CODEWITHSUNDEEP
swaggerswagger-uiswagger-editor
Jonathan Huet
Jonathan Huet

Reputation: 505

Swagger: Add description with ref

I want to add a description to an object property that his definition is referenced. Something like that:

      newCreditCard:
        type: object
        properties:
          billingPhone:
            description: Phone number of the card holder
            $ref: "#/definitions/PhoneNumber"

But the editor warns that the description property will be skipped:

Extra JSON Reference properties will be ignored: description

I have found a less elegant workaround that works for the editor, but not for the Swagger UI (not sure that is may due to the recent update to 3.0.2 version of the Swagger UI)

      newCreditCard:
        type: object
        properties:
          billingPhone:
            description: Phone number of the card holder
            allOf:
            - $ref: "#/definitions/PhoneNumber"

How do you do it in your Swaggers specification?

Thanks for the help!

Upvotes: 39

Views: 18855

Answers (4)

Eric Mutta
Eric Mutta

Reputation: 1437

For anyone using Swashbuckle with ASP.NET, you can use the following code to have the $ref construct put under the allOf (just like the :

// do this wherever you are calling AddSwaggerGen()
ArgBuilder.Services.AddSwaggerGen(opts => {
  opts.UseAllOfToExtendReferenceSchemas(); // add this line.
});

Now if you have a model with two properties of the same type, the individual descriptions for each field will show up in Swagger UI (e.g. below, both FooHeader and BarHeader are properties of type HttpHeader and their descriptions show up):

enter image description here

Upvotes: 4

Amir Sasson
Amir Sasson

Reputation: 11493

although it is not according to JSON standards. if you are using Swashbuckle to generate your swagger. i took advantage over the "Extensions" property of schema. and managed to create a swagger JSON, with $ref and extended properties.


var refSchema = new OpenApiSchema
{      
     //Reference = new OpenApiReference { ExternalResource = referenceLink, Type = ReferenceType.Link }, this wont work and omit all your other properties
    Extensions = new Dictionary<string, IOpenApiExtension>
    {
        { "$ref" , new OpenApiString(referenceLink) } // adding ref as extension cause according to JSON standards $ref shouldnt have any other properties
    },
    Description = prop.Value.Description,
    ReadOnly = prop.Value.ReadOnly,
    Required = prop.Value.Required,
    Type = prop.Value.Type,
    Example = prop.Value.Example,
};

Upvotes: 0

Arun Killu
Arun Killu

Reputation: 14233

If you add anything to the same level of $ref it will be ignored .

json $ref definition https://datatracker.ietf.org/doc/html/draft-pbryan-zyp-json-ref-03#section-3

correct way is to provide the description in the referenced object.

Upvotes: 0

Dave Delay
Dave Delay

Reputation: 1320

You could simply move the description property to the definition of PhoneNumber. Your original post does not show how you have defined PhoneNumber, but this snippet validates without warnings: 

definitions:
  phoneNumber:
    type: string
    description: Phone number of the card holder

  newCreditCard:
    type: object
    properties:
      billingPhone:
        $ref: "#/definitions/phoneNumber"

If this answer is not what you are looking for, please restate the question. We need to know what you are trying to accomplish.

Upvotes: 0

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