python使用socket实现多线程端口扫描

末蓝、 2022-06-15 10:48 488阅读 0赞

使用socket主要是判断是否端口能否联通  
socket[<<详情参考>>][Link 1]

1.socket.connect() 检测端口，若端口不能联通，则会抛出异常，若无异常，则端口存活，使用try-except异常捕获处理，进而进行端口扫描

这里将使用100个线程，每个端口连接都要检查是否超过65535

#coding=utf8
    import socket
    import threading
    
    
    def scan(host,cport):
        for x in range(660):
           p=x+660*cport
           try:
             if p < 65535:
                 s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
                 s.connect((host, p))
                 s.settimeout(0.1)
                 print "local open port : %d"%p
                 s.close()
             else :
                 break
           except Exception:
               pass
    
    list = []
    for i in range(100):
            t = threading.Thread(target=scan,args=('localhost',i,))
            list.append(t)
            t.start()
    
    for threadinglist in list:
            threadinglist.join()

输出：

local open port : 22
    local open port : 111
    local open port : 34172
    local open port : 49716
    local open port : 51078

2.socket.connect\_ex() 若联通则返回为0，否则返回为非0  
依据connect\_ex返回值判断端口是否存活

import socket 
    import threading 
    from Queue import Queue 
    
    def scan(port): 
      s = socket.socket() 
      s.settimeout(1) 
      if s.connect_ex(('localhost', port)) == 0: 
         print "local open port : %d"%port
      s.close() 
    
    def worker(): 
      while not q.empty(): 
        port = q.get() 
        try: 
          scan(port) 
        finally: 
          q.task_done() 
    
    if __name__ == '__main__': 
      q = Queue() 
      map(q.put,xrange(1,65535)) 
      threads = [threading.Thread(target=worker) for i in xrange(100)] 
      map(lambda x:x.start(),threads) 
      q.join()

输出：

local open port : 22
    local open port : 25
    local open port : 111
    local open port : 631
    local open port : 35156
    local open port : 38929
    local open port : 51078

[Link 1]: http://blog.csdn.net/comprel/article/details/72824784