是否可以使我的自定义验证注释在持久时被忽略

Question

我使用 Spring 4 和 Hibernate 5

我有带有自定义验证器的密码字段的用户类。

我需要在表单绑定时验证它，长度为 8 个字符，包括小写和大写字母以及数字。

当用户输入密码时，它是有效的，但在我编码时它无效。

那么有没有办法让我的自定义验证注释在持久化时被忽略呢？

我知道我可以为未加密的密码制作不同的字段，或者制作数据传输对象，对其进行验证，然后将其数据发送给用户。但我对注释参数化的可能性感兴趣。

@Entity
@Table(name = "user")
public class User { 
//other fields
@NotNull
@NotEmpty
@ValidPassword
@Column(name = "password", nullable = false, length = 60)
private String password;
//getters and setters 
}

我的验证人

@Target({ TYPE, FIELD, ANNOTATION_TYPE })
@Retention(RUNTIME)
@Constraint(validatedBy = PasswordValidator.class)
@Documented
public @interface ValidPassword {
String message() default "Password is too short! Must be 8 digits and    include lowercase, uppercase letters and numbers.";
Class<?>[] groups() default {};
Class<? extends Payload>[] payload() default {};
}

和

public class PasswordValidator implements ConstraintValidator<ValidPassword, String> {
private Pattern pattern;
private Matcher matcher;
private static final String PATTERN = "((?=.*\d)(?=.*[a-z])(?=.*[A-Z]).{8,})";
@Override
public void initialize(ValidPassword constraintAnnotation) {
}
@Override
public boolean isValid(String password, ConstraintValidatorContext context) {
    return (validate(password));
}
private boolean validate(String password) {
    pattern = Pattern.compile(PATTERN);
    matcher = pattern.matcher(password);
    return matcher.matches();
}
}

控制器方法

@RequestMapping(value = "/registeruser", method = RequestMethod.POST)
public String registerUser(@ModelAttribute("user") @Valid User user, BindingResult result, Model model) {
    if (result.hasErrors()) {
        model.addAttribute("errorSummary", result.getFieldErrors().stream()
                .map(e -> e.getField() + " error - " + e.getDefaultMessage() + " ").collect(Collectors.toList()));
        model.addAttribute("user", user);
    } else {
        User registered = null;
        registered = createUserAccount(user, result);
        if (registered == null) {
            model.addAttribute("errorSummary", "User with this email already registered!");
            model.addAttribute("user", user);
            return "registration";
        }
        model.addAttribute("flashMessage", "User registered successfully!");
    }
    return "registration";
}

用户服务实现方法(我编码密码的地方(

@Transactional
@Override
public User registerNewUserAccount(User user) throws EmailExistsException {
    if (emailExist(user.getEmail())) {
        throw new EmailExistsException("There is an account with that email address:" + user.getEmail());
    }
    if (user.getPassword() == null) {
        user.setPassword(new BigInteger(130, new SecureRandom()).toString(32));
        System.out.println("+++++++++++++++" + user.getPassword());
    }
    user.setPassword(passwordEncoder.encode(user.getPassword()));
    user.setUserRole(new HashSet<UserRole>(1));
    user.getUserRole().add(new UserRole(user, Constants.RoleType.USER.name()));
    save(user);
    return user;
}

Answer 1

默认情况下，将对所有约束进行验证。或者，您可以指定分组约束

您可以通过创建接口来创建组：

interface FormValidationGroup{}

并像这样注释password字段：

@ValidPassword(groups = FormValidationGroup.class)
private String password;

提及

groups参数的自定义约束注释的文档。

休眠验证程序现在应忽略password字段，除非您指定要验证的组。要指定用于验证 Spring MVC 处理程序方法参数的组，请使用 已验证注释 而不是 Valid 。例如：

String registerUser(@ModelAttribute @Validated(FormValidationGroup.class) User user,
        BindingResult result, Model model) {
    if (result.hasErrors()) {