我是春天的新手。我想通过Spring Boot项目使用POSTMAN通过发布请求在MySQL数据库中添加一行。
我发送带有json正文的发布请求，该声明为表提供了必要的值。但是当我检查表时，我看到在表中插入了NULL值。您能帮我防止在表中插入空值吗？
CODE
AuthConroller.java类
import com.hackernewsclient.com.hackernewsclient.dto.ResgisterRequest;import com.hackernewsclient.com.hackernewsclient.service.AuthService;import org.springframework.beans.factory.annotation.Autowired;import org.springframework.http.HttpStatus;import org.springframework.http.ResponseEntity;import org.springframework.web.bind.annotation.*;@RestController@RequestMapping("/api/auth")public class AuthController {
@Autowired
private AuthService authService;
@PostMapping("/signup")
public ResponseEntity signup(@RequestBody ResgisterRequest resgisterRequest) {
authService.signup(resgisterRequest);
return new ResponseEntity(HttpStatus.OK);
}}
User.java实体类
import javax.persistence.*;@Entity@Table(name = "user")public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@Column
private String userName;
@Column
private String password;
@Column
private String email;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}}
AuthService.java类
import com.hackernewsclient.com.hackernewsclient.dto.ResgisterRequest;import com.hackernewsclient.com.hackernewsclient.model.User;import com.hackernewsclient.com.hackernewsclient.repository.UserRepository;import org.springframework.beans.factory.annotation.Autowired;import org.springframework.stereotype.Service;@Servicepublic class AuthService {
@Autowired
private UserRepository userRepository;
public void signup(ResgisterRequest resgisterRequest) {
User user = new User();
user.setUserName(resgisterRequest.getUsername());
user.setPassword(resgisterRequest.getPassword());
user.setEmail(resgisterRequest.getEmail());
userRepository.save(user);
}
}
UserRepository.java接口
import com.hackernewsclient.com.hackernewsclient.model.User;import org.springframework.data.jpa.repository.JpaRepository;import org.springframework.stereotype.Repository;@Repositorypublic interface UserRepository extends JpaRepository<User,Long> {}
PostRequest正文的RegisterRequest.java数据对象类
public class ResgisterRequest {
private String user_name;
private String password;
private String email;
public String getUsername() {
return user_name;
}
public void setUsername(String username) {
username = username;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
password = password;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
email = email;
}}
POSTMAN的屏幕快照-我的请求尝试（其状态为成功）
在MySQL中的用户表值的屏幕快照phpmyadmin
Intellij Log
Hibernate: insert into user (email, password, user_name) values (?, ?, ?)
您在ResgisterRequest二传手中遇到问题。username = username;将值分配给同一变量。它应该是this.user_name = user_name;
class ResgisterRequest {
private String user_name;
private String password;
private String email;
public String getUser_name() {
return user_name;
}
public void setUser_name(String user_name) {
this.user_name = user_name;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}}首先，您可以重新截图吗？它是不可见的。
请更正ResgisterRequest.java中的设置器。该方法中的值应为：this.password = password; 同样。请更正所有设置者。
在ResgisterRequest.java中将变量名称保留为userName，然后再次生成getter / setter。您可以从eclipse源菜单中生成getter和setter来避免此类错误。
您可以在AuthController方法中设置调试并检查resgisterRequest的值吗？