皮纳克斯Apache:仅向经过身份验证的用户传递附件

发布于 2024-08-12 03:13:18 字数 1809 浏览 9 评论 0原文

我希望 Apache 和 Pinax 只向经过身份验证的用户提供附件。

我找到了这个帖子,但我无法做到工作。

我的 Apache-conf-file:

WSGIPythonPath /usr/local/bin/python

<VirtualHost *:80>
    ServerName      www.domain.com
    ServerAlias     domain.com


    WSGIDaemonProcess k-production python-path=/path/to/app/pinax-env/lib/python2.6/site-packages
    WSGIProcessGroup k-production

    Alias /site_media  /path/to/app/cp/site_media    
    <Directory /path/to/app/cp/site_media>
    Order deny,allow
    Allow from all
    </Directory>

    WSGIScriptAlias /site_media/media/attachments /path/to/app/cp/deploy/pinax.wsgi
    <Directory /path/to/app/cp/site_media/media/attachments>
    Deny from all
    </Directory>

    XSendFile On
    XSendFileAllowAbove On

    WSGIScriptAlias / /path/to/app/cp/deploy/pinax.wsgi
    <Directory /path/to/app/cp/deploy>
        Order deny,allow
        Allow from all
    </Directory>

</VirtualHost>

和我的(仍然粗略的)视图,应该被调用:

@login_required 
def sendfile(request, slug): 

    app, content_object_id, img = slug.split('/')
    project_file = get_object_or_404(Attachment, attachment_file = 'attachments/'+slug) 
    response = HttpResponse() 
    response['X-Sendfile'] =  os.path.join(settings.MEDIA_ROOT,   'attachments/'+slug) 
    content_type = 'application/octet-stream' 
    response['Content-Type'] = content_type 
    response['Content-Disposition'] = 'attachment; filename="%s"' %  os.path.basename(os.path.join(settings.MEDIA_ROOT,   'attachments/'+slug))
    return response 

无论用户是否登录,Apache 都会抛出 403。

通过开发服务器,我可以访问视图,但不会传输任何数据。

出了什么问题?

I want that Apache and Pinax only deliver Attachments to authenticated users.

I found this post, but i can't make it work.

My Apache-conf-file:

WSGIPythonPath /usr/local/bin/python

<VirtualHost *:80>
    ServerName      www.domain.com
    ServerAlias     domain.com


    WSGIDaemonProcess k-production python-path=/path/to/app/pinax-env/lib/python2.6/site-packages
    WSGIProcessGroup k-production

    Alias /site_media  /path/to/app/cp/site_media    
    <Directory /path/to/app/cp/site_media>
    Order deny,allow
    Allow from all
    </Directory>

    WSGIScriptAlias /site_media/media/attachments /path/to/app/cp/deploy/pinax.wsgi
    <Directory /path/to/app/cp/site_media/media/attachments>
    Deny from all
    </Directory>

    XSendFile On
    XSendFileAllowAbove On

    WSGIScriptAlias / /path/to/app/cp/deploy/pinax.wsgi
    <Directory /path/to/app/cp/deploy>
        Order deny,allow
        Allow from all
    </Directory>

</VirtualHost>

and my (still rough) view, that should get called:

@login_required 
def sendfile(request, slug): 

    app, content_object_id, img = slug.split('/')
    project_file = get_object_or_404(Attachment, attachment_file = 'attachments/'+slug) 
    response = HttpResponse() 
    response['X-Sendfile'] =  os.path.join(settings.MEDIA_ROOT,   'attachments/'+slug) 
    content_type = 'application/octet-stream' 
    response['Content-Type'] = content_type 
    response['Content-Disposition'] = 'attachment; filename="%s"' %  os.path.basename(os.path.join(settings.MEDIA_ROOT,   'attachments/'+slug))
    return response 

Apache throws a 403 no matter if the user is logged in.

Via the develoment-server i can access the view, but no data will get transmitted.

What is wrong?

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评论(2

ぺ禁宫浮华殁 2024-08-19 03:13:18

我试图做几乎完全相同的事情,结果证明解决方案不是使用 WSGIScriptAlias,而是使用定义 wsgi 处理程序的目录的常规别名。对于视图,我基本上只是围绕 django.views.static.serve 编写了一个包装器。

我的 apache conf 最终看起来像这样:

# myproject
<VirtualHost *:8080>
    #DocumentRoot /var/www/myproject/public
    ServerName myproject
    ErrorLog /var/www/myproject/logs/apache_error_log
    CustomLog /var/www/myproject/logs/apache_access_log common

    AliasMatch ^/(media/uploads/protected/.*) /var/www/myproject/src/myproject-trunk/server/django.wsgi/$1
    Alias /media/ /var/www/myproject/public/media/
    Alias / /var/www/myproject/src/myproject-trunk/server/django.wsgi/

    <Directory /var/www/myproject/src/myproject-trunk/server>
        Options ExecCGI
        AddHandler wsgi-script .wsgi
        # WSGIApplicationGroup %{GLOBAL}
        Order allow,deny
        Allow from all
    </Directory>

    <Directory /var/www/myproject/public/media>
        Order deny,allow
        Allow from all
    </Directory>
</VirtualHost>

I was trying to do pretty much exactly the same thing, and the solution turned out to be not using WSGIScriptAlias, and instead using a regular Alias to a directory that defined a wsgi handler. For the view I basically just wrote a wrapper around django.views.static.serve.

My apache conf ended up looking like this:

# myproject
<VirtualHost *:8080>
    #DocumentRoot /var/www/myproject/public
    ServerName myproject
    ErrorLog /var/www/myproject/logs/apache_error_log
    CustomLog /var/www/myproject/logs/apache_access_log common

    AliasMatch ^/(media/uploads/protected/.*) /var/www/myproject/src/myproject-trunk/server/django.wsgi/$1
    Alias /media/ /var/www/myproject/public/media/
    Alias / /var/www/myproject/src/myproject-trunk/server/django.wsgi/

    <Directory /var/www/myproject/src/myproject-trunk/server>
        Options ExecCGI
        AddHandler wsgi-script .wsgi
        # WSGIApplicationGroup %{GLOBAL}
        Order allow,deny
        Allow from all
    </Directory>

    <Directory /var/www/myproject/public/media>
        Order deny,allow
        Allow from all
    </Directory>
</VirtualHost>
韵柒 2024-08-19 03:13:18

首先尝试专注于开发服务器 - 因为它的设置更简单,因此更不容易出错。

也许可以尝试这个:

@login_required  def sendfile(request, slug): 
    ## these are never used    
    # app, content_object_id, img = slug.split('/')
    # project_file = get_object_or_404(Attachment, attachment_file = 'attachments/'+slug)

    response = HttpResponse() 
    response['X-Sendfile'] =  os.path.join(settings.MEDIA_ROOT,   'attachments/'+slug)

    import pdb; pdb.set_trace() 
    # your development server will stop here
    # you can now inspect the your context, e.g.
    # >> p response['X-Sendfile']
    # this should print the value of response['X-Sendfile']
    # >> c
    # this will continue program execution
    # for more commands see http://www.python.org/doc/2.4/lib/debugger-commands.html

    content_type = 'application/octet-stream' 
    response['Content-Type'] = content_type 
    # Content-Disposition filename is only for suggesting a name for the file
    # when the user tries to download it, e.g.:
    # response['Content-Disposition'] = 'attachment; filename='localfile.txt'
    response['Content-Disposition'] = 'attachment; filename="%s"' %  os.path.basename(os.path.join(settings.MEDIA_ROOT, 'attachments/'+slug))
    return response

但是通过使用开发服务器,您将不会获得任何文件服务,因为 Apache 将执行文件服务。
您只需确保发送到 apache 的数据是正确的即可。
或者使用(不推荐用于您的生产服务器)

f = open(os.path.join(settings.MEDIA_ROOT, 'attachments/'+slug), 'rb')
response = HttpResponse(f.read())

而不是

response = HttpResponse() 
response['X-Sendfile'] =  os.path.join(settings.MEDIA_ROOT, 'attachments/'+slug)

一些链接:

Try to concentrate on the development server first - since it is a simpler setup and thus less error prone.

Perhaps try this:

@login_required  def sendfile(request, slug): 
    ## these are never used    
    # app, content_object_id, img = slug.split('/')
    # project_file = get_object_or_404(Attachment, attachment_file = 'attachments/'+slug)

    response = HttpResponse() 
    response['X-Sendfile'] =  os.path.join(settings.MEDIA_ROOT,   'attachments/'+slug)

    import pdb; pdb.set_trace() 
    # your development server will stop here
    # you can now inspect the your context, e.g.
    # >> p response['X-Sendfile']
    # this should print the value of response['X-Sendfile']
    # >> c
    # this will continue program execution
    # for more commands see http://www.python.org/doc/2.4/lib/debugger-commands.html

    content_type = 'application/octet-stream' 
    response['Content-Type'] = content_type 
    # Content-Disposition filename is only for suggesting a name for the file
    # when the user tries to download it, e.g.:
    # response['Content-Disposition'] = 'attachment; filename='localfile.txt'
    response['Content-Disposition'] = 'attachment; filename="%s"' %  os.path.basename(os.path.join(settings.MEDIA_ROOT, 'attachments/'+slug))
    return response

But by using the dev server you won't get any files served, since Apache will do the file serving.
You can just ensure, the data sent to apache is correct.
Or use (not recommended for your production server)

f = open(os.path.join(settings.MEDIA_ROOT, 'attachments/'+slug), 'rb')
response = HttpResponse(f.read())

instead of

response = HttpResponse() 
response['X-Sendfile'] =  os.path.join(settings.MEDIA_ROOT, 'attachments/'+slug)

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