Python判斷telnet通不通
阿新 • • 發佈:2019-01-24
這個跟ping那個差不多,ping的那個指令碼就是通過這個改了下,大體一致,不過telnet的不需要判斷返回的字串。快一些
這裡具體需要telnet的ip是需要自己向定義好的陣列中寫的
我這裡加了一個定時,是7200秒,也就是2小時
行了,上程式碼吧:
#!/usr/bin/env python # coding: utf8 import telnetlib import time import codecs import os # telnet host def telnetdo(host, statusFile): status1 = 'telnet success' status2 = 'telnet faild' for ipAdd in host: # get now time nowTime = time.strftime('%Y-%m-%d %H:%M:%S',time.localtime(time.time())) try: t = telnetlib.Telnet(ipAdd, port=23, timeout=1) writeToText(nowTime, ipAdd, status1, statusFile) except: writeToText(nowTime, ipAdd, status2, statusFile) # write status information to txt def writeToText(nowTime, ipAdd, status, statusFile): s_text = 'TIME:' + nowTime + '\t' + 'IP:' + ipAdd + '\t' + 'STATUS:' + status + '\r\n' if '0' == judgeFile(statusFile): with open(statusFile, 'a') as f: f.write(s_text) f.close() if '1' == judgeFile(statusFile): with open(statusFile, 'w') as f: f.write(s_text) f.close() # Determine whether statusFile exists # 0: exists # 1: no exists def judgeFile(statusFile): if os.path.exists(statusFile): return '0' else: return '1' if __name__ == "__main__": host = ['192.168.1.254', '192.168.1.100'] # write file statusFile = '/root/telnetStatus.txt' i = 7200 while i: telnetdo(host, statusFile) time.sleep(2) i = i - 1
結果會存在/root下面