Question

我是使用Java和Spring启动的新手，我试图验证将存储在DB中的json输入，验证效果很好，问题是响应格式，我认为它对于客户端来说是不可读的Angular JS，需要知道如何更改输出格式：

{
"error": {
    "code": 400,
    "message": "Validation failed for classes [com.company.customerhelpcenter.domain.Ticket] during persist time for groups [javax.validation.groups.Default, ]\nList of constraint violations:[\n\tConstraintViolationImpl{interpolatedMessage='Descripción no puede ser null', propertyPath=description, rootBeanClass=class com.company.customerhelpcenter.domain.Ticket, messageTemplate='Descripción no puede ser null'}\n\tConstraintViolationImpl{interpolatedMessage='not a well-formed email address', propertyPath=email, rootBeanClass=class com.company.customerhelpcenter.domain.Ticket, messageTemplate='{org.hibernate.validator.constraints.Email.message}'}\n\tConstraintViolationImpl{interpolatedMessage='Número de teléfono no puede ser null', propertyPath=phone, rootBeanClass=class com.company.customerhelpcenter.domain.Ticket, messageTemplate='Número de teléfono no puede ser null'}\n]"
}}

到这个

{
"error": {
    "description" : "Description cannot be null",
    "email": "Email cannot be null",
    "phone": "Phone cannot be null"
} }

这是我的控制器

@PostMapping
public ResponseEntity<?> create(@RequestBody Ticket ticket)
{
    return new ResponseEntity<>(this.ticketService.create(ticket), HttpStatus.CREATED);
}

服务：

public Ticket create(Ticket ticket)
{
    return this.ticketRepository.save(ticket);
}

存储库：

@Repository
public interface TicketRepository extends JpaRepository<Ticket, Integer>, Serializable
{}

实体

@Entity
@Table(name = "ticket")
public class Ticket implements Serializable
{
    private static final long serialVersionUID = 1L;

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

    @Basic(optional = false)
    @NotNull(message = "Descripción no puede ser null")
    @Size(min = 1, max = 255)
    @Column(name = "description")
    private String description;

    @Basic(optional = false)
    @NotNull(message = "Número de teléfono no puede ser null")
    @Size(min = 1, max = 20)
    @Column(name = "phone")
    private String phone;

    @Basic(optional = false)
    @NotNull(message = "Email no puede ser null")
    @Size(min = 1, max = 80)
    @Email
    @Column(name = "email")
    private String email;

    @Basic(optional = false)
    @NotNull
    @Column(name = "locked")
    private boolean locked;

    @JoinColumn(name = "ticket_type_id", referencedColumnName = "id")
    @ManyToOne(optional = false, fetch = FetchType.EAGER)
    @Exclude
    private TicketType ticketType;

    @OneToMany(mappedBy = "ticket")
    private Set<TicketHistory> ticketHistories = new HashSet<TicketHistory>();


    public Integer getId()
    {
        return this.id;
    }

    public String getDescription()
    {
        return this.description;
    }

    public String getPhone()
    {
        return this.phone;
    }

    public String getEmail()
    {
        return this.email;
    }

    public TicketType getTicketType()
    {
        return this.ticketType;
    }

    public boolean isLocked()
    {
        return this.locked;
    }

    public Set<TicketHistory> getTicketHistories()
    {
        return this.ticketHistories;
    }
}

Answer 1

您的例外是ConstraintValidationException，这意味着代码通过Controller传递但由于验证而无法将数据保存在数据库中。我的建议是在此之前验证数据，在获取Controller之前验证数据。所以，就像这样：

@ControllerAdvice
public class GlobalExceptionHandler {

    @ExceptionHandler({MethodArgumentNotValidException.class})
    public ResponseEntity<Map<String, Object>> yourExceptionHandler(MethodArgumentNotValidException e) {
        Map<String, Object> response = new HashMap<String, Object>();
        Map<String, String> errors = new HashMap<String, String>();

        BindingResult bindingResult = e.getBindingResult();
        List<FieldError> fieldErrors = bindingResult.getFieldErrors();
        for (FieldError fieldError : fieldErrors) {
            errors.put(fieldError.getField(), fieldError.getDefaultMessage());
        }

        response.put("error", errors);
        return new ResponseEntity<Map<String, Object>>(response, HttpStatus.BAD_REQUEST);
    }
}

您的控制器应该有@Valid注释：

@PostMapping
public ResponseEntity<?> create(@RequestBody @Valid Ticket ticket) {
    return new ResponseEntity<>(this.ticketService.create(ticket), HttpStatus.CREATED);
}

然后，您的验证工具（例如@NotNull和@Size等）的验证将会得到验证。

但是，如果您不想像这样进行验证，并且只想处理ConstraintValidationException，则可以执行此操作：

@ControllerAdvice
public class GlobalExceptionHandler {

    @ExceptionHandler({ ConstraintViolationException.class})
    public ResponseEntity<Map<String, Object>> yourExceptionHandler(ConstraintViolationException e) {
        Map<String, Object> response = new HashMap<String, Object>();
        Map<String, String> errors = new HashMap<String, String>();
        Set<ConstraintViolation<?>> constraintViolations = e.getConstraintViolations();

        for (ConstraintViolation<?> constraintViolation : constraintViolations) {
            errors.put(constraintViolation.getPropertyPath().toString() , constraintViolation.getMessage());
        }

        response.put("error", errors);
        return new ResponseEntity<Map<String, Object>>(response, HttpStatus.BAD_REQUEST);
    }
}

错误验证消息格式Spring启动

1 个答案: