我有一个Python脚本,它实现了一个内置的web服务器:
class http_server(BaseHTTPRequestHandler):
def log_message(self, format, *args):
# prevent the BaseHTTPServer log messages, we use our own logging instead
return
def do_GET(self):
log("responding to http request from %s: %s" % (self.client_address[0], self.path))
text_string = "Hello World"
self.send_response(200)
self.send_header("Content-type", "text/plain")
self.send_header("Content-Length", len(text_string))
self.end_headers()
self.wfile.write(text_string)
def start_server():
try:
httpd = SocketServer.TCPServer(("", 8888), http_server)
httpd.serve_forever()
except Exception as e:
cleanup(None, None)
print "Error starting internal http server: %s" % repr(e)
sys.exit(1)
# main script
# does various things, initializes curses, etc.
start_server()
这很好,但问题是python脚本还使用在另一个线程中运行的curses实现了屏幕状态显示。当HTTP服务器中发生错误时(例如"对等方重置连接"等),指示所述错误的python回溯会在我漂亮的curses显示屏上显示。
我曾尝试在BaseHTTPRequestHandler
类的do_GET
部分周围添加try...exception
块,但没有效果。
如何在该代码中静默Python回溯消息?
尝试重写BaseServer
:的handle_error
方法
class MyServer(SocketServer.TCPServer):
def handle_error(self, request, client_address):
pass
然后在start_server
函数中使用MyServer
。