(bsnmpd),uid 0:退出信号6(核心转储)/语法错误:单词意外(期待“那么”)

我正在使用FreeBSD 9.2-RELEASE:

# uname -a FreeBSD f9.alexus.org 9.2-RELEASE FreeBSD 9.2-RELEASE #0 r255898: Thu Sep 26 22:50:31 UTC 2013 [email protected].freebsd.org:/usr/obj/usr/src/sys/GENERIC amd64 # 

每隔一段时间bsnmpd(1)为我产生以下信息:

 # bzip2 -cd /var/log/all.log.0.bz2 | grep bsnmpd Oct 12 21:01:44 f9 kernel: pid 62584 (bsnmpd), uid 0: exited on signal 6 (core dumped) # 

真的想弄清楚是什么造成的,但同时:

 # crontab -l | grep @hourly @hourly `which service` bsnmpd status >/dev/null ; if ( $? != 0 ) `which service` bsnmpd start ; endif # 

我不断收到电子邮件w /以下消息:

 Syntax error: word unexpected (expecting "then") 

testing(通过shell):

 # `which service` bsnmpd status >/dev/null ; if ( $? != 0 ) `which service` bsnmpd start ; endif # /etc/rc.d/bsnmpd stop Stopping bsnmpd. Waiting for PIDS: 60671. # /etc/rc.d/bsnmpd status bsnmpd is not running. # `which service` bsnmpd status > /dev/null ; if ( $? != 0 ) `which service` bsnmpd start ; endif Starting bsnmpd. # /etc/rc.d/bsnmpd status bsnmpd is running as pid 61042. # 
  1. 我怎样才能debugging是什么导致bsnmpd(1)退出在第一个地方?
  2. 我的cronjob有什么问题?

你会希望cron工作看起来更像这样:

 /usr/sbin/service bsnmpd status >/dev/null ; if [ $? != 0 ] ; then /usr/sbin/service bsnmpd start ; fi 

无论如何,让我们来看看为什么bsnmpd是核心倾销。 看看你是否可以findbsnmpd.core文件,然后运行/usr/bin/gdb /usr/sbin/bsnmpd bsnmpd.core然后运行bt并粘贴输出。