criteria.CriteriaQuery;
import javax.persistence.criteria.Root;
import javax.transaction.Transactional;

@Transactional
public class AccountDAOImpl implements AccountDAO {

    @Autowired
    private EntityManager em;    

    @Override
    public ACCOUNTS findAccount(String userName) {    
        final CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
        CriteriaQuery<ACCOUNTS> crit = criteriaBuilder.createQuery(ACCOUNTS.class);
        Root<ACCOUNTS> root = crit.from(ACCOUNTS.class);
        crit.where(criteriaBuilder.equal(root.get("user_name"), userName))
            .distinct(true);
        return em.createQuery(crit).getSingleResult();      
    }

}

Answer 2

您可以使用JPA的CriteriaBuilder API，如下所示

// Create entity manager factory first
// Make sure the package name you specify `com.db.entity` matched what you specify in `pesistence.xml`
EntityManagerFactory entityManagerFactory = Persistence
        .createEntityManagerFactory("com.db.entity");
EntityManager em = entityManagerFactory.createEntityManager();


CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<User> cq = cb.createQuery(Accounts.class);
cq.where(cb.equal(join.get("name"), "ordinary"));
TypedQuery<Accounts> tq = em.createQuery(cq);
List<Accounts> accounts = tq.getResultList();

return accounts;

不推荐使用Hibernate的旧版org.hibernate.Criteria API

2 个答案: