[英]How to use PyMongo with Flask Blueprints?
在蓝图中拾取mongo对象的正确方法是什么?
这是我的父母login.py
:
app.config.from_object('config')
from flask.ext.pymongo import PyMongo
from child import child
from child2 import child2
app = Flask(__name__)
app.register_blueprint(child2.child2)
app.register_blueprint(child.child)
在我的child.py
from app import app
from flask.ext.pymongo import PyMongo
mongo = PyMongo(app)
child = Blueprint('child', __name__)
child2.py
与 child 的结构相同:
from app import app
from flask.ext.pymongo import PyMongo
mongo = PyMongo(app)
child2 = Blueprint('child2', __name__)
这是我收到的错误消息:
raise Exception('duplicate config_prefix "%s"' % config_prefix)
Exception: duplicate config_prefix "MONGO"
我在蓝图中尝试了以下内容
mongo = app.data.driver
但这会引发错误。 这是完整的追溯:
Traceback (most recent call last):
File "login.py", line 12, in <module>
from child import child
File "/home/xxx/xxx/child/child.py", line 13, in <module>
mongo = PyMongo(app) #blueprint
File "/home/xxx/xxx/lib/python3.4/site-packages/flask_pymongo/__init__.py", line 97, in __init__
self.init_app(app, config_prefix)
File "/home/xxx/xxx/lib/python3.4/site-packages/flask_pymongo/__init__.py", line 121, in init_app
raise Exception('duplicate config_prefix "%s"' % config_prefix)
Exception: duplicate config_prefix "MONGO"
(xxx)xxx@linux:~/xxx$ python login.py
Traceback (most recent call last):
File "login.py", line 12, in <module>
from courses import courses
File "/home/xxx/xxx/child/child.py", line 13, in <module>
mongo = PyMongo(app) #blueprint
File "/home/xxx/xxx/lib/python3.4/site-packages/flask_pymongo/__init__.py", line 97, in __init__
self.init_app(app, config_prefix)
File "/home/xxx/xxx/lib/python3.4/site-packages/flask_pymongo/__init__.py", line 121, in init_app
raise Exception('duplicate config_prefix "%s"' % config_prefix)
Exception: duplicate config_prefix "MONGO"
一旦我的应用程序创建了连接,我应该如何在我的蓝图中选择它?
所以问题是如何在每个蓝图中构建到数据库的连接字符串。 这是文件结构:
login.py
config.py
/child/child.py
/child2/child2.py
这是config.py
MONGO_DBNAME = 'xxx'
MONGO_URL = os.environ.get('MONGO_URL')
if not MONGO_URL:
MONGO_URL = "mongodb://xxx:xxxx@xxxx.mongolab.com:55822/heroku_xxx";
MONGO_URI = MONGO_URL
我已经在答案中尝试了以下建议,但这没有用。 请参阅我在该预期答案下方的评论。
正如 Emanuel Ey 所建议的那样,在蓝图中执行导入的方法的问题之一是它会导致循环导入。 经过多次尝试,事实证明(我能找到的)唯一方法是创建一个名为database.py
的单独文件,该文件连接到数据库,然后我可以通过蓝图导入此连接,如下所示:
child.py
from database import mongo
courses = Blueprint('courses', __name__)
和我的database.py
from flask.ext.pymongo import PyMongo
mongo = PyMongo()
和应用程序 login.py 但必须初始化数据库
from database import mongo
app = Flask(__name__)
app.config.from_object('config')
mongo.init_app(app) # initialize here!
from child import child
from child import2 child2
app.register_blueprint(child.child)
app.register_blueprint(child2.child2)
您要初始化 PyMongo 驱动程序两次,一次在child.py
,第二次在child2.py
。
尝试在设置您的应用程序对象的文件中初始化 PyMongo 连接,然后将其导入子项:
登录.py:
app.config.from_object('config')
from flask.ext.pymongo import PyMongo
from child import child
from child2 import child2
app = Flask(__name__)
mongo = PyMongo(app)
# Register blueprints
def register_blueprints(app):
# Prevents circular imports
app.register_blueprint(child2.child2)
app.register_blueprint(child.child)
register_blueprints(app)
在 child.py 中
from app import app, mongo
child = Blueprint('child', __name__)
child2.py:
from app import app, mongo
child2 = Blueprint('child2', __name__)
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.