标准查询中的 SELECT答案

【问题标题】：SELECT in Criteria Queries标准查询中的 SELECT
【发布时间】：2012-08-24 17:59:17
【问题描述】：

例如，我有这个课程：

public class Person {
    private Integer id;  
    private String name;  
    private Address address;  
}

public class Address {
    private Integer id;  
    private String street;  
    private String number;      

    ...
}

我想在 JPA 中为此 SQL 构建 CriteriaQuery

select p.name, d.street from Person p join Address d on d = p.address;

对于这种特殊情况，我是否必须为 Person 类和 Address 类创建构造函数？

谢谢。

【问题讨论】：

标签： select jpa criteria multi-select

【解决方案1】：

您的where 条件无效：

select p.name, d.street from Person p join Address d on d = p.address;

应该是这样的：

select p.name, d.street from Person p join Address d on d.person_id = p.id;

或：

select p.name, d.street from Person p join Address d on d.id = p.address_id;

以及 Person 类的构造函数示例：

//cTor I
public Person(Integer id, String name, Address address){
    this.id = id;
    this.name = name;
    this.address = address;
}

//cTor II
public Person(Integer id, String name, Integer addressId, String street, String number){
    this.id = id;
    this.name = name;
    this.address = new Address(addressId, street, number);
}

【讨论】：

好的，谢谢您的回答。现在如何使用此 JPQL 制定标准？

【解决方案2】：

来自 JPQL select p.name, d.street from Person p join Address d on d = p.address; 目标标准

带有构造函数的选项：

首先在您的实体中标记关系 @OneToOne/@OneToMany 等等。使用@JoinColumn 即

    public class Person {
           private Integer id;  
           private String name;  

           @JoinColumn(name="id", referencedColumnName = "id")
           private Address address;  
}

稍后为您的结果类型创建一个类

      public class Result(){
           String name;
            String street;

         /* you must have constructor here for the multi select later */
         public Result(String name,String street) {
            this.name = name;
            thi.street = street
         }
      }

然后做（em 是实体管理器）

      CriteriaBuilder cb = em.getCriteriaBuilder();
      CriteriaQuery<PlaceSearch> query = cb.createQuery(PlaceSearch.class);
  Root<Person> personRoot = query.from(Person .class);
  Join<Person, Address> address= personRoot .join(Person_.address);

      / * without meta model 
  Join<Person, Address> address= personRoot .join(person.get("address"));
      */
/* we know Result has a construct of (String,String) . this is important for the                        multi select here */

query.multiselect(person.get(Person_.name),address.get(Address_.street);
List<Result> result = em.

createQuery(query).getResultList();

【讨论】：