设想：用一个shell查看一个后台运行的PHP程序是否非正常退出

如果退出，利用守护进程自动restart.

类似mysql的safe_mysqld

共3个程序

shstart.sh 负责启动

sh.sh 守护进程主程序

shstop.sh 安全退出守护进程和php程序主体

/**************************/

shstart.sh

#!/usr/local/bin/bash

/home/phpshell/sh.sh &

shpid="$!";

echo "$shpid" > /home/phpshell/sh.sid

/**************************/

sh.sh

#!/usr/local/bin/bash

PHP="/usr/local/bin/php"

PROGRAM="/www/time.php"

#start dameo

$PHP $PROGRAM &

chpid="$!";

echo "$chpid" > /home/phpshell/php.sid

echo "child pid is $chpid"

echo "status is $?"

while [ 1 ]

do

wait $chpid

exitstatus="$?"

echo "child pid=$chpid is gone, $exitstatus" >> /home/phpshell/phperror.log

echo `date` >> /home/phpshell/phperror.log

echo "**************************" >>/home/phpshell/phperror.log

sleep 10

$PHP $PROGRAM &

chpid="$!";

echo "$chpid" > /home/phpshell/php.sid

echo "next child pid is $chpid"

echo "next status is $?"

echo "userkill is $userkill"

done

/******************************/

shstop.sh

#!/usr/local/bin/bash

chpid="`cat sh.sid`";

kill $chpid;

echo "kill sh.sh done!"

ps ax|grep php |grep -v grep|awk '{print $1}'|xargs kill

sleep 2

ps ax|grep php |grep -v grep|awk '{print $1}'|xargs kill

echo "kill php done"

利用这个原理，可以实现很多后台程序的吊线问题