用一个shell查看一个后台运行的php程序是否非正常退出

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