我的烧瓶应用程序中有这棵树：
<DataTemplate x:Key="ImageItemTemplate">
<Border Padding="10" Width="325" Height="350"
BorderBrush="{DynamicResource SignificantInformationColorBrush}"
BorderThickness="0" Margin="5" Background="Transparent" Focusable="True" MouseLeftButtonDown="ImageBorderOnMouseLeftButtonDown">
<!--<Border.InputBindings>
<MouseBinding MouseAction="LeftClick"
Command="{Binding Path=DataContext.SelectImageCommand, RelativeSource={RelativeSource FindAncestor, AncestorType=UserControl}}"
CommandParameter="{Binding}" />
</Border.InputBindings>-->
<StackPanel>
<Border BorderBrush="LightGray"
BorderThickness="1"
CornerRadius="1" Height="300" Width="300" Background="Transparent">
<StackPanel Background="Black">
<Canvas Width="300" Height="300" VerticalAlignment="Center"
HorizontalAlignment="Center" Background="Transparent">
<Image Width="300" Stretch="UniformToFill" StretchDirection="DownOnly"
Source="{Binding BitmapImage}"
Height="300">
<Image.InputBindings>
<MouseBinding MouseAction="LeftDoubleClick"
Command="{Binding Path=DataContext.OpenOriginalCommand, RelativeSource={RelativeSource FindAncestor, AncestorType=UserControl}}"
CommandParameter="{Binding}" />
</Image.InputBindings>
</Image>
</Canvas>
</StackPanel>
</Border>
<StackPanel Orientation="Horizontal" Margin="1 10 0 0">
<TextBlock Margin="5 0 0 0" TextAlignment="Center"
FontSize="11" Text="{Binding ImageInfo.SopInstanceUid}" Background="Transparent" />
</StackPanel>
</StackPanel>
</Border>
</DataTemplate>
这是来自模型目录的 -- api
-- migrations
-- model
-- __init__.py
-- Persons.py
-- Comments.py
-- other_classes.py
-- resources
-- __init__.py
-- app.py
-- util.py
：
__init_.py这是util.py
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app.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()
Persons.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 )
当我执行 # 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) )
时，未检测到模型文件夹中的类：
python3.6 util.py db migrate我已经查看了其他问题，但找不到任何线索。 API的结构是否正确？我有很多表，并将它们拆分为文件。当我将它们直接放在util.py中时，迁移就可以了。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在文件夹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.
中的单个文件中。
请帮助
谢谢
编辑 我也尝试过：
/model它没有检测到模型类。
我认为问题在于我在util中没有使用与模型中相同的db变量。 甚至当我在模型类中进行操作时：
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)
它什么都不会改变
我看到了周围的例子，但其中没有一个有那么多模型。当我们有太多的表时，将所有表放在一个文件中不是一个好习惯。拆分它们并将它们放在文件夹中是一个好主意，但似乎不起作用。
答案 0 :(得分：1)
我认为您的做法正确。首先，您需要在整个应用程序中使用相同的db。我会将其中一个保留在util.py中，然后删除所有其他。
在那之后，您需要确保在运行util.py时导入了模型。我发现您正在尝试聪明的方法，将所有模型导入到model包中。我赞成对所有模型进行显式导入，而不要对这种类型的自动导入进行导入，因此我的建议是您只需逐个导入模型。
确保模型可以识别的最后一件事是删除或移走SQLite数据库。删除数据库文件后运行db migrate命令时，应该进行包含所有内容的迁移。
答案 1 :(得分：0)
您不会导入模型
db = SQLAlchemy(app)
# import the models
migrate = Migrate(app, db)
如果这不能帮助您查看此问题。