我已经写在python的web服务器获取网址,客户端使用网络浏览器发送的请求是这样的:如何从Python客户端请求
http://localhost:13555/ChessBoard_x16_y16.bmp
在服务器端,当我在打印客户端 - 请求是这样的:
GET /ChessBoard_x16_y16.bmp HTTP/1.1
Host: localhost:13555
Connection: keep-alive
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0
.8
User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like
Gecko) Chrome/43.0.2357.81 Safari/537.36
Accept-Encoding: gzip, deflate, sdch
Accept-Language: en-US,en;q=0.8
GET /favicon.ico HTTP/1.1
Host: localhost:13555
Connection: keep-alive
User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like
Gecko) Chrome/43.0.2357.81 Safari/537.36
Accept: */*
Referer: http://localhost:13555/ChessBoard_x16_y16.bmp
Accept-Encoding: gzip, deflate, sdch
Accept-Language: en-US,en;q=0.8
但我只想要得到&打印实际的URL,如:
Referer: http://localhost:13555/ChessBoard_x16_y16.bmp
请告诉我该怎么做?
的URL应该从得到的**' GET'行**,而不是favicon上的Referer头。 –
为什么这个问题几乎完全重复[this one](http://stackoverflow.com/questions/30552358/how-to-get-integer-values-from-a-url-request)? – MattDMo