在Python3.7中使用_init__u.py文件导入*

首先是我的目录结构：

Root
   - models 
        car.py 
        __init__.py 
   hello.py 

在

\uuuu init\uuuuu.py

中，我有以下内容：

__all__ = ["car"]

在

hello.py

中，我尝试从models文件夹导入所有内容：

from models import *

car = Car()

这给了我一个错误：

Traceback (most recent call last):
  File "hello.py", line 4, in <module>
    car = Car()
NameError: name 'Car' is not defined

回溯（最近一次呼叫最后一次）：
文件“hello.py”，第4行，在
汽车
NameError:未定义名称“Car”

---

我做错了什么

您必须指定要导入

\uuuuu init\uuuuuuuuy.py

文件的属性

中的类。请参见下面的示例：

from car import *
__all__ = ["Car"]

您必须指定要导入到
\uuuuuu init\uuuuuuuuy.py
文件的属性
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu>中的类。请参见下面的示例：

from car import *
__all__ = ["Car"]

如果您想在执行从模型导入的
操作后，在
hello.py
中直接访问类
Car
，请在
\uuuu init\uuuuuuuuuuuuuuuuuuuuuupy
文件中，将
从模型.Car导入Car

另一方面，
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。您可以将
hello.py
更改为如下所示，您当前的
\uuuuuu init\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuupy
包含
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

from models import *
car_obj = car.Car() # Reference module.class instead of just the class

如果您想在执行从模型导入的
操作后，在
hello.py
中直接访问类
Car
，请在
\uuuu init\uuuuuuuuuuuuuuuuuuuuuupy
文件中，将
从模型.Car导入Car

另一方面，
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。您可以将
hello.py
更改为如下所示，您当前的
\uuuuuu init\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuupy
包含
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

from models import *
car_obj = car.Car() # Reference module.class instead of just the class

从python中：

如果包的
\uuuuu init\uuuuuuuupy.py
代码定义了名为
\uuuuuuu all\uuuuuuuuuuuu
的列表，则当遇到来自包导入*
的
时，它被视为应导入的模块名称列表

这意味着您的
hello.py
刚刚将
car
模块导入到它的命名空间中，而不是
car
类。因此，这是可行的

from models import *
auto = car.Car()

从python中：

如果包的
\uuuuu init\uuuuuuuupy.py
代码定义了名为
\uuuuuuu all\uuuuuuuuuuuu
的列表，则当遇到来自包导入*
的
时，它被视为应导入的模块名称列表

这意味着您的
hello.py
刚刚将
car
模块导入到它的命名空间中，而不是
car
类。因此，这是可行的

from models import *
auto = car.Car()

你错过了一步

尝试：

或者尝试：

from models.car import *

car = Car()

你错过了一步

尝试：

或者尝试：

from models.car import *

car = Car()

\uuuuuuuuuuuuuuuuuuuuuuuuu
仅控制在当前范围内使用
*
导出的内容

在您的情况下，
Car
不在您的
\uuuu init\uuuuuuuuupy
的范围内。所以这是毫无意义的

要解决此问题，您需要将
Car
导入
\uuuu init\uuuuuu.py
的范围，仅此而已

我知道您认为只要使用
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu<代码>\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

在您的情况下，
Car
不在您的
\uuuu init\uuuuuuuuupy
的范围内。所以这是毫无意义的

要解决此问题，您需要将
Car
导入
\uuuu init\uuuuuu.py
的范围，仅此而已

我知道您认为只要使用
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu<代码>\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
除了控制当前范围内的导出之外，其他什么都不做
现在我得到以下信息：从模型导入*属性错误：模块“模型”没有属性“汽车”，这些行到哪里去了？我在init.py文件中已经有了all=[“Car”]。@johndoe:这是你的
\uuu init\uuuuuuuuuuupy
文件的内容，别忘了，它是
Car
而不是
Car
像你的例子中那样。现在我得到以下信息：从模型导入*AttributeError:module'models'没有属性'Car'？我在init.py文件中已经有了所有=[“Car”]。@johndoe:这是你的
\uuu init\uuuuuuuuuupy
文件的内容，别忘了，它是
Car
而不是
Car
像你的例子一样。FWIW我通常建议避免导入
*
。没有必要，而且。这也是我们给出的建议：“应该避免通配符导入（
from import*
），因为通配符导入会使名称空间中存在的名称不清楚，从而使读者和许多自动化工具都感到困惑。”FWIW我通常建议避免导入
*
。没有必要，而且。这也是给出的建议：“应该避免通配符导入（
来自导入*
），因为它们使名称空间中存在的名称不清楚，从而混淆了读者和许多自动化工具。”