java - Hibernate 的旧版 org.hibernate.Criteria API 已弃用

Question

当我使用 Hibernate 5 时，它显示错误为:

Apr 28, 2018 12:24:45 PM org.hibernate.internal.SessionImplcreateCriteria WARN: HHH90000022: Hibernate's legacyorg.hibernate.Criteria API is deprecated; use the JPAjavax.persistence.criteria.CriteriaQuery instead

@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();
    }

}

接口(interface)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 + "]";
    }
}

当我搜索时，它说使用 CreateBuilder 来避免弃用(因为我得到的示例仅显示 List 中的所有数据)，但我正在使用添加条件这里有限制。

请帮助我如何在使用条件时使用 createBuilder ，就像我在代码中使用 add(Restrictions.eq("user_name", userName)); 或者是否存在还有其他解决方案可以避免弃用吗？

Answer 1

此日志是警告而不是错误:

Apr 28, 2018 12:24:45 PM org.hibernate.internal.SessionImpl createCriteria WARN: HHH90000022: Hibernate's legacy org.hibernate.Criteria API is deprecated; use the JPA javax.persistence.criteria.CriteriaQuery instead

它鼓励您在使用 Hibernate 时绕过 JPA API。

问题的根源在于，您操作的是 Session (Hibernate)，而不是 EntityManager(JPA)。
执行相反的操作会导致实际代码发生许多变化，因为大多数导入以及处理方式都会有所不同。
你不能只改变一件事。

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();      
    }

}