我的烧瓶应用程序中有这棵树:
-- api
-- migrations
-- model
-- __init__.py
-- Persons.py
-- Comments.py
-- other_classes.py
-- resources
-- __init__.py
-- app.py
-- util.py
This is __init_.py
从模型目录:
from os.path import dirname, basename, isfile
import glob
modules = glob.glob(dirname(__file__)+"/*.py")
__all__ = [ basename(f)[:-3] for f in modules if isfile(f) and not f.endswith('__init__.py')]
from .Persons import Persons
这是 util.py
from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from flask_script import Manager
from flask_migrate import Migrate, MigrateCommand
import os
import model
def create_app():
app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = os.environ['DATABASE_ENGINE'] + '://' + \
os.environ['DATABASE_USERNAME'] + ':' + \
os.environ['DATABASE_PASSWORD'] + '@' + \
os.environ['DATABASE_SERVER'] + '/api_rest?charset=utf8'
app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = True
return app
app = create_app()
db = SQLAlchemy(app)
migrate = Migrate(app, db)
manager = Manager(app)
manager.add_command('db', MigrateCommand)
if __name__ == '__main__':
manager.run()
这是应用程序.py
from util import app
from flask import Flask, jsonify
from flask_restful import reqparse, abort, Api, Resource
@app.errorhandler(404)
def not_found(e):
return jsonify({'message' : 'Not Found'}), 404
@app.errorhandler(500)
def internal_server_error(e):
return jsonify({'message' : 'Internal Server Error'}), 500
api = Api(app)
class Overview(Resource):
def get(self):
return {'hello': 'world'}
api.add_resource(Overview, '/v1/api/overview')
if __name__ == '__main__':
app.run( host = '0.0.0.0', port = 5000, debug = True, threaded = True )
人物.py
# coding=utf-8
import sys
import os
from flask_sqlalchemy import SQLAlchemy
sys.path.append(os.path.dirname(os.getcwd()))
db = SQLAlchemy()
class Persons(db.Model):
id = db.Column( db.Integer, primary_key = True )
name = db.Column( db.String(255) )
firstname = db.Column( db.String(255) )
lastname = db.Column( db.String(255) )
当我做python3.6 util.py db migrate
,未检测到模型文件夹内的类:
INFO [sqlalchemy.engine.base.Engine] SHOW VARIABLES LIKE 'sql_mode'
INFO [sqlalchemy.engine.base.Engine] ()
INFO [sqlalchemy.engine.base.Engine] SELECT DATABASE()
INFO [sqlalchemy.engine.base.Engine] ()
INFO [sqlalchemy.engine.base.Engine] show collation where `Charset` = 'utf8' and `Collation` = 'utf8_bin'
INFO [sqlalchemy.engine.base.Engine] ()
INFO [sqlalchemy.engine.base.Engine] SELECT CAST('test plain returns' AS CHAR(60)) AS anon_1
INFO [sqlalchemy.engine.base.Engine] ()
INFO [sqlalchemy.engine.base.Engine] SELECT CAST('test unicode returns' AS CHAR(60)) AS anon_1
INFO [sqlalchemy.engine.base.Engine] ()
INFO [sqlalchemy.engine.base.Engine] SELECT CAST('test collated returns' AS CHAR CHARACTER SET utf8) COLLATE utf8_bin AS anon_1
INFO [sqlalchemy.engine.base.Engine] ()
INFO [alembic.runtime.migration] Context impl MySQLImpl.
INFO [alembic.runtime.migration] Will assume non-transactional DDL.
INFO [sqlalchemy.engine.base.Engine] DESCRIBE `alembic_version`
INFO [sqlalchemy.engine.base.Engine] ()
INFO [sqlalchemy.engine.base.Engine] SELECT alembic_version.version_num
FROM alembic_version
INFO [sqlalchemy.engine.base.Engine] ()
INFO [sqlalchemy.engine.base.Engine] DESCRIBE `alembic_version`
INFO [sqlalchemy.engine.base.Engine] ()
INFO [sqlalchemy.engine.base.Engine] SHOW FULL TABLES FROM `dsiapi_rest`
INFO [sqlalchemy.engine.base.Engine] ()
INFO [alembic.env] No changes in schema detected.
我查看了其他问题,但找不到任何线索。 API的结构是否正确?我有很多表,我把它们分成文件。当我将它们直接放入 util.py 中时,迁移就可以进行。但我不想将所有内容都放在一个文件中。这就是为什么我需要将每个表放入文件夹内的单个文件中/model
.
请帮忙
谢谢
EDIT我也尝试过这个:
MODELS_DIRECTORY = "models"
EXCLUDE_FILES = ["__init__.py"]
def import_models():
for dir_path, dir_names, file_names in os.walk(MODELS_DIRECTORY):
for file_name in file_names:
if file_name.endswith("py") and not file_name in EXCLUDE_FILES:
file_path_wo_ext, _ = os.path.splitext((os.path.join(dir_path, file_name)))
module_name = file_path_wo_ext.replace(os.sep, ".")
importlib.import_module(module_name)
它没有检测到模型类。
我认为问题是我在 util 中没有使用与模型中相同的 db 变量。
即使我在模型类中执行此操作:
from util import db
它不会改变任何东西
我见过周围的例子,但没有一个有那么多的模型。当我们有太多表时,将所有表放入一个文件中并不是一个好习惯。将它们拆分并将它们放入文件夹中是一个好主意,但它似乎不起作用。