Spring boot:java.sql.SQLSyntaxErrorException:你的SQL语法有错误;检查手册

时间:2021-08-01 18:00:43

标签: java spring-boot

我正在编写控制器类并收到此错误: java.sql.SQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'match, user_id_like, user_id_liked, user_liked_id) values (0, null, 'e89a0596-b1' at line 1

这是我的代码: UserController.java

@Autowired
UserService userService;

@PostMapping("/like/{userLikedId}")
public String likeUser(@PathVariable(name = "userLikedId") String userLikedId,
                       @RequestBody UserInfo userInfo) {
    System.out.println(userInfo.getUserId());
    userService.likeUser(userInfo.getUserId(), userLikedId);
    return "Liked other user!";
}

UserService.java

@Autowired
UserLikedRepository userLikedRepository;
@Autowired
UserRepository userRepository;
public UserLiked likeUser(String userLikeId, String userLikedId){
    UserLiked userLiked = new UserLiked();
    userLiked.setUserLikeId(userLikeId);
    userLiked.setUserLikedId(userLikedId);
    userLiked.setId(UUID.randomUUID().toString());
    System.out.println(userLiked.toString());

    userLikedRepository.save(userLiked);
    return userLiked;
}

邮递员:

http://localhost:8080/user/like/e89a0596-b121-4e40-8c64-844c868c2fae
raw: {
    "userLikeId": "user tesst"
}

控制台:

null
UserLiked(Id=57e176b9-3c5d-41b0-af0a-5f3bd14e0cac, userLikeId=null, userLikedId=e89a0596-b121-4e40-8c64-844c868c2fae, match=0)

为什么会出现此错误以及如何解决?谢谢。

0 个答案:

没有答案