@ApiParam modifications?

[drm...@g.rit.edu]

Theres a couple of things I was wondering if are possible using swagger. First, can you override the type of an ApiParam? Also, can you remove the model/model schema fields in the parameters list? So that if you have a Car class I still want model schema on the end point but it bloats the parameter list and is redundant a lot of times and would be cool if I could remove it. Thanks

[tony tam]

Hi, as of 1.3.3-SNAPSHOT, you can mark a Model parameter as hidden:

@ApiModelProperty(hidden=true)

// your getter/setter

The Type override will be supported in 1.3.3 release for both models and @ApiParam annotations.

[drm...@g.rit.edu]

Ok cool, Another thing I'm wondering about is that it seems that in the Model Schema/Model that whenever it parses parameter that is an object, it appends it's model which is something that I do not want. Anything that can be done about that?

[tony tam]

Well, that's the whole idea.  What good would it be to have a "User" object with an unknown "Address" object buried inside it?  Swagger will scan the dependencies and append them to the model as well.  Is it getting them wrong?

[Twinkle]

Hi Tony, i have a requirement where i dont want to use all the parameters of nested object. How can i modify the body schema so that it would contains only necessary fields.

[tony tam]

Can you please give a simple example?

On Oct 23, 2015, at 10:07 PM, Twinkle <jassyt...@gmail.com> wrote:

Hi Tony, i have a requirement where i dont want to use all the parameters of nested object. How can i modify the body schema so that it would contains only necessary fields.

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[Twinkle]

Thanks for you quick reply :)

Assume user model schema is like below :

{
  "firstName": "string",
  "lastName": "string",
  "ids": [
    "string"
  ],
  "birthDetails": {
    "place": "string",
    "date": "string"
  },
  "address": {
    "homeAAddress": {
      "postalCode": "string",
      "place": "string"
    },
    "officeAAddress": {
      "postalCode": "string",
      "place": "string"
    }
  }
}

birthDetails and homeAAddress fields are not required in request. How i can avoid or hide them in model schema
Note : not use @ApiModel annotation

So that i would get :
{
  "firstName": "string",
  "lastName": "string",
  "ids": [
    "string"
  ],
  "address": {
    "homeAAddress": {
      "postalCode": "string",
      "place": "string"
    }
  }
}

as my model schema while using tryOut option

On Saturday, October 24, 2015 at 9:08:13 AM UTC+4, tony tam wrote:

Can you please give a simple example?

On Oct 23, 2015, at 10:07 PM, Twinkle <jassyt...@gmail.com> wrote:

Hi Tony, i have a requirement where i dont want to use all the parameters of nested object. How can i modify the body schema so that it would contains only necessary fields.

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[tony tam]

You can use @ApiModelProperty(hidden = true) on the property.  Won’t that work?

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[Twinkle]

Model schema is getting generated automatically. I am not using @ApiModel on my models.

Please check below my sample resource :
    @POST
    @Path("/create")
    @ApiOperation(notes = "create user", value = "create user")
    public Response createUser(UserDTO user) {
        // ...
        return responseBuilder.build();
    }

Can i hide above mentioned parameters here ??

[tony tam]

You can use the SwaggerSpecFilter mechanism to filter away properties.  See here:

https://github.com/swagger-api/swagger-core/blob/f95ea0c6f35f1ce9e72a677b410ec27d7886199e/modules/swagger-core/src/main/java/io/swagger/core/filter/SwaggerSpecFilter.java#L28

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[Twinkle]

Please find my Filter code below :

    @Override
    public boolean isParamAllowed(Parameter parameter, Operation operation,
            ApiDescription api, Map<String, List<String>> params,
            Map<String, String> cookies, Map<String, List<String>> headers) {

        // do not allow the documentation to be generated on parameters that
        // have their access set to "internal"
        if (parameter.getAccess() != null
                && parameter.getAccess().equalsIgnoreCase("internal")) {
            LOGGER.debug("The following parameter has been hidden from the Swagger API documentation: "
                    + parameter.getDescription());
            return false;
        } else {
            return true;
        }
    }


If I have only one parameter object, e.g. UserDTO then how I could put "access" on one property of user ?? Please help.