Question

我使用在Tomcat上运行的Spring MVC + Hibernate + MySql和Intellij Editor

我想像这样查询像MySql这样的HQL

SELECT * from Customer c LEFT JOIN Person p ON p.idPerson = c.idCustomer

它没有按照这样的方式返回错误

HTTP Status 500 - Request processing failed; nested exception is org.hibernate.hql.internal.ast.QuerySyntaxException: Path expected for join! [from com.springapp.mvc.model.Customer c LEFT JOIN Person p ON p.idPerson = c.idCustomer]

这是我的数据库

    CREATE TABLE `Customer` (
  `idCustomer` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(45) DEFAULT NULL,
  PRIMARY KEY (`idCustomer`),
  CONSTRAINT `customer to person` FOREIGN KEY (`idCustomer`) REFERENCES `Person` (`idPerson`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

    CREATE TABLE `Person` (
  `idPerson` int(11) NOT NULL AUTO_INCREMENT,
  `country` varchar(45) DEFAULT NULL,
  PRIMARY KEY (`idPerson`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

从customer.idCustomer到person.idPerson的关系是OneToOne

这是我的代码

model.Customer.java

@Entity
@Table(name = "Customer")
public class Customer {
    private int idCustomer;
    private String name;

    @Id
    @Column(name = "idCustomer")
    public int getIdCustomer() {
        return idCustomer;
    }

    public void setIdCustomer(int idCustomer) {
        this.idCustomer = idCustomer;
    }

    @Basic
    @Column(name = "name")
    public String getName() {
        return name;
    }

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

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        Customer customer = (Customer) o;

        if (idCustomer != customer.idCustomer) return false;
        if (name != null ? !name.equals(customer.name) : customer.name != null) return false;

        return true;
    }

    @Override
    public int hashCode() {
        int result = idCustomer;
        result = 31 * result + (name != null ? name.hashCode() : 0);
        return result;
    }
}

model.Person.java

@Entity
@Table(name = "Person")
public class Person {
    private int idPerson;
    private String country;

    @Id
    @Column(name = "idPerson")
    public int getIdPerson() {
        return idPerson;
    }

    public void setIdPerson(int idPerson) {
        this.idPerson = idPerson;
    }

    @Basic
    @Column(name = "country")
    public String getCountry() {
        return country;
    }

    public void setCountry(String country) {
        this.country = country;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        Person person = (Person) o;

        if (idPerson != person.idPerson) return false;
        if (country != null ? !country.equals(person.country) : person.country != null) return false;

        return true;
    }

    @Override
    public int hashCode() {
        int result = idPerson;
        result = 31 * result + (country != null ? country.hashCode() : 0);
        return result;
    }
}

这是我的实施 dao.CustomerDAOImpl.java

public List<Customer> listCustomers() {
    Session session = this.sessionFactory.getCurrentSession();
    List<Customer> customersList = session.createQuery("from Customer c LEFT JOIN Person p ON p.idPerson = c.idCustomer").list();

    for (Customer c : customersList) {
        logger.info("Customer List::" + c);
    }

    return customersList;
}

我是新的java开发人员，英文不好有人请帮帮我谢谢

Answer 1

Hibernate / JPA构建某种数据库抽象层。因此，它可以与Entity（而不是表格）和关系（@ManyToOne ...）一起使用实体，而不是明确地使用Joining Ids来获取表格列。

@Entity
Person {
   @Id
   Long id;
   ...
   @OneToOne
   @JoinColumn(name = "customer_fk")   
   Customer customer;
   //NO other customer_fk mapping!!!
}


@Entity
Customer {
  @Id
  Long id; 
}

现在你可以SELECT c from Person p LEFT JOIN p.customer c

如何使用Spring + Hibernate查询连接表

1 个答案: