提问人:Nimasha Madhushani 提问时间:1/20/2023 最后编辑:Christian BaumannNimasha Madhushani 更新时间:1/20/2023 访问量:460
“status”:Spring Boot CRUD 中 POST 请求的 404 错误
"status": 404 Error for POST Request in Spring Boot CRUD
问:
我开发了Spring Boot CRUD应用程序。我连接的数据库是 PostgreSQL。 工作正常,GET 请求可以检索空的对象数组。但是在 中,POST 请求给出了 404 错误。@GetMapping
@PostMapping
📌SpringRecapApplication.java
package com.example.SpringRecap;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.boot.autoconfigure.security.servlet.SecurityAutoConfiguration;
import org.springframework.web.bind.annotation.*;
import java.util.List;
@RestController
@SpringBootApplication(exclude = SecurityAutoConfiguration.class)
//@RequestMapping("api/v1/customers")
//@EnableWebMvc
@RequestMapping(name = "api/v1/customers" ,method = RequestMethod.POST)
public class SpringRecapApplication {
//dependency injection
private static CustomerRepository customerRepository;
public SpringRecapApplication(CustomerRepository customerRepository) {
this.customerRepository = customerRepository;
}
public static void main(String[] args) {
SpringApplication.run(SpringRecapApplication.class, args);
}
@GetMapping
public List<Customer> getCustomer() {
return customerRepository.findAll();
}
record NewCustomerRequest(
String name,
String email,
Integer age
) {
@PostMapping
public void addCustomer(@RequestBody NewCustomerRequest newCustomerRequest) {
Customer customer = new Customer();
customer.setAge(newCustomerRequest.age());
customer.setName(newCustomerRequest.name());
customer.setEmail(newCustomerRequest.email());
customerRepository.save(customer);
}
}
}
customerRepository.save(customer);
不允许使依赖注入成为最终结果。( ).IDEA建议将其设置为静态。但它没有用。当我使用时,收到 405 错误。然后我通过执行以下操作解决了该问题,private static CustomerRepository customerRepository;
@RequestMapping("api/v1/customers")
@RequestMapping(name = "api/v1/customers" ,method = RequestMethod.POST)
📌客户存储库.java
package com.example.SpringRecap;
import org.springframework.data.jpa.repository.JpaRepository;
public interface CustomerRepository extends JpaRepository<Customer,Integer> {
}
📌客户 .java
package com.example.SpringRecap;
import jakarta.persistence.*;
import java.util.Objects;
@Entity
public class Customer {
@Id
@SequenceGenerator(
name = "customer_id_sequence",
sequenceName = "customer_id_sequence",
allocationSize = 1
)
@GeneratedValue(
strategy = GenerationType.SEQUENCE,
generator = "customer_id_sequence"
)
private Integer id;
private String name;
private String email;
private Integer age;
public Customer(Integer id, String name, String email, Integer age) {
this.id = id;
this.name = name;
this.email = email;
this.age = age;
}
public Customer() {
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public Integer getAge() {
return age;
}
public void setAge(Integer age) {
this.age = age;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Customer customer = (Customer) o;
return Objects.equals(id, customer.id) && Objects.equals(name, customer.name) && Objects.equals(email, customer.email) && Objects.equals(age, customer.age);
}
@Override
public int hashCode() {
return Objects.hash(id, name, email, age);
}
@Override
public String toString() {
return "Customer{" +
"id=" + id +
", name='" + name + '\'' +
", email='" + email + '\'' +
", age=" + age +
'}';
}
}
邮差:
如果需要更多信息来获得解决方案,请发表评论。
答:
3赞
times29
1/20/2023
#1
代码的问题在于,您将 POST 端点指定为 DTO 的一部分,而不是控制器的一部分。由于 DTO 不是 Spring 托管的 Bean,因此 Spring 不会将 URL 映射到您的端点。无论如何,您应该将终结点移动到单独的类中。例:
@RestController
@RequestMapping("api/v1/customers")
public class CustomerController {
private final CustomerRepository customerRepository;
public SpringRecapApplication(CustomerRepository customerRepository) {
this.customerRepository = customerRepository;
}
@GetMapping
public List<Customer> getCustomer() {
return customerRepository.findAll();
}
@PostMapping
public void addCustomer(@RequestBody NewCustomerRequest newCustomerRequest) {
Customer customer = new Customer();
customer.setAge(newCustomerRequest.age());
customer.setName(newCustomerRequest.name());
customer.setEmail(newCustomerRequest.email());
customerRepository.save(customer);
}
// Helper classes
record NewCustomerRequest(String name, String email, Integer age) { }
}
最好将 DTO 也移动到单独的类中。我建议将 DTO 放在包中,将控制器放在包中。dto
controller
两个旁注:你不应该通过你的 API 暴露你的实体。应将 DTO 用于传入和传出数据。看看龙目岛和mapstruct,它们使这变得非常容易。
评论
0赞
Nimasha Madhushani
1/20/2023
我想你已经提到龙目岛@Data注释来删除 getters setter,...等。
0赞
phaen
1/20/2023
我更喜欢使用 DTO 的记录 - 自 Java17 IMO 以来,龙目岛的使用被高估了
0赞
Nimasha Madhushani
1/20/2023
我尝试过 DTO。它正在工作。但@Autowired应该用于连接 CustomerRepository。在构造函数中,我不得不使用 void。但是,构造函数通常没有返回类型。我假设 void 在那里工作,因为 void 不返回任何内容。@倍29
0赞
Nimasha Madhushani
1/20/2023
谢谢,@times29的支持。现在它正在工作。正如我在上面的评论中提到的,我对你的代码做了一个小改动
0赞
times29
1/20/2023
@NimashaMadhushani,如果你的类中只有构造函数,则不需要注释。另外,请查看这篇文章,了解为什么你应该更喜欢构造函数注入而不是其他机制: reflectoring.io/constructor-injection@Autowired
评论