Swagger Parameters and Complex Types

Question

In the following Swagger definition, I need the parameter labelValue to be of type LabelValueObject, so that it will be validated and correctly deserialized. However, I can't figure out the syntax! How can that be done?

swagger: "2.0" paths: /competition: post: parameters: - name: labelValue in: formData type: array items: type: string ### this has to be a LabelValueObject ### responses: default: description: Error schema: $ref: "#/definitions/AnyResponse" definitions: AnyResponse: properties: any: type: string LabelValueObject: properties: label: type: string value: type: string required: - label - value

You're right. I could set it to body and use a schema $ref I guess... I'm new to Swagger, so I missed that. If that's where you were going with your question, please post an answer. — Merott
– Merott, Commented Aug 28, 2016 at 0:37

Arnaud Lauret · Accepted Answer · 2016-08-28 18:27:59Z

The only way to pass an object as a parameter is to put it in the body (in: body) and then define this object in schema (inline definition or reference to an predefined object with $ref). Here's a full example:

swagger: "2.0" info: title: A dummy title version: 1.0.0 paths: /competition: post: parameters: - name: labelValue in: body schema: $ref: '#/definitions/LabelValueObject' responses: default: description: Error schema: $ref: "#/definitions/AnyResponse" definitions: AnyResponse: properties: any: type: string LabelValueObject: properties: label: type: string value: type: string required: - label - value

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