2017-03-17 70 views
1

我有一個類「SubModel」,它繼承了類「SuperModel」。我的REST請求返回其中一個類的對象。現在我想在我的Swagger UI中記錄那個SubModel的對象或SuperModel的對象將被返回。我搜索了這個用例,發現了@ApiModel註解,但他們不工作,有沒有人知道我在這裏做錯了什麼?Swagger does not show inherited objects

@ApiModel(value = "SuperModel", discriminator = "foo", subTypes = { 
SubModel.class }) 
public class SuperModel 
{ 
    @ApiModelProperty(required = true) 
    private String foo; 

    public String getFoo() { 
     return this.foo; 
    } 

    public void setFoo(String foo) { 
     this.foo = foo; 
    } 
} 

@ApiModel(value = "SubModel") 
public class SubModel extends SuperModel 
{ 
    private int number; 

    public int getNumber() { 
     return this.number; 
    } 

    public void setNumber(int number) { 
     this.number = number; 
    } 
} 

我在揚鞭UI看到的只是:

SuperModel { 
    foo (string) 
} 

我發現的例子在這個網站:https://github.com/swagger-api/swagger-core/wiki/Annotations#apiresponses-apiresponse但它不工作:-(

任何想法非常感謝!

+0

[Swagger Inheritance and Composition]的可能重複(http://stackoverflow.com/questions/27862407/swagger-inheritance-and-composition) –

回答