python之Django Assets 管道

我正在使用 django-pipeline 来管理我的静态 Assets 并编译我的 coffeescript 和 SCSS 文件。我在配置此 Django 应用程序时遇到问题。我想要实现的是让 django-pipeline 从 assets/子目录中读取我的 coffeescript 和 SCSS 文件，并将它们编译并压缩到我告诉 Django 提供静态文件的 public/子目录中。

作为引用，我的 Django 文件结构如下所示:

DjangoApp 
    assets 
        coffeescript 
        scss 
    bin 
    configs 
    db 
    logs 
    DjangoApp 
    public 
        css 
        images 
        js 
    requirements 

我当前的 Django 设置如下所示(为简洁起见省略了不相关的内容):

# Absolute path to the directory static files should be collected to. 
# Don't put anything in this directory yourself; store your static files 
# in apps' "static/" subdirectories and in STATICFILES_DIRS. 
# Example: "/var/www/example.com/static/" 
STATIC_ROOT = '' 
 
# URL prefix for static files. 
# Example: "http://example.com/static/", "http://static.example.com/" 
STATIC_URL = '/public/' 
 
# Additional locations of static files 
STATICFILES_DIRS = ( 
    # Put strings here, like "/home/html/static" or "C:/www/django/static". 
    # Always use forward slashes, even on Windows. 
    # Don't forget to use absolute paths, not relative paths. 
    PROJECT_PATH + '/public', 
) 
 
# Asset compilers. 
PIPELINE_COMPILERS = ( 
    'pipeline.compilers.coffee.CoffeeScriptCompiler', 
    'pipeline.compilers.sass.SASSCompiler' 
) 
 
PIPELINE_CSS = { 
    'ie': { 
        'source_filenames': ( 
            'scss/ie.scss', 
        ), 
        'output_filename': 'css/ie.min.css' 
    }, 
    'print': { 
        'source_filenames': ( 
            'scss/print.scss', 
        ), 
        'output_filename': 'css/print.min.css' 
    }, 
    'screen': { 
        'source_filenames': ( 
            'scss/screen.scss', 
            'scss/screen-responsive-768px.scss', 
            'scss/screen-responsive-992px.scss', 
            'scss/screen-responsive-1200px.scss' 
        ), 
        'output_filename': 'css/screen.min.css' 
    } 
} 
 
PIPELINE_JS = { 
    'application': { 
        'source_filenames': ( 
          'coffeescript/application.coffee', 
        ), 
        'output_filename': 'js/application.min.js', 
    } 
} 

也许这个应用程序不能做我想做的事。我应该寻找其他东西吗？

请您参考如下方法:

我已经玩了一段时间了，运气不好。然而，我已经从 python gears 或 django-gears 中找到了我想要的东西。考虑到我想要实现的是复制 Ruby sprockets，这将是完美的。