Question

我正在尝试将pyvirtualdisplay作为WSGI应用程序的一部分运行。我安装了Xfvb，我的000-default.conf文件：

<VirtualHost *:80>
    # The ServerName directive sets the request scheme, hostname and port that
    # the server uses to identify itself. This is used when creating
    # redirection URLs. In the context of virtual hosts, the ServerName
    # specifies what hostname must appear in the request's Host: header to
    # match this virtual host. For the default virtual host (this file) this
    # value is not decisive as it is used as a last resort host regardless.
    # However, you must set it for any further virtual host explicitly.
    #ServerName www.example.com

    ServerAdmin webmaster@localhost
    DocumentRoot /var/www/html
    WSGIDaemonProcess flaskapp user=ubuntu threads=5
    WSGIScriptAlias / /var/www/html/flaskapp/flaskapp.wsgi
    <Directory flaskapp>
        WSGIProcessGroup flaskapp
        WSGIApplicationGroup %{GLOBAL}
        Require all granted
    </Directory>
    # Available loglevels: trace8, ..., trace1, debug, info, notice, warn,
    # error, crit, alert, emerg.
    # It is also possible to configure the loglevel for particular
    # modules, e.g.
    #LogLevel info ssl:warn

    ErrorLog ${APACHE_LOG_DIR}/error.log
    CustomLog ${APACHE_LOG_DIR}/access.log combined

    # For most configuration files from conf-available/, which are
    # enabled or disabled at a global level, it is possible to
    # include a line for only one particular virtual host. For example the
    # following line enables the CGI configuration for this host only
    # after it has been globally disabled with "a2disconf".
    #Include conf-available/serve-cgi-bin.conf
</VirtualHost>

# vim: syntax=apache ts=4 sw=4 sts=4 sr noet

我的代码：

from flask import Flask
from pyvirtualdisplay import Display

app = Flask(__name__)

@app.route('/')
def crawl_sc():
    display = Display(visible=0, size=(1920, 999))
    display.start()
    display.stop()
    return 'ran'
if __name__ == '__main__':
    app.run(debug=True)

当我通过WSGI运行它时，它将不断加载，但是error.log中没有错误。我正在将AWS EC2与Ubuntu 16.04和Apache 2.4.18，mod_wsgi 4.6.4和Python 3.6一起使用。

Answer 1

因此，在与mod_wsgi的git repo所有者联系后，我发现我已经使用上面的000-default.conf在嵌入式模式下运行了它。在嵌入式中，它不使用主解释器。为了解决这个问题，我将以上内容更改为：

 <Directory /var/www/html/flaskapp>
    WSGIProcessGroup flaskapp
    WSGIApplicationGroup %{GLOBAL}
    Require all granted
 </Directory>

这使用了主解释器，现在脚本运行了：）

在WSGI上运行Pyvirtualdisplay

1 个答案: