才学-简单端口爆破小脚本-白红宇

才学-简单端口爆破小脚本

阅读量：6451 次

发布时间：2019-06-23

本文共 1668 字，大约阅读时间需要 5 分钟。

简单的脚本

#coding:utf-8from socket import *import sysimport threadingimport time#python scan_port.py 
     
     
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       lock = threading.RLock()class INFO:    HEADER = '\033[95m'    OKBLUE = '\033[94m'    OKGREEN = '\033[92m'    WARNING = '\033[93m'    ARNING = '\033[93m'    FAIL = '\033[91m'    ENDC = '\033[0m'    BOLD = '\033[1m'    UNDERLINE = '\033[4m'def scan_port(host, port):    s = socket(AF_INET, SOCK_STREAM)    s.settimeout(5)    try:        result = s.connect_ex((host, port))        #print result        if result == 0:            lock.acquire()            print INFO.OKBLUE+u"[+]---%s, %d 端口开放---[+] "%(host, port)+INFO.ENDC            lock.release()        else:            print INFO.WARNING+u"[+]---%s, %d 端口没有开放---[+] " % (host, port)+INFO.ENDC        #s.close()        #    except Exception, e:        raise eif __name__ == '__main__':    #scan_port("demo.testfire.net", 80)    start_time =time.time()    if len(sys.argv) != 3:        print "python scan_port.py 
        
        
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          , and port must 1-65535 int" sys.exit() host = sys.argv[1] ports = sys.argv[2].split('-') #print ports #port = ports[0] #print len(sys.argv) if len(ports) ==2 and (1 <= int(ports[0]) <= 65535) and (1 <= int(ports[1]) <=65535): for p in range(int(ports[0]),int(ports[1])+1): t = threading.Thread(target=scan_port,args=(host,p)) t.start() #print "Finished in %.2fs" % (time.time() - start_time) #scan_port(host, p) elif len(ports) ==1 and len(sys.argv) == 3 and (1 <= int(ports[0]) <= 65535): t = threading.Thread(target=scan_port,args=(host, int(ports[0]))) t.start() else: print "python scan_port.py 
           
           
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             , and port must 1-65535 int" sys.exit()