因此,我正在尝试构建一个餐饮计划程序,但当我尝试提交用户名时,它会显示标题中显示的错误。我尝试向我的类添加额外的映射,我为User类创建了一个新的存储库,但我不确定为什么程序没有像它应该的那样自动递增user_ID值。
下面是我的令人不快的mySQL表:
CREATE TABLE `user_table` (
`user_id` int NOT NULL AUTO_INCREMENT,
`user_name` varchar(30),
PRIMARY KEY (`user_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci;
/*!40101 SET character_set_client = @saved_cs_client */;下面是User类:
@Entity
@Table(name="user_table")
public class User {
@Id
@GeneratedValue(strategy= GenerationType.IDENTITY)
@Column(name="user_ID")
private int userID;
@Column(name="user_name")
private String userName;
public User() {
super();
// TODO Auto-generated constructor stub
}
public User(int userID, String userName) {
super();
this.userID = userID;
this.userName = userName;
}
public int getUserID() {
return userID;
}
public void setUserID(int userID) {
this.userID = userID;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
@Override
public String toString() {
return "User [userID=" + userID + ", userName=" + userName + "]";
}
}用户类附加到的存储库:
@Repository
public interface MealPlannerRepositoryUser extends JpaRepository<User, Long>{
}以及我的控制器类中的相关方法:
@Autowired
MealPlannerRepository repo;
@Autowired
MealPlannerRepositoryUser userRepo;
@GetMapping("/inputUser")
public String addNewUser(Model model) {
User u = new User();
Meal m = new Meal();
model.addAttribute("newUser", u);
model.addAttribute("newMeal", m);
return "inputUser";
}
@PostMapping("/inputUser")
public String addNewUser(@ModelAttribute User u, Meal m, Model model) {
userRepo.save(u);
repo.save(m);
model.addAttribute("newUser", repo.findAll());
model.addAttribute("newMeal", repo.findAll());
return "results";
}我一定是漏掉了什么,但我没办法了。
发布于 2019-11-13 02:10:26
我不确定这是否区分大小写,但是User类中的注释@ column ( 'user_id‘)与您的SQL查询中的列名’user_id‘不匹配。
发布于 2019-11-14 14:36:26
实体类user_id与列名user_ID不匹配,请在实体类user_id中进行更改
https://stackoverflow.com/questions/58824345
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