0

这是一件我无法理解的奇怪事情。

这是我的 django 文件目录层次结构

project/
       apache/ django.wsgi
       project/ __init__.py, settings.py, urls.py ..
       services/
               __init__.py
               apis/
                    __init__.py
                    fparser.py
               wservice.py
       ...

       profile.py

因此，在开发服务器甚至在 heroku (gunicorn) 上一切正常，但在 apache (localhost) 上不起作用

当我打开页面时：

它的显示

Exception Type: ImportError at /
Exception Value: cannot import name website_feed_address

这website_feed_address位于 profile.py 中import error，发现于fparser.py

我应该如何解决？

编辑：

django.wsgi

import os, sys

sys.path.append('d:/code/projects-dev/project')


os.environ['DJANGO_SETTINGS_MODULE'] = 'project.settings'

import django.core.handlers.wsgi

application = django.core.handlers.wsgi.WSGIHandler()

而问题中上面提到的项目目录层次结构位于d:/code/projects-dev/

编辑 2

这些是 apache 日志错误

[Sun Jul 08 23:14:04 2012] [notice] Parent: Received restart signal -- Restarting the server.
httpd.exe: Could not reliably determine the server's fully qualified domain name, using 124.123.136.220 for ServerName
[Sun Jul 08 23:14:04 2012] [warn] mod_wsgi: Compiled for Python/2.7.
[Sun Jul 08 23:14:04 2012] [warn] mod_wsgi: Runtime using Python/2.7.2.
[Sun Jul 08 23:14:05 2012] [notice] Child 5912: Exit event signaled. Child process is ending.
[Sun Jul 08 23:14:05 2012] [warn] RSA server certificate CommonName (CN) `127.0.0.1' does NOT match server name!?
[Sun Jul 08 23:14:05 2012] [notice] Apache/2.2.22 (Win32) mod_wsgi/3.3 Python/2.7.2 mod_ssl/2.2.22 OpenSSL/0.9.8t configured -- resuming normal operations
[Sun Jul 08 23:14:05 2012] [notice] Server built: Jan 28 2012 11:16:39
[Sun Jul 08 23:14:05 2012] [notice] Parent: Created child process 2120
httpd.exe: Could not reliably determine the server's fully qualified domain name, using 124.123.136.220 for ServerName
[Sun Jul 08 23:14:05 2012] [warn] RSA server certificate CommonName (CN) `127.0.0.1' does NOT match server name!?
httpd.exe: Could not reliably determine the server's fully qualified domain name, using 124.123.136.220 for ServerName
[Sun Jul 08 23:14:05 2012] [warn] mod_wsgi: Compiled for Python/2.7.
[Sun Jul 08 23:14:05 2012] [warn] mod_wsgi: Runtime using Python/2.7.2.
[Sun Jul 08 23:14:05 2012] [warn] RSA server certificate CommonName (CN) `127.0.0.1' does NOT match server name!?
[Sun Jul 08 23:14:05 2012] [notice] Child 2120: Child process is running
[Sun Jul 08 23:14:06 2012] [notice] Child 2120: Acquired the start mutex.
[Sun Jul 08 23:14:06 2012] [notice] Child 5912: Released the start mutex
[Sun Jul 08 23:14:06 2012] [notice] Child 2120: Starting 64 worker threads.
[Sun Jul 08 23:14:06 2012] [notice] Child 2120: Starting thread to listen on port 443.
[Sun Jul 08 23:14:06 2012] [notice] Child 2120: Starting thread to listen on port 80.
[Sun Jul 08 23:14:07 2012] [notice] Child 5912: Terminating 126 threads that failed to exit.
[Sun Jul 08 23:14:07 2012] [notice] Child 5912: All worker threads have exited.
[Sun Jul 08 23:14:07 2012] [notice] Child 5912: Child process is exiting

编辑 3

这就是 profile.py、fparser.py 的样子

profile.py只包含一些变量，元组就像settings.py. website_feed_address在这种情况下，它只是导入一个变量。

这是fparser.py

from profile import website_feed_address
import feedparser


class FParser(object):
      def __init__(self):
            self.pFeed = feedparser.parse(website_feed_address)
      # rest of the code goes...

我只想在 FParser 类中使用 website_feed_address 而不将其作为对象参数。有没有更好的方法？..或者，我可以像这样使用它吗？

4

2 回答 2

1

最有可能的问题是您在项目中使用了与包相关的导入语句，但 Apache 不知道正确的 python 路径。例如，由于您遇到此特定的导入错误，我假设fparser.py您正在执行类似from profile import website_feed_address. 您可以尝试在 django.wsgi 中明确包含您的项目包：

sys.path.append('d:/code/projects-dev/project')
sys.path.append('d:/code/projects-dev/project/project')

一个更好的建议是为您的项目设置一个virtualenv。这样做的好处是允许您的项目包含自己的站点包环境，以便您安装的任何依赖项都将与项目环境一起使用。然后在 Apache 中你可以告诉 mod_wsgi python 环境在哪里：

# or wherever you decide to create the virtualenv
WSGIPythonHome d:/code/projects-dev/project/project

通过 Apache 设置 pythonpath

您还可以尝试在 Apache conf 中为您的 wsgi 应用程序设置 PYTHONPATH ：

WSGIPythonPath "d:/code/projects-dev/project/project"

*请注意该链接中区分 wsgi 嵌入模式和守护程序模式的地方。

解决进口问题的完全替代方法

虽然我之前的建议是解决 PYTHONPATH 并导入您的profile模块，但基于您只是使用它来传递一个常量这一事实，我建议您只使用 settings.py，因为这就是它的用途（声明常数）：

设置.py

WEBSITE_FEED_ADDRESS = "http://foo.com"

解析器.py

from django.conf import settings

...
    self.pFeed = feedparser.parse(settings.WEBSITE_FEED_ADDRESS)

您可以确定 django 环境将始终为您提供设置模块。常见的方法是让应用程序添加到设置中，以提供默认常量。

于 2012-07-08T16:40:14.303 回答
0

如果您将 profile.py 移动到项目文件夹中，它将位于一个包中。然后你可以这样做：

from project.profile import website_feed_address
于 2012-07-08T16:27:53.330 回答