Flask Apache：如何从烧瓶应用程序根目录（而不是__init.__.py）中的python脚本访问静态文件

Question

我已经用Apache部署了Flask应用程序。 / var / www / FlaskApp /下的目录结构如下

./FlaskApp/__init__.py
./FlaskApp/oracledb.py
./FlaskApp/templates
./FlaskApp/static
./FlaskApp/static/css
./FlaskApp/static/conf/inv_conf.yml

inv_conf.yml包含数据库连接参数。 oracledb.py包含在 init .py

中导入的DB类。

class OracleDB:
    def __init__(self):

        with open('static/conf/inv_conf.yaml', 'r') as file:
          conf = yaml.load(file)

init .py导入

之类的类

from oracledb import OracleDB
app = Flask(__name__)
db = OracleDB()

当我访问应用程序时，从apache获取错误信息

[Sat May 25 11:17:53.158072 2019] [:error] [pid 84236]   File "/var/www/FlaskApp/FlaskApp/oracledb.py", line 7, in __init__
[Sat May 25 11:17:53.158074 2019] [:error] [pid 84236]     with open('static/conf/inv_conf.yaml', 'r') as file:
[Sat May 25 11:17:53.158078 2019] [:error] [pid 84236] IOError: [Errno 2] No such file or directory: 'static/conf/inv_conf.yaml'

但是如果我将inv_conf.yml放在系统根目录'/'下，则可以正常工作。能否请您帮我解决相对路径的问题以及如何解决此问题？

我的httpd.conf就像

<VirtualHost *:80>
    ServerName <server name>

    WSGIDaemonProcess FlaskApp user=apache  group=apache
    WSGIScriptAlias /inventory /var/www/FlaskApp/FlaskApp.wsgi
    Alias /static /var/www/FlaskApp/FlaskApp/static

    <Directory /var/www/FlaskApp/FlaskApp>
        WSGIProcessGroup FlaskApp
        WSGIApplicationGroup %{GLOBAL}
        Order allow,deny
        Allow from all
    </Directory>
</VirtualHost>

wsgi文件是

[root@slc10xxi FlaskApp]# cat FlaskApp.wsgi
#!/usr/bin/python
import sys
import logging

logging.basicConfig(stream=sys.stderr)
sys.path.insert(0, '/var/www/FlaskApp/')
from FlaskApp import app as application