Python實現簡單的HTTP請求發送
代碼如下
import socket
def clean_host(host):
"""Remove leading http:// and trailing /path_to_somewhere in host name"""
host = host.strip()
prefixes = ['http://']
for prefix in prefixes:
if host.lower().startswith(prefix):
host = host[len(prefix):]
slash = host.find('/')
if slash>0:
host = host[:slash]
return host
def get_ipv4_list(host, port=80):
"""Return the list of ips corresponding to given host and port number"""
info = socket.getaddrinfo(host, port)
return [item[4][0] for item in info]
def send_request(ip, data, remote_port=80, local_port=2333, bufsize=1024):
"""Send data to give ip and receive response from the server"""
sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.connect((ip, remote_port))
sock.send(data)
recv_data = []
while True:
buf = sock.recv(bufsize)
if not buf:
break
recv_data += [buf]
sock.close()
return ''.join(recv_data)
def read_text(promp='> ', promp2='~ '):
"""Read in text from stdin terminted by EOF"""
text = []
try:
while True:
if not text:
text = [raw_input(promp)+'\n']
else:
text += [raw_input(promp2)+'\n']
except EOFError:
return ''.join(text)
return ''
def http_app():
remote_host = raw_input('Input remote host name (for example www.baidu.com): ')
remote_port = 80
buf = raw_input('Input remote port number (default 80): ')
if buf.isdigit() and 0<=int(buf)<=65535:
remote_port = int(buf)
print remote_host, 'at port', remote_port
print 'Resolving host...'
ip_list = get_ipv4_list(clean_host(remote_host), port=remote_port)
ip = ip_list[0]
print 'Using ip address', ip
while True:
data = read_text()
if not data:
break
print '[REQUEST]'
print data
response = send_request(ip, data, remote_port=remote_port)
print '[REPONSE]'
print response
if __name__ == '__main__':
print 'Running http_app. Press Ctrl+C to quit'
while True:
try:
http_app()
except KeyboardInterrupt:
if raw_input('Quit? ').lower() == 'n':
break
不過不知道是代碼哪里有問題導致返回的全都是Bad Request :(
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