Spring JPA - 部分查找EmbeddedId [英] Spring JPA - Find By EmbeddedId partially

问题描述

下面的代码仅用于演示目的。

My Entity bean看起来像这样

  @Entity 
类员工{
 
 @EmbeddedId 
私人EmployeeKey employeeKey; 
 
 private String firstName; 
 private String lastName; 
 
 //其他字段
 // Getter和Setters 
} 
  

可嵌入的类：

  @Embeddable 
类EmployeeKey实现了Serializable {
 
 private int employeeId; 
 private String branchName; 
 private String departmentName; 
 
 // Getter and Setters 
} 
  

---

我可以编写 JPARepository 接口方法来查找EmbeddedId的Employees，它也返回结果。

 接口EmployeeRepository扩展了JpaRepository< Employee，EmployeeKey> {
列表<员工> findByEmployeeKey（EmployeeKey employeeKey）; 
 
 b 
 
 
 
 $ b 
 问题：  
假设，在查询我只有 employeeId 和 branchName 时，我不想将过滤器放在 departmentName  
 
 
 
 
在这种情况下，我如何编写我的Repository方法
 
 
为这种情况构建？
      

@Ketrox的回答是绝对正确的，并且工作正常。但在我真实的情况下，我有6个字段进行搜索，并导致120+字符长的方法名称。 （ ）

 列表< Employee> findByEmployeeKeyField1AndEmployeeKeyField2AndEmployeeKeyField3AndEmployeeKeyField4AndEmployeeKeyField5AndEmployeeKeyField6（String field1，String field2，String field3，String field4，String field5，String field6）; 
  

这当然不够好，不足以使 codenarc 不高兴。

---

最后，我用例子找到了这个结果，真是令人愉快的解决方案。

储存库：

  //跳过行
 import org.springframework.data.domain.Example 
 //跳过的行
接口EmployeeRepository扩展JpaRepository< Employee，EmployeeKey> {
 List< Employee> findAll（示例<员工>员工）; 
} 
  

用法：

  //准备所有可用键搜索的Employee键（在我的例子中为6）
 EmplyeeKey键= new EmplyeeKey （）; 
 key.setField1（field1_value）; 
 key.setField2（field2_value）; 
 //设置剩余的4个字段
 
 //创建新员工并设置搜索关键字
 Employee employee = new Employee（）; 
 employee.setEmployeeKey（key）; 
 
 
 //通过传递上述Employee对象的示例来调用findAll 
 List< Employee> result = employeeRepository.findAll（Example.of（employee））; 
  

Below code is for demo purpose only.

My Entity bean looks like this

@Entity
class Employee {

    @EmbeddedId
    private EmployeeKey employeeKey;

    private String firstName;
    private String lastName;

    // Other fields
    // Getter and Setters
}

The Embeddable class:

@Embeddable
class EmployeeKey implements Serializable {

    private int employeeId;
    private String branchName;
    private String departmentName;

    //Getter and Setters
}

---

I can write JPARepository interface method to find Employees by the EmbeddedId that returns me results as well.

interface EmployeeRepository extends JpaRepository<Employee, EmployeeKey> {
        List<Employee> findByEmployeeKey(EmployeeKey employeeKey);
}

---

Question: Suppose, while querying I have employeeId and branchName only, and I don't want to put filter on departmentName

• In such cases how can I write my Repository method
• Does JPA have something in-build for such scenario?

解决方案

Here is how it worked for me.

@Ketrox's answer is absolutely correct and works fine. But in my real scenario I had 6 fields to search by and which resulted in an 120+ characters long method name. (Something like below)

List<Employee> findByEmployeeKeyField1AndEmployeeKeyField2AndEmployeeKeyField3AndEmployeeKeyField4AndEmployeeKeyField5AndEmployeeKeyField6(String field1, String field2, String field3, String field4, String field5, String field6);

Which is certainly not good enough to read and more than good enough to make codenarc unhappy.

---

Finally I used find by example and that turned out to be really pleasant solution.

Repository:

//skipped lines    
import org.springframework.data.domain.Example
//skipped lines
interface EmployeeRepository extends JpaRepository<Employee, EmployeeKey>{
    List<Employee> findAll(Example<Employee> employee);
}

Usage:

// Prepare Employee key with all available search by keys (6 in my case)
EmplyeeKey key = new EmplyeeKey();
key.setField1("field1_value");
key.setField2("field2_value");
//Setting remaining 4 fields

// Create new Employee ans set the search key
Employee employee = new Employee();
employee.setEmployeeKey(key);


// Call the findAll by passing an Example of above Employee object
List<Employee> result = employeeRepository.findAll(Example.of(employee));

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