Spring Boot：使用dblink在多個數據庫中對同一實體建模

Question

我有一個數據庫MAIN_DB ，它具有到其他數據庫的EMP_DB1 EMP_DB2 ： EMP_DB1 ， EMP_DB2 ， EMP_DB3和EMP_DBx 。 為了簡單起見，我們假設另外兩個數據庫。

EMP_DB1和EMP_DB2都具有具有相同表結構的表EMPLOYEE ：

create table employee(
id number primary key,
first_name varchar2(40),
last_name varchar2(40));

MAIN_DB能夠通過它們的dblink查詢EMP_DB1和EMP_DB2 ：

select * from employee@emp_db1;
1  John  Smith

select * from employee@emp_db2;
2  Jane  Doe

我想讓RestController監聽/employee/{id}並根據id查詢相應的EMP_DBx 。

當前工作（不理想）的示例：

com.apitest.repository.PersonRepository

package com.apitest.repository;

import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.repository.NoRepositoryBean;

@NoRepositoryBean
interface PersonRepository<T> extends JpaRepository<T, Long> {
}

com.apitest.repository.EmployeeRepository

package com.apitest.repository;

import com.apitest.model.Employee;
import org.springframework.stereotype.Repository;

@Repository
public interface EmployeeRepository extends PersonRepository<Employee> {
    Employee findById(Long id);
}

com.apitest.repository.Employee2Repository

package com.apitest.repository;

import com.apitest.model.Employee2;
import org.springframework.stereotype.Repository;

@Repository
public interface Employee2Repository extends PersonRepository<Employee2> {
    Employee2 findById(Long id);
}

com.apitest.model.Person

package com.apitest.model;

...

@MappedSuperclass
public abstract class Person {
    private long id;
    private String firstName;
    private String lastName;

    @Id
    @Column(name = "ID")
    public long getId() {
        return id;
    }

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

    @Basic
    @Column(name = "FIRST_NAME")
    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    @Basic
    @Column(name = "LAST_NAME")
    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    public String toString() {
        return "id: " + id + ", " +
                "first_name: " + firstName + ", " +
                "last_name: " + lastName;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        Person employee = (Person) o;
        return id == employee.id &&
                Objects.equals(firstName, employee.firstName) &&
                Objects.equals(lastName, employee.lastName);
    }

    @Override
    public int hashCode() {
        return Objects.hash(id, firstName, lastName);
    }
}

com.apitest.model.Employee

package com.apitest.model;

import javax.persistence.Entity;

@Entity(name = "EMPLOYEE@EMP_DB1")
public class Employee extends Person {

}

com.apitest.model.Employee2

package com.apitest.model;

import javax.persistence.Entity;

@Entity(name = "EMPLOYEE@EMP_DB2")
public class Employee2 extends Person {

}

com.apitest.controller.EmployeeController

package com.apitest.controller;

...

@RestController
@RequestMapping("/employee")
public class EmployeeController {
    private final EmployeeRepository employeeRepository;
    private final Employee2Repository employee2Repository;

    @Autowired
    public EmployeeController(EmployeeRepository employeeRepository, Employee2Repository employee2Repository) {
        this.employeeRepository = employeeRepository;
        this.employee2Repository = employee2Repository;
    }

    // logic will be different, this is just an example
    @RequestMapping(path = "/{id}", method = RequestMethod.GET)
    public ResponseEntity<?> getByUserid(@PathVariable("id") Long id) {
        if (id % 2 == 1) {
            Employee employee = employeeRepository.findById(id);
            return new ResponseEntity<>(ResponseEntity.ok(employee), HttpStatus.OK);
        }
        else {
            Employee2 employee2 = employee2Repository.findById(id);
            return new ResponseEntity<>(ResponseEntity.ok(employee2), HttpStatus.OK);
        }
    }
}

有沒有辦法設置Entity名稱的參數？

@Entity(name = "EMPLOYEE@{db_link_name}")
public class Employee extends Person

如果沒有，如何使用@Query查詢dblink？

EmployeeRepository ：

public interface EmployeeRepository extends JpaRepository<Employee, Long> {
    @Query("select e from employee@emp_dbx e e.id = ?1")
    Employee findById(Long id);
}

Spring不喜歡查詢字符串中的@dblink_name 。 我可以創建同義詞來解決此問題，但是無法參數化同義詞名稱：

create or replace synonym employee_emp_db1 for employee@emp_db1;
create or replace synonym employee_emp_db2 for employee@emp_db2;

public interface EmployeeRepository extends JpaRepository<Employee, Long> {
    @Query("select e from ?1 e e.id = ?2")
    Employee findById(String dblinkName, Long id);
}

QuerySyntaxException: unexpected token: ? near line 1, column 15 [select e from ?1 e e.id = ?2]

所需行為：

GET /employee/1
{
    "headers": {},
    "body": {
        "id": 1,
        "firstName": "John",
        "lastName": "Smith"
    },
    "statusCode": "OK",
    "statusCodeValue": 200
}

GET /employee/2
{
    "headers": {},
    "body": {
        "id": 2,
        "firstName": "Jane",
        "lastName": "Doe"
    },
    "statusCode": "OK",
    "statusCodeValue": 200
}

我的問題歸結為什么是春季最佳的最佳設計，我該如何實現呢？

Answer 1

您應該實現多租戶架構。 這將是用於您的用例的正確設計模式。 查看此鏈接，使您的spring / spring-boot應用成為多租戶。 https://fizzylogic.nl/2016/01/24/make-your-spring-boot-application-multi-tenant-aware-in-2-steps/