swagger中分组方法的flask api规范标记
我试图大摇大摆地将我的应用程序服务分组。在我的项目中,我使用FlaskApiSpec为我的Python Flask应用程序生成Swagger文档 代码中有一部分,我是如何使用Flask Blueprint模式为我的应用程序生成swagger文档的: 应用程序/init.pyswagger中分组方法的flask api规范标记,flask,swagger,flask-apispec,Flask,Swagger,Flask Apispec,我试图大摇大摆地将我的应用程序服务分组。在我的项目中,我使用FlaskApiSpec为我的Python Flask应用程序生成Swagger文档 代码中有一部分,我是如何使用Flask Blueprint模式为我的应用程序生成swagger文档的: 应用程序/init.py from flask import Flask from flask_sqlalchemy import SQLAlchemy from flask_apispec.extension import FlaskApiSpec
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/characters/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)
但是我不明白怎么…我也试了很长时间。。。 对我来说,这是可行的:只是装饰你的功能
@docs(tags=['characters'])
是的,您可以使用以下方法在同一标记下分配多个方法:
@docs(标记=['characters'])
对要在标记名下显示的所有方法使用相同的标记名。还可以在标记中分别为每个方法添加说明
试试这个:
@doc(description='Describing GET method', tags=['charavters'])
def get():
<code here>
@doc(description='Describing POST method', tags=['charavters'])
def post():
<code here>
@doc(description='description GET method',tags=['charavters']
def get():
@文档(description='description POST method',tags=['charavters'])
def post():
@doc(description='Describing GET method', tags=['charavters'])
def get():
<code here>
@doc(description='Describing POST method', tags=['charavters'])
def post():
<code here>