java - Spring，从异常中获取多个模型违规

Question

我正在尝试验证用户注册。场景很常见:如果用户名或电子邮件存在，则应用程序必须通知用户哪个字段被违反(唯一约束)。到目前为止，我已经完成了以下操作: 该方法处理注册。基本上我尝试提交事务(插入)，如果失败，则用 try catch 捕获异常。

@RequestMapping(value = "/registration", method = RequestMethod.POST)
public ModelAndView createNewUser(@Valid User user, BindingResult bindingResult) {
    ModelAndView modelAndView = new ModelAndView();
    try{
        userService.saveUser(user);
        modelAndView.addObject("successMessage", "User has been registered successfully");
        modelAndView.addObject("user", user);
        modelAndView.setViewName("registration");
    }catch(DataIntegrityViolationException e){
        bindingResult
        .rejectValue("email", "error.user",
                "There is already a user registered with the email provided");
    }
    return modelAndView;
}

这是我的模型类:

@Entity
@Table(name = "users")
public class User implements Serializable{

private static final long serialVersionUID = 1L;

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name="user_id")
private Long user_id;

@Column(name="username")
@NotEmpty(message = "*Username is required")
@Length(min = 5, max = 20,  message = "*Your username can have min 5 and max 20 characters")
private String username;

@Column(name="password")
@NotEmpty(message = "*Password is reqired")
@Transient
private String password;

@Column(name="email")
@Email(message = "*Please provide a valid Email")
@NotEmpty(message = "*Email is required")
private String email;

@Column(name="enabled")
private int enabled;

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;
}

public int getEnabled() {
    return enabled;
}

public void setEnabled(int enabled) {
    this.enabled = enabled;
}

public Long getUser_id() {
    return user_id;
}

public void setUser_id(Long user_id) {
    this.user_id = user_id;
}
}

这是我的注册页面:

<form autocomplete="off" action="#" th:action="@{/registration}"
                th:object="${user}" method="post" class="form-horizontal"
                role="form">
                <h2>Registration Form</h2>
                <div class="form-group">
                    <div class="col-sm-9">
                    <label th:if="${#fields.hasErrors('username')}" th:errors="*{username}"
                            class="validation-message"></label>
                    <input type="text" th:field="*{username}" placeholder="Username"
                            class="form-control" /> 
                    </div>
                </div>

                <div class="form-group">
                    <div class="col-sm-9">
                        <input type="text" th:field="*{email}" placeholder="Email"
                            class="form-control" /> <label
                            th:if="${#fields.hasErrors('email')}" th:errors="*{email}"
                            class="validation-message"></label>
                    </div>
                </div>
                <div class="form-group">
                    <div class="col-sm-9">
                        <input type="password" th:field="*{password}"
                            placeholder="Password" class="form-control" /> <label
                            th:if="${#fields.hasErrors('password')}" th:errors="*{password}"
                            class="validation-message"></label>
                    </div>
                </div>

                <div class="form-group">
                    <div class="col-sm-9">
                        <button type="submit" class="btn btn-primary btn-block">Register User</button>
                    </div>
                </div>
                <span th:utext="${successMessage}"></span>
            </form>

问题。现在的问题是，在我的 createNewUser 方法中，我确实 try catch 。当发生唯一违规时，catch block 会处理它。但如果存在多个违规行为:用户名和电子邮件怎么办？如何从异常中获取哪个字段被违反并将其与正确的表单字段关联？

Answer 1

我想说您需要添加自己的自定义 validator 来检查唯一的用户名和电子邮件。

代码摘自下面的示例。 @Unique(service = UserService.class, fieldName = "email", message = "{email.unique.violation}") 私有(private)字符串电子邮件；

注释和 validator

@Target({ METHOD, FIELD, ANNOTATION_TYPE })
@Retention(RUNTIME)
@Constraint(validatedBy = UniqueValidator.class)
@Documented
public @interface Unique {
    String message() default "{unique.value.violation}";
    Class<?>[] groups() default {};
    Class<? extends Payload>[] payload() default {};
    Class<? extends FieldValueExists> service();
    String serviceQualifier() default "";
    String fieldName();
}

和

public class UniqueValidator implements ConstraintValidator<Unique, Object> {
    @Autowired
    private ApplicationContext applicationContext;

    private FieldValueExists service;
    private String fieldName;

    @Override
    public void initialize(Unique unique) {
        Class<? extends FieldValueExists> clazz = unique.service();
        this.fieldName = unique.fieldName();
        String serviceQualifier = unique.serviceQualifier();

        if (!serviceQualifier.equals("")) {
            this.service = this.applicationContext.getBean(serviceQualifier, clazz);
        } else {
            this.service = this.applicationContext.getBean(clazz);
        }
    }

    @Override
    public boolean isValid(Object o, ConstraintValidatorContext constraintValidatorContext) {
        return !this.service.fieldValueExists(o, this.fieldName);
    }
}

参见the example