bash - 在 bash 中杀死运行超过指定时间的进程？

我在/etc/init.d目录中有Oracle的关闭脚本
在“停止”命令上，它会执行以下操作:

su oracle -c "lsnrctl stop >/dev/null"
su oracle -c "sqlplus sys/passwd as sysdba @/usr/local/PLATEX/scripts/orastop.sql >/dev/null"

..
问题是当lsnrctl或sqlplus没有响应时-在这种情况下，此“停止”脚本永无休止，并且服务器无法关闭。唯一的方法-是“ kill -9”。

我想重写脚本，以便在5分钟后(例如)如果命令未完成-应该终止它。

我怎样才能做到这一点？你能给我一个例子吗？
我正在使用Linux RHEL 5.1 + bash。

最佳答案

如果能够使用 3rd 方工具，我会利用您可以从脚本调用的 3rd 方预先编写的帮助程序之一(doalarm 和 timeout 都被 the BashFAQ entry on the subject 提及)。

如果不使用此类工具自己编写此类内容，我可能会执行以下操作:

function try_proper_shutdown() {
  su oracle -c "lsnrctl stop >/dev/null"
  su oracle -c "sqlplus sys/passwd as sysdba @/usr/local/PLATEX/scripts/orastop.sql >/dev/null"
}

function resort_to_harsh_shutdown() {
  for progname in ora_this ora_that ; do
    killall -9 $progname
  done
  # also need to do a bunch of cleanup with ipcs/ipcrm here
}

# here's where we start the proper shutdown approach in the background
try_proper_shutdown &
child_pid=$!

# rather than keeping a counter, we check against the actual clock each cycle
# this prevents the script from running too long if it gets delayed somewhere
# other than sleep (or if the sleep commands don't actually sleep only the
# requested time -- they don't guarantee that they will).
end_time=$(( $(date '+%s') + (60 * 5) ))
while (( $(date '+%s') < end_time )); do
  if kill -0 $child_pid 2>/dev/null; then
    exit 0
  fi
  sleep 1
done

# okay, we timed out; stop the background process that's trying to shut down nicely
# (note that alone, this won't necessarily kill its children, just the subshell we
# forked off) and then make things happen.
kill $child_pid
resort_to_harsh_shutdown

关于bash - 在 bash 中杀死运行超过指定时间的进程？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/1516458/