Question

要求是从数据库中获取记录，其中在运行时选择了特定列和表名。

我正在使用Java，Hibernate和Mysql。可以使用JDBC。但我想用hibernate来做。

发布提示，如果有人知道

提前致谢

Answer 1

Hibernate的功能多于JDBC。最好的选择是Hibernate支持Object Oriented Principles。您可以对您的实体使用Polymorphism，Inheritance，Encapsulation。

您必须拥有BaseEntity类，并且所有其他类必须从此类扩展。

public class BaseEntity {

}

然后所有其他类都扩展了这个类。假设您有另外两个类或实体。

User.class

import java.io.Serializable;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;

@Table(name = "USERS")
@Entity
public class User extends BaseEntity implements Serializable {

    @Id
    @Column(name = "ID")
    private int id;

    @Column(name = "NAME")
    private String name;

    @Column(name = "SURNAME")
    private String surname;

    public User() {
    }

    public User(int id, String name, String surname) {
        this.id = id;
        this.name = name;
        this.surname = surname;
    }

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getSurname() {
        return surname;
    }

    public void setSurname(String surname) {
        this.surname = surname;
    }

}

Info.class

import java.io.Serializable;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;

@Table(name = "INFO")
@Entity
public class Info extends BaseEntity implements Serializable {

    @Column(name = "TEXT")
    private String text;

    @Id
    @Column(name = "ID")
    private int id;

    public Info() {
    }

    public Info(String text, int id) {
        this.text = text;
        this.id = id;
    }

    public String getText() {
        return text;
    }

    public void setText(String text) {
        this.text = text;
    }

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

}

另外，您必须为mapping每个类添加configuration.hbm.xml标记您的DAO如下所示。

public List getList(BaseEntity baseEntity, String[] columnNames) {
    Criteria criteria = session.createCriteria(baseEntity.getClass(), "be");
    ProjectionList projectionList = Projections.projectionList();
    for (String columnName : columnNames) {
        projectionList.add(Projections.property("be." + columnName), columnName); // adding column for selection
    }
    criteria.setProjection(projectionList);
    criteria.setResultTransformer(new AliasToBeanResultTransformer(baseEntity.getClass())); // for mapping result to entity
    return criteria.list();
}

public static void main(String[] args) {
    String[] columnNames = {"name", "surname"}; // columnNames which you want to select
    List<Info> users = new GeneralDAO().getList(new User(), columnNames); // you can pass any class User, Info or etc.
}

Answer 2

好吧，Hibernate可以做的不仅仅是JDBC，你正在寻找的东西可以用Hibernate这样编写：

Query query = session.createSQLQuery(
"select s.stock_code from stock s where s.stock_code = :stockCode")
.setParameter("stockCode", "7277");
List result = query.list();

正如您所看到的，表名和列可以动态定义，这至少与JDBC相同，甚至可能更简单。您不必使用参数，显然可以在代码中模块化地构建查询。

参考更多细节 https://www.mkyong.com/hibernate/hibernate-native-sql-queries-examples/

如何在运行时在hibernate中获取表名和列名的表记录？

2 个答案: