Swagger-codegen: [Python] JSON reference in $ref isn't acknowledged

Created on 6 Dec 2017  路  5Comments  路  Source: swagger-api/swagger-codegen

Description

It looks like I am having a couple issues.

  1. $ref: '#/definitions/Tenant/properties/name' gets rendered in the Tenant model with type Tenantpropertiesname
  2. The rendered Tenant model only has a name property. The Tenant definition below has 6 total properties which do not make it into the rendered model.
  3. Replacing $ref: '#/definitions/Tenant/properties/name' with a simple type: string fixes the type issue but still leaves a relatively empty Template model.
@property
    def name(self):
        """
        Gets the name of this Tenant.

        :return: The name of this Tenant.
        :rtype: Tenantpropertiesname
        """
        return self._name
Swagger-codegen version

2.2.3

Swagger declaration file content or url
patch:
      description: Update a tenant record.
      operationId: update
      tags:
      - tenants
      parameters:
      - $ref: "#/parameters/id"
      - in: body
        name: tenant
        description: The tenant properties to update
        schema:
          type: object
          required:
            - name
          properties:
            name:
              $ref: '#/definitions/Tenant/properties/name'

Tenant:
    type: object
    required:
      - name
      - handle
    properties:
      id:
        type: integer
        format: int32
        readOnly: true
      name:
        type: string
      handle:
        type: string
        pattern: '^[a-zA-Z0-9-]{0,62}$'
      createdAt:
        type: string
        format: date-time
        readOnly: true
      updatedAt:
        type: string
        format: date-time
        readOnly: true
      deletedAt:
        type: string
        format: date-time
        readOnly: true
Command line used for generation

swagger-codegen generate --input-spec fetched/tenant-service.yaml --lang python --output sdk

Steps to reproduce
Related issues/PRs

I am not able to find a related issue.

Suggest a fix/enhancement

Unsure.

Python Invalid spec

Most helpful comment

Please workaround the problem for the time being by specifying the property type directly instead of using $ref.

All 5 comments

          $ref: '#/definitions/Tenant/properties/name'

I don't think you can do that. You can only reference a model not a specified property of a model.

@wing328 Then shouldn't swagger-codegen fail on validation if this is not valid?

@wing328 That's not a requirement in the spec. In fact, the spec says:

The Reference Object is a JSON Reference that uses a JSON Pointer as its value. For this specification, only canonical dereferencing is supported.

The { $ref: "#/some/deeply/nested/object" } syntax is perfectly valid syntax according to the JSON reference RFC.

In addition to the _de rigueur_ definition in the spec, the _de facto_ standards enforced by swagger-codegen validate and other official and non-official tooling all point to deep JSON pointers as valid references.

This issue should be re-opened. The point of a formal specification is to remove the ambiguity caused by assuming "I don't think you can do that."

Just reopened it. Please open an issue in the Swagger Parser project to report this.

Please workaround the problem for the time being by specifying the property type directly instead of using $ref.

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