Python脚本检测一个url地址
说明:检测URL地址,检测依据返回的url的code地址,返回不正常,则报警
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#!/usr/bin/envpython
#coding:utf-8
#
#Author Summer
#
#Create date:2015-07-24
import smtplib
import string
import urllib
import time
def email():
HOST="smtp.163.com"
FROM="***********@163.com"
TO="***********@139.com"
PASS="yjl520105"
SUBJECT="阿里聚石塔服务器Nginx代理服务报警"
TEXT="IP地址为 121.41.17.164 聚石塔服务器不能正常代理，请立即登录该主机检测。10分钟无法解决，请理解切换IP为121.41.173.43地址，该操作需要在数据库更改。"
BODY=string.join((
"From: %s" %FROM,
"To： %s" %TO,
"Subject: %s" %SUBJECT,
"",
TEXT
),"\r\n")
Server=smtplib.SMTP()
Server.connect(HOST,25)
Server.login(FROM,PASS)
Server.sendmail(FROM,TO,BODY)
Server.quit()
def url():
now=time.strftime("%Y-%m-%d %H:%M",time.localtime(time.time()))
m=open("/opt/shell/check_ali/curl_success.txt","a")
n=open("/opt/shell/check_ali/curl_failure.txt","a")
status_code=urllib.urlopen("http://121.41.17.164/router/rest").code
tab=''
text = str(now) + " " + str(status_code) +'\n'
if status_code != 200:
email()
n.write(text)
else:
m.write(text)
url()
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