java - 将 JSON 数组发布到 Spring Boot 2 JPA 中

Question

我想这样发帖:

[{
        "employeeid": "1111",
        "employeename": "YOA"

},
{
        "employeeid": "2222",
        "employeename": "OYA"
}]

我的 Controller 是这样的:

@PostMapping("/api/employee/save")
public Employee createEmployee(@Valid @RequestBody List<Employee> employee) {
        return employeeService.save(employee);
}

型号:

@Entity
@Table(name="EMPLOYEE")
public class Employee implements Serializable{
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "ID")
int id;
@Column(name = "EMP_ID")
int employeeid;
@Column(name = "EMPLOYEE_NAME")
String employeename;

//GETTER AND SETTER
}

当我发布数据时，我得到的错误如下:

{
    "timestamp": "2018-07-13T03:36:25.898+0000",
    "status": 500,
    "error": "Internal Server Error",
    "message": "Invalid property 'id' of bean class [java.util.ArrayList]: Could not find field for property during fallback access!",
    "path": "/api/employee/save"
}

服务:

public interface employeeService{

    Employee save(List<Employee> employee);

}

服务 import/export :

@Service("employeeService")
public class EmployeeServiceImpl implements EmployeeService {

    @Autowired
    EmployeeRepository employeeRepository;

    @Override
    public Employee save(List<Employee> employee) {
     return employeeRepository.save(employee);
}

}

存储库:

public interface EmployeeRepository extends Repository <Employee, Long>
    Employee save(List<Employee> employee);
}

错误描述是这样的:bean 类 [java.util.ArrayList] 的属性“id”无效:在回退访问期间找不到属性字段!

使用 saveAll() 时，我收到错误消息:未找到 Employee 类型的属性 saveAll

有谁可以帮助我吗？我修改了我的问题。

问候，我

Answer 1

问题出在这一行

public Employee createEmployee(@Valid @RequestBody List<Employee> employee) {
        return employeeService.save(employee);  //  Problem 
} 

employeeService.save 在您的案例中只能获取一个实体对象Employee

有两种方法

1.

public Boolean createEmployee(@Valid @RequestBody List<Employee> lstEmployee) {
    try{
        for(Employee emp : lstEmployee){
            employeeService.save(employee);
        }
        return true;
    }catch(Exception e){}
    return false;
}

2.

使用saveAll代替

public Employee createEmployee(@Valid @RequestBody List<Employee> employee) {
        return employeeService.saveAll(employee);  
}

编辑 1:

添加服务类后，看起来您正在手动实现

选项 1:

我建议您直接在 Controller 类中使用EmployeeRepository。通过手动覆盖这些方法，您实际上并不是享受使用存储库的好处

@Autowired
EmployeeRepository employeeRepository;

@PostMapping("/api/employee/save")
public Employee createEmployee(@Valid @RequestBody List<Employee> lstEmployee) {
        return employeeRepository.saveAll(lstEmployee);
}

选项 2:

更进一步，像这样改变你的实现。对象可能存在一些错误，但它应该给您一个想法

public interface employeeService{

    Employee save(Employee employee);

    public <S extends User> List<S> saveAll(Iterable<S> entites);



}

@Service("employeeService")
public class EmployeeServiceImpl implements EmployeeService {

    @Autowired
    EmployeeRepository employeeRepository;

    @Override
    public Employee save(Employee employee) {
        return employeeRepository.save(employee);
    }

   @Override
    public List<Employee> saveAll(List<Employee> employee) {
        return employeeRepository.saveAll(employee);
    }

}