我有一个使用pcreate创建的Pyramid-Fanstatic应用程序:
pcreate -s starter -s pyramid_fanstatic
如果我使用/ bin / pserve-fanstatic启动服务器,一切正常。但是,当我使用Apache WSGI模块加载应用程序时,会出现如下链接:
<link rel="stylesheet" type="text/css" href="/fanstatic/services/bootstrap/bootstrap.min.css" />
返回Apache找不到404。
这是我的WSGI应用程序:
import os
activate_this = os.path.join('/opt/services/services/bin/activate_this.py')
execfile(activate_this, dict(__file__=activate_this))
os.environ['PYTHON_EGG_CACHE'] = '/opt/services/python-eggs'
from pyramid.paster import get_app, setup_logging
ini_path = '/opt/services/services/development.ini'
setup_logging(ini_path)
application = get_app(ini_path, 'main')
这是我的Apache conf文件:
<VirtualHost *:80>
ServerAdmin a.orth@cgiar.org
ServerName data.ilri.org
# Pass authorization info on (needed for rest api).
WSGIPassAuthorization On
WSGIDaemonProcess services python-path=/opt/services/services display-name=services processes=2 threads=15
WSGIScriptAlias /services /opt/services/services/services/services.wsgi process-group=services application-group=%{GLOBAL}
<Location /services>
WSGIProcessGroup services
</Location>
ErrorLog /var/log/httpd/services.error.log
CustomLog /var/log/httpd/services.custom.log combined
</VirtualHost>
在加载应用程序之前,我可以看到/ bin / pserve-fanstatic包含我的资源目录:
"""A script aware of static resource"""
import pyramid.scripts.pserve
import pyramid_fanstatic
import os
dirname = os.path.dirname(__file__)
dirname = os.path.join(dirname, 'resources')
pyramid.scripts.pserve.add_file_callback(
pyramid_fanstatic.file_callback(dirname))
pyramid.scripts.pserve.main()
但即使我在init.py中包含这样的行,Apache也无法找到狂热的资源。
我还将此添加到我的ini文件中:
[filter:fanstatic]
use = egg:fanstatic#fanstatic
[pipeline:main]
pipeline = fanstatic services
我还应该检查/做什么?
答案 0 :(得分:1)
您似乎没有告诉Apache提供您的静态文件。参见:
答案 1 :(得分:1)
固定!!
我必须将这些行添加到[app:main]中的ini文件中:
fanstatic.publisher_signature = fanstatic
fanstatic.use_application_uri = true