我想弄清楚如何将错误输出到浏览器,而不是一直通过ssh和“tail-f error\u log”来logginf
在dev实例中,我们刚刚做了debug =True
,现在在生产服务器中?
这是我的配置:
DocumentRoot /home/nikos/public_html
<Directory /home/nikos/public_html>
Require all granted
</Directory>
Alias /static /home/nikos/public_html/static
<Directory /home/nikos/public_html/static>
Options +Indexes
</Directory>
WSGIPassAuthorization On
WSGIDaemonProcess clientele user=nikos group=nikos home=/home/nikos/public_html
WSGIScriptAlias /clientele /home/nikos/public_html/clientele.py process-group=clientele application-group=%{GLOBAL}
WSGIDaemonProcess downloads user=nikos group=nikos home=/home/nikos/public_html
WSGIScriptAlias /downloads /home/nikos/public_html/downloads.py process-group=downloads application-group=%{GLOBAL}
WSGIDaemonProcess www user=nikos group=nikos home=/home/nikos/public_html
WSGIScriptAliasMatch ^/(?!phpmyadmin) /home/nikos/public_html/www.py process-group=www application-group=%{GLOBAL}
debug flag可能会执行您想要的操作:
但我想重申:你不应该这样做。这是一个security hole
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