请问一下如何改变Swagger的参数输入。很难说明白，进来看下图片

[yang147896523]

最近学Restful 的技术，搭了一个 JPA+hibernate+jboss+swagger，然后这个是我的customer的实体类

@Entity
@NamedQueries({
    @NamedQuery(name = Customer.FIND_ALL, query = "SELECT c FROM Customer c ORDER BY c.lastName ASC, c.firstName ASC"),
    @NamedQuery(name = Customer.FIND_BY_EMAIL, query = "SELECT c FROM Customer c WHERE c.email = :email")
})
@XmlRootElement
@Table(name = "customer", uniqueConstraints = @UniqueConstraint(columnNames = "email"))
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class,property = "id")  
public class Customer implements Serializable {

	
	private static final long serialVersionUID = 1L;

    public static final String FIND_ALL = "Customer.findAll";
    public static final String FIND_BY_EMAIL = "Customer.findByEmail";

    @Id
    @GeneratedValue(strategy = GenerationType.TABLE)
    @ApiModelProperty(hidden = true)
    private Long id;

    @OneToMany(mappedBy = "customer", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
    private Set<Booking> booking = new HashSet<Booking>();

    set....get...

这个是Booking的实体类。 customer和booking一对多的关联，并且设置了customer_id在booking表是外键，关联customer的id

@Entity
@NamedQuery(name = Booking.FIND_ALL, query = "SELECT c FROM Booking c ORDER BY c.id ASC")
@XmlRootElement
@Table(name = "booking")
public class Booking implements Serializable {

	
	private static final long serialVersionUID = 1L;

	public static final String FIND_ALL = "Booking.findAll";
	
	@Id
	@GeneratedValue(strategy = GenerationType.TABLE)
	private Long id;
	
	@NotNull
	@ManyToOne
	@JoinColumn(name = "customer_id")
	@ApiModelProperty(readOnly = true)
	private Customer customer;

        set....get....

这个是Restful的POST接口代码，我是以Booking作为参数进行输入参数

@SuppressWarnings("unused")
	@POST
	@ApiOperation(value = "Add a new Booking to the database")
	public Response createBooking(
			@ApiParam(value = "JSON representation of Booking object to be added to the database", required = true)
			Booking booking) {
       .....
}

按照以上的代码，在Swagger上显示的是


可以看到，当我要求输入的参数是Booking类型的时候，Swagger显示需要输入Booking里的Customer类型的所有值，因为我只需要customer的id就可以了，所以如何不改变Booking参数的输入，而且还能将这个Customer输入的值改成只需要输入customer的id就可以，而且这个customer_id 是 在booking与customer进行关联的列名。
如下图显示的结果。是不是有什么其他注释可以做到这个效果？

[Insist_on_progress]

@PostMapping("save")
@ApiOperation("发布内容")
@CrossOrigin
public JSONResponse save(Content content) {
int save = service.save(content);
if (save > 0) {
return ResultUtil.success("发布成功");
}
return ResultUtil.success("发布失败");
}