[Django基础] gunicorn启动django时静态文件的加载

目前在用nginx+gunicorn对django进行部署

当我用gunicorn -w 4 -b 127.0.0.1:8080 myproject.wsig:application启动django时访问主页却发现所有static文件夹下的静态文件都找不到,部分如下:

Not Found: /static/js/app.min.js
Not Found: /static/js/custom.js
Not Found: /static/img/user1.jpg
Not Found: /static/plugins/bootstrap/js/bootstrap.min.js
Not Found: /static/plugins/morris/raphael-2.1.0.min.js
Not Found: /static/plugins/jquery-ui-1.11.0.custom/jquery-ui.js
Not Found: /static/js/sb-admin.js
Not Found: /static/js/app.min.js

用python manage.py runserver启动时是没有问题的(可以见上一篇文章django解决静态文件依赖问题以及前端引入方式),

搜了一下午终于找到了解决方法(详见),即在urls中添加以下代码

from django.contrib.staticfiles.urls import staticfiles_urlpatterns

# ... the rest of your URLconf goes here ...

urlpatterns += staticfiles_urlpatterns()

---------------------------------------------不知道要分割啥的分割线------------------------------------------
看一下源码,该方法返回一个static类

from django.conf import settings
from django.conf.urls.static import static
from django.contrib.staticfiles.views import serve

urlpatterns = []

def staticfiles_urlpatterns(prefix=None):
    """
    Helper function to return a URL pattern for serving static files.
    """
    if prefix is None:
        prefix = settings.STATIC_URL #这里 static_url = '/static/'
    return static(prefix, view=serve)

# Only append if urlpatterns are empty
if settings.DEBUG and not urlpatterns:
    urlpatterns += staticfiles_urlpatterns()

再看一下这个static

def static(prefix, view=serve, **kwargs):
    """
    Helper function to return a URL pattern for serving files in debug mode.

    from django.conf import settings
    from django.conf.urls.static import static

    urlpatterns = [
        # ... the rest of your URLconf goes here ...
    ] + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
    """
    # No-op if not in debug mode or an non-local prefix
    if not settings.DEBUG or (prefix and '://' in prefix):
        return []
    elif not prefix:
        raise ImproperlyConfigured("Empty static prefix not permitted")
   #
Serve static files below a given point in the directory structure.
  return [ url(r'^%s(?P<path>.*)$' % re.escape(prefix.lstrip('/')), view, kwargs=kwargs), ]

这里view=server就不细看了,其中心思想即把目录结构中的静态文件挂载到一个给定的路径上,这样前端html中的link或者src属性就能根据这一路径找到目标文件。

DONE.

 

posted @ 2018-09-14 15:44  赞美上帝  阅读(1777)  评论(0编辑  收藏  举报