找不到类型的验证器：java.lang.Integer

Posted 2023-02-27

【中文标题】找不到类型的验证器：java.lang.Integer

【英文标题】：No validator could be found for type: java.lang.Integer

【发布时间】：2016-04-16 00:01:47

【问题描述】：

尝试处理我的创建/编辑表单时出现以下错误：

HTTP 状态 500 - 请求处理失败；嵌套异常是 javax.validation.UnexpectedTypeException：HV000030：没有验证器可以 可以找到类型：java.lang.Integer。

我的控制者：

/**
 * This method will provide the medium to update an existing Game.
 *
 * @param id bla
 * @param model bla
 * @return bla
 */
@RequestMapping(value = "/edit?id=id", method = RequestMethod.GET)
public String initEditGame(
        @PathVariable int id,
        ModelMap model
) 
    Game game = gameService.findById(id);

    model.addAttribute("game", game);
    model.addAttribute("edit", true);
    return "host/games/createOrUpdate";


/**
 * This method will be called on form submission, handling POST request for
 * updating game in database. It also validates the user input
 *
 * @param game bla
 * @param id bla
 * @param result bla
 * @param model bla
 * @param locale blie
 * @return bloe
 */
@RequestMapping(value = "/edit?id=id", method = RequestMethod.POST)
public String processEditGame(
        @ModelAttribute("game") @Valid Game game,
        BindingResult result,
        ModelMap model,
        @PathVariable int id,
        Locale locale
) 

    if (result.hasErrors()) 
        return "host/games/createOrUpdate";
    

    gameService.update(game);

    model.addAttribute("success", "Game " + game.getName() + " updated successfully");
    return "host/games/createGameSucces";

我的模型课：

@Entity
@Table(name = "GAME")
public class Game implements Serializable 

    @Id
    @Column(name = "ID")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer id;

    @NotEmpty
    @Column(name = "NAME", unique = true, nullable = false)
    private String name;

    @NotEmpty
    @Column(name = "DESCRIPTION", nullable = false)
    private String description;

    @NotNull
    @Length(min = 4, max = 4)
    @Column(name = "CODE", unique = true, nullable = false)
    private Integer code;

    @NotNull
    @Temporal(javax.persistence.TemporalType.DATE)
    @Column(name = "CREATED")
    @DateTimeFormat(pattern = "yyyy/MM/dd")
    private Date created;

    @NotNull
    @Column(name = "STATE", nullable = false)
    private String state = State.ACTIVE.getState();

    /**
     *
     */
    public Game() 
    

    /**
     *
     * @param name bla
     * @param description bleh
     */
    public Game(String name, String description) 
        this.name = name;
        this.description = description;
        this.code = generateRoomNumber();
    

有什么建议吗？

【参考方案1】：

我认为问题在于：

@NotNull
@Length(min = 4, max = 4)
@Column(name = "CODE", unique = true, nullable = false)
private Integer code;

@Length 注解只能用于 String 而不是 Integer

【讨论】：

啊哈，我会试试这个。但是我怎样才能验证这个长度的整数呢？

我把你的问题放在谷歌上，这是第一个链接：@Range(min=, max=)