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Python判斷telnet通不通

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