超时程序管理

遇到一个新的场景：

从两个线上tt拷贝数据到两个线下tt，按照前面写法，两个拷贝的进程是串行执行的，这不满足业务的需求，任何处理呢？

最开始考虑cmd1;cmd2同时提交，但这种方式依然是串行执行，怎么解呢？

最直接的方式和就是两个命令后台执行cmd1&;cmd2&，然后监控进程运行时间，超时就去kill进程，那么如何精确获取进程pid呢？

调研了好久，最后给出解决方案:

def waitPid(name, timeOut):

for line in psutil.process_iter():#获取当前进程list

lists= line.cmdline() #获取进程的命令行，类型为list

Contains="true"

for key in name:#name为list，起进程时所用变量

if key not in lists:

Contains="false"

break

if Contains=="true":

print "pid:", line.pid

time.sleep(timeOut)

if line in psutil.process_iter(): #需要判断进程是否还存在，避免不存在的进程kill时抛异常

cmd = "kill -9 %s"%line.pid #kill进程

print "cmd:%s"%cmd

Execute(cmd)

具体使用：

namelist=["com.aliyun.timetunnel.demo.TTReadAndWrite", "member_cart", "0324153555NBJ3YZX0", "e62375c5-6bcb-4bde-900b-0a38c2f6b218", "1490842758", "10000", "member_cart_blink_mufeng", "5fa49387-e8da-461f-a835-abf03f9b9d4c"]

waitPid(namelist,1)#先kill掉相同任务的进程，否则可能新提交的进程处于sleep状态无法执行

cmd = "java -cp utils/blink_test-1.0-shaded.jar com.aliyun.timetunnel.demo.TTReadAndWrite member_cart 0324153555NBJ3YZX0 e62375c5-6bcb-4bde-900b-0a38c2f6b218 1490842758 10000 member_cart_blink_mufeng 5fa49387-e8da-461f-a835-abf03f9b9d4c >aa.log 2>&1 &"

os.system(cmd)

waitPid(namelist,timeOut)

工作中常常遇到这种场景：

　　例如导数据时上游无数据，任务夯住；执行某个命令时命令夯住等，此时需要加超时判断，如果超过指定时间就退出程序，那怎么写呢？

shell写法：

python写法：

代码如下：

function timeout()

{

command=$*

#echo "command:" $command

waitfor=300

( $command ; echo "success" ) &

commandpid=$!

( sleep $waitfor ; kill -9 $commandpid > /dev/null 2>&1 ) &

watchdog=$!

sleeppid=$PPID

wait $commandpid > /dev/null 2>&1

kill $sleeppid > /dev/null 2>&1

}

python写法：

import time

from subprocess import Popen, PIPE

import commands

import datetime

def sys_command_outstatuserr(cmd, timeout=120):

#print "cmd:%s"%cmd

p = Popen(cmd, stdout=PIPE, stderr=PIPE, shell=True)

t_beginning = time.time()

print "t_begging:%s"%t_beginning

seconds_passed = 0

while True:

if p.poll() is not None:

res = p.communicate()

exitcode = p.poll() if p.poll() else 0

#print "out:%s, exitcode:%d, err:%s"%(res[0], exitcode, res[1])

return res[0], exitcode, res[1]

seconds_passed = time.time() - t_beginning

print "seconds_passed:%s"%seconds_passed

if timeout and seconds_passed > timeout:

p.terminate()

out, exitcode, err = '', 128, 'Timeout'

#print "out:%s, exitcode:%d, err:%s"%(out, exitcode, err)

return out, exitcode, err

time.sleep(10)