Python Flask自动生成的SwaggerOpenAPI 3.0[已关闭]。

Posted 2023-02-17

已关闭。 这个问题不符合 堆栈溢出指南. 目前不接受回答。

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已关闭 5天前.

改进这个问题

我试图为我现有的Flask应用程序生成swagger文档，我尝试了用 Flask-RESTPlus 最初，发现现在项目很丰富，于是在分叉的项目上进行了检查。flask-restx https:/github.compython-restxflask-restx。 但我仍然不认为他们支持openapi 3.0。

我对如何选择适合我需要的包有点困惑。我想解决一个问题，我们不想手动为我们的API创建swagger doc，而是想用一个包自动生成。

import os
import requests
import json, yaml


from flask import Flask, after_this_request, send_file, safe_join, abort
from flask_restx import Resource, Api, fields
from flask_restx.api import Swagger

app = Flask(__name__)
api = Api(app=app, doc='/docs', version='1.0.0-oas3', title='TEST APP API',
          description='TEST APP API')


response_fields = api.model('Resource', 
    'value': fields.String(required=True, min_length=1, max_length=200, description='Book title')
)


@api.route('/compiler/', endpoint='compiler')
# @api.doc(params='id': 'An ID')
@api.doc(responses=403: 'Not Authorized')
@api.doc(responses=402: 'Not Authorized')
# @api.doc(responses=200: 'Not Authorized')
class DemoList(Resource):
    @api.expect(response_fields, validate=True)
    @api.marshal_with(response_fields, code=200)
    def post(self):
        """
        returns a list of conferences
        """
        api.payload["value"] = 'Im the response ur waiting for'
        return api.payload


@api.route('/swagger')
class HelloWorld(Resource):
    def get(self):
        data = json.loads(json.dumps(api.__schema__))
        with open('yamldoc.yml', 'w') as yamlf:
            yaml.dump(data, yamlf, allow_unicode=True, default_flow_style=False)
            file = os.path.abspath(os.getcwd())
            try:
                @after_this_request
                def remove_file(resp):
                    try:
                        os.remove(safe_join(file, 'yamldoc.yml'))
                    except Exception as error:
                        log.error("Error removing or closing downloaded file handle", error)
                    return resp

                return send_file(safe_join(file, 'yamldoc.yml'), as_attachment=True, attachment_filename='yamldoc.yml', mimetype='application/x-yaml')
            except FileExistsError:
                abort(404)


# main driver function
if __name__ == '__main__':
    app.run(port=5003, debug=True)

上面的代码是我在不同的包上尝试的组合，但它可以生成swagger 2.0的文档，但我想生成openapi 3.0的文档。

有谁能给我推荐一个支持openapi 3.0的好包来生成swagger yaml或json。

答案

我找到了一个生成openapi 3.0文档的包。

https:/apispec.readthedocs.ioenlatestinstall.html。

这个包的作用很整齐。详细使用方法请找下面的代码。

from apispec import APISpec
from apispec.ext.marshmallow import MarshmallowPlugin
from apispec_webframeworks.flask import FlaskPlugin
from marshmallow import Schema, fields
from flask import Flask, abort, request, make_response, jsonify
from pprint import pprint
import json


class DemoParameter(Schema):
    gist_id = fields.Int()


class DemoSchema(Schema):
    id = fields.Int()
    content = fields.Str()


spec = APISpec(
    title="Demo API",
    version="1.0.0",
    openapi_version="3.0.2",
    info=dict(
        description="Demo API",
        version="1.0.0-oas3",
        contact=dict(
            email="[e-mail]"
            ), 
        license=dict(
            name="Apache 2.0",
            url='http://www.apache.org/licenses/LICENSE-2.0.html'
            )
        ),
    servers=[
        dict(
            description="Test server",
            url="https://resources.donofden.com"
            )
        ],
    tags=[
        dict(
            name="Demo",
            description="Endpoints related to Demo"
            )
        ],
    plugins=[FlaskPlugin(), MarshmallowPlugin()],
)

spec.components.schema("Demo", schema=DemoSchema)

# spec.components.schema(
#     "Gist",
#     
#         "properties": 
#             "id": "type": "integer", "format": "int64",
#             "name": "type": "string",
#         
#     ,
# )
#
# spec.path(
#     path="/gist/gist_id",
#     operations=dict(
#         get=dict(
#             responses="200": "content": "application/json": "schema": "Gist"
#         )
#     ),
# )
# Extensions initialization
# =========================
app = Flask(__name__)


@app.route("/demo/<gist_id>", methods=["GET"])
def my_route(gist_id):
    """Gist detail view.
    ---
    get:
      parameters:
      - in: path
        schema: DemoParameter
      responses:
        200:
          content:
            application/json:
              schema: DemoSchema
        201:
          content:
            application/json:
              schema: DemoSchema
    """
    # (...)
    return jsonify('foo')


# Since path inspects the view and its route,
# we need to be in a Flask request context
with app.test_request_context():
    spec.path(view=my_route)
# We're good to go! Save this to a file for now.
with open('swagger.json', 'w') as f:
    json.dump(spec.to_dict(), f)

pprint(spec.to_dict())
print(spec.to_yaml())

希望对大家有所帮助! :)