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我尝试大摇大摆地对我的应用服务进行分组。在我的项目中，我使用 FlaskApiSpec 为我的 Python-Flask 应用程序生成 Swagger 文档。

有一部分代码，我如何使用 Flask Blueprint 模式为我的应用程序生成 swagger 文档：

应用程序/初始化.py

from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from flask_apispec.extension import FlaskApiSpec
from apispec.ext.marshmallow import MarshmallowPlugin
from apispec import APISpec

db = SQLAlchemy()
docs = FlaskApiSpec()

def create_app():
   app = Flask(__name__)

   app.config.update({
        'APISPEC_SPEC': APISpec(
            title = 'D&D Master Screen',
            version = 'v0.1',
            openapi_version = '2.0',
            plugins = [MarshmallowPlugin()],
        ),
        'APISPEC_SWAGGER_URL': '/api/swagger/',
        'APISPEC_SWAGGER_UI_URL': '/api/swagger-ui/'
    })
    # Import Blueprints
    from .characters.views import characters_bp
    from .main.views import main_bp

    # Registrate Blueprints
    app.register_blueprint(characters_bp, url_prefix='/api/characters')
    app.register_blueprint(main_bp, url_prefix='/')

    # Initialize Plugins
    db.init_app(app)
    docs.init_app(app)

    return app

应用程序/字符/views.py

from flask import Blueprint
# from application.characters import characters_bp
from flask import jsonify
from flask_apispec import use_kwargs, marshal_with
from application import docs
from application.models import  db, db_add_objects, db_delete_objects, Character
from application.schemas import CharacterSchema, ErrorSchema, CharacterIdSchema

characters_bp = Blueprint('characters_bp', __name__)


@characters_bp.route("/", methods=['GET'])
@marshal_with(CharacterSchema(many = True), code=200)
def characters():
    characters = Character.get_all()
    if characters is None:
        return {"message": "Characters not found"}, 404
    return characters


@characters_bp.route("/<character_id>", methods=['GET'])
@marshal_with(CharacterSchema, code=200)
@marshal_with(ErrorSchema, code=404)
def character(character_id):
    character = Character.get(character_id)
    if character is None:
        return {"message": str("Character with id=" + character_id + " not found")}, 404
    return character
    

@characters_bp.route("/", methods=['POST'])
@use_kwargs(CharacterSchema)
@marshal_with(CharacterIdSchema, code=200)
def create_character(**kwargs):
    new_character = Character(**kwargs)
    db_add_objects(new_character)
    return {"id": new_character.id}
    

@characters_bp.route("/<character_id>", methods=['DELETE'])
@marshal_with(ErrorSchema, code=200)
@marshal_with(ErrorSchema, code=404)
def delete_character(character_id):
    character = Character.get(character_id)
    if character is None:
        return {"message": str("Character with id=" + character_id + " not found")}, 404
    db_delete_objects(character)
    return jsonify({"message": "success"})



# Swagger docs for Characters Module services
blueprint_name = characters_bp.name
docs.register(character, blueprint = blueprint_name)
docs.register(characters, blueprint = blueprint_name)
docs.register(create_character, blueprint = blueprint_name)
docs.register(delete_character, blueprint = blueprint_name)

结果是 在此处输入图像描述

我想将我的 /api/characters 方法大摇大摆地归为一组，并正确命名。我试着在网上找了很多东西，学到了一些关于标签的东西是Swagger。但我不明白，如何在 FlaskApiSpec 中使用此功能

我想可以在此处的某处添加标签：

docs.register(character, blueprint = blueprint_name)"

但不明白怎么...

4

2 回答 2

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我也尝试了很长时间......对我来说这很有效：只需装饰你的功能

 @docs(tags=['characters'])
于 2020-11-25T18:07:34.690 回答
0

是的，您可以使用以下方法在同一标签下分配多个方法：

@docs(tags=['characters'])

对要在标签名称下显示的所有方法使用相同的标签名称。您还可以在标签中分别为每个方法添加描述。

尝试这个：

@doc(description='Describing GET method', tags=['charavters'])
def get():
    <code here>


@doc(description='Describing POST method', tags=['charavters'])
def post():
    <code here>
于 2021-05-18T07:30:12.343 回答