How to create api using post request with json as payload using swagger

Question

I try to create api for post request using swagger but cannot determine what parameters I should configure in swagger in order to create this API.

I used postman in order to get all information about the response and implement in swagger.

'/api/v1/labels':
post:
      tags:
        - devices
      summary: 'create new label'
      description: 'create new label to a given device'
      operationId: createNewLabel
      produces:
        - application/json
      parameters:
        - name: x-access-token
          description: 'cognito token'
          in: header
          type: string
          required: true
        - in: body
          name: body
          description: add label details
          required: true
          schema:
            $ref: '#/definitions/labelRequest'
      responses:
        '201':
          schema:
            $ref: '#/definitions/labelRequest'
          description: Created

and this is the definition I wrote:

labelRequest:
    type: object
    items:
      $ref: '#/definitions/labelRequestBody'

  labelRequestBody:
    type: object
    properties:
      device_ids:
        type: array
        items:
          type: string
          example: ["9bc11e25-4db2-4780-b761-390e3806082a"]
          format: uuid
      name:
        type: string
        example: new_label

This is the API call:

https:///api/v1/labels

with those headers:

x-access-token

and with this body as payload:

{
    "device_ids":["ea4b9daa-07cd-4cd6-981f-c86e1e81f04c"],
    "name":"labelName"
}

expected result is: 201 (created)

Helen · Accepted Answer · 2019-06-19 09:54:19Z

1

You are almost there, just change the labelRequest schema to:

definitions:
  labelRequest:
    type: object
    properties:
      device_ids:
        type: array
        items:
          type: string
          # Array item example should NOT be enclosed in []
          example: "9bc11e25-4db2-4780-b761-390e3806082a"
          format: uuid
      name:
        type: string
        example: new_label

answered Jun 19, 2019 at 9:54

Helen

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