Python端口扫描简单程序
短信预约 -IT技能 免费直播动态提醒
本文实例为大家分享了Python端口扫描的实现代码,供大家参考,具体内容如下
获取本机的IP和端口号:
import socket
def get_my_ip():
try:
csock = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
csock.connect(('8.8.8.8', 80))
(addr, port) = csock.getsockname()
csock.close()
return addr,port
except socket.error:
return "127.0.0.1"
def int_to_ip(int_ip):
return socket.inet_ntoa(struct.pack('I', socket.htonl(int_ip)))
def ip_to_int(ip):
return socket.ntohl(struct.unpack("I", socket.inet_aton(str(ip)))[0])
(ip,port)=get_my_ip()
print "ip=%s port=%d" %(ip,port)
PortScan.py
#!/usr/bin/python
# -*- coding: utf-8 -*-
import optparse
from socket import *
from threading import *
screenLock = Semaphore(value=1)
def connScan(tgtHost, tgtPort):
try:
connSkt = socket(AF_INET, SOCK_STREAM)
connSkt.connect((tgtHost, tgtPort))
connSkt.send('ViolentPythonrn')
results = connSkt.recv(100)
screenLock.acquire()
print '[+] %d/tcp open' % tgtPort
print '[+] ' + str(results)
except:
screenLock.acquire()
print '[-] %d/tcp closed' % tgtPort
finally:
screenLock.release()
connSkt.close()
def portScan(tgtHost, tgtPorts):
try:
tgtIP = gethostbyname(tgtHost)
except:
print "[-] Cannot resolve '%s': Unknown host" %tgtHost
return
try:
tgtName = gethostbyaddr(tgtIP)
print 'n[+] Scan Results for: ' + tgtName[0]
except:
print 'n[+] Scan Results for: ' + tgtIP
setdefaulttimeout(1)
for tgtPort in tgtPorts:
t = Thread(target=connScan,args=(tgtHost,int(tgtPort)))
t.start()
def main():
parser = optparse.OptionParser('usage %prog '+
'-H <target host> -p <target port>')
parser.add_option('-H', dest='tgtHost', type='string',
help='specify target host')
parser.add_option('-p', dest='tgtPort', type='string',
help='specify target port[s] separated by comma')
(options, args) = parser.parse_args()
tgtHost = options.tgtHost
tgtPorts = str(options.tgtPort).split(',')
if (tgtHost == None) | (tgtPorts[0] == None):
print parser.usage
exit(0)
portScan(tgtHost, tgtPorts)
if __name__ == '__main__':
main()
以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持编程网。
免责声明:
① 本站未注明“稿件来源”的信息均来自网络整理。其文字、图片和音视频稿件的所属权归原作者所有。本站收集整理出于非商业性的教育和科研之目的,并不意味着本站赞同其观点或证实其内容的真实性。仅作为临时的测试数据,供内部测试之用。本站并未授权任何人以任何方式主动获取本站任何信息。
② 本站未注明“稿件来源”的临时测试数据将在测试完成后最终做删除处理。有问题或投稿请发送至: 邮箱/279061341@qq.com QQ/279061341