^{Originally posted by hashlash March 11, 2022}

The Issue

I have a model with SafeString as its help_text (inspired from Django's password validation help text):

from django.db import models
from django.utils.html import format_html, format_html_join


def list_help_text_html(help_texts):
    help_items = format_html_join(
        "", "<li>{}</li>", ((help_text,) for help_text in help_texts)
    )
    return format_html("<ul>{}</ul>", help_items) if help_items else ""


class ExampleModel(models.Model):
    field = models.TextField(
        help_text=list_help_text_html([
            'info a',
            'info b',
        ]),
    )

I followed the DRF's doc about Generating a dynamic schema with SchemaView and A minimal example with Swagger UI, but the Swagger UI fails to render the generated schema.

Parser error on line 59
unknown tag !<tag:yaml.org,2002:python/object/new:django.utils.safestring.SafeString>

The generated schema (full version):

openapi: 3.0.2
info:
  # redacted
paths:
  # redacted
components:
  schemas:
    Example:
      type: object
      properties:
        id:
          # redacted
        field:
          type: string
          description: !!python/object/new:django.utils.safestring.SafeString
          - <ul><li>info a</li><li>info b</li></ul>
      required:
      - field

Here's a repo for demonstrating the bug. The Swagger UI can be accessed on /swagger-ui/.

Package versions:

django==3.2.12
djangorestframework==3.13.1
pyyaml==6.0
uritemplate==4.1.1

Findings

I found that the schema generation for the help_text lies on:

and:

But since SafeString is a subclass of str and it override the __str__() method to return itself, those DRF's method will still return a SafeString object.

class SafeString(str, SafeData):
    ...

    def __str__(self):
        return self

Workaround

I've with 3 possible workarounds, and so far I think the third is the best:

• Adding the SafeString with an empty string (implementation, demo)

  >>> from django.utils.safestring import SafeString
  >>> s = SafeString('abc')
  >>> s + ''
  'abc'
  >>> type(s + '')
  <class 'str'>
  

• Use str.__str__() (implementation, demo)

  >>> from django.utils.safestring import SafeString
  >>> s = SafeString('abc')
  >>> str.__str__(s)
  'abc'
  >>> type(str.__str__(s))
  <class 'str'>
  

• Use represent_str() to represent SafeString as str (implementation, demo)

  class OpenAPIRenderer(BaseRenderer):
      ...
      def render(self, data, media_type=None, renderer_context=None):
          class Dumper(yaml.Dumper):
              ...
          Dumper.add_representer(SafeString, Dumper.represent_str)
          return yaml.dump(data, default_flow_style=False, sort_keys=False, Dumper=Dumper).encode('utf-8')

PS: I also asked about the "string conversion" on Django's forum.