使用lsof的脚本在shell上运行良好而不在cron上运行

I have a small script do count open files on Linux an save results into a flat file. I intend to run it on Cron every minute to gather results later. Script follows:

我有一个小脚本在Linux上计算打开文件，将结果保存到平面文件中。我打算每分钟在Cron上运行它以便稍后收集结果。脚本如下：

/bin/echo "Timestamp: ` date +"%m-%d-%y %T"` Files: `lsof | grep app | wc -l`"

And the crontab is this:

而crontab是这样的：

*/1 * * * * /usr/local/monitor/appmon.sh >> /usr/local/monitor/app_stat.txt

If I run from shell ./script.sh it works well and outputs as:

如果我从shell ./script.sh运行它运行良好并输出为：

Timestamp: 01-31-13 09:33:59 Files: 57

But on the Cron output is:

但是在Cron输出上是：

Timestamp: 01-31-13 09:33:59 Files: 0

Not sure if any permissions are needed or similar. I have tried with sudo on lsof without luck as well.

不确定是否需要任何权限或类似。我已经尝试过sus on lsof也没有运气。

Any hints?

任何提示？

1 个解决方案

#1

from your working cmd-line, do

从你的工作cmd-line，做

which lsof
which grep
which wc
which date

Take the full paths for each of these commands and add them into your shell script, producing something like

获取每个命令的完整路径并将它们添加到shell脚本中，生成类似的内容

/bin/echo "Timestamp: `/bin/date +"%m-%d-%y %T"` Files: `/usr/sbin/lsof | /bin/grep app | /bin/wc -l`"

OR you can set a PATH var to include the missing values in your script, i.e.

或者您可以设置PATH var以在脚本中包含缺失值，即

 PATH=/usr/sbin:${PATH}

Also unless you expect your script to be run from a true Bourne Shell environment, join the early 90's and use the form $( cmd ... ) for cmd-substitution, rather than backticks. The Ksh 93 book, published in 1995 remarks that backticks for command substitution are deprecated ;-)

除非您希望您的脚本从真正的Bourne Shell环境运行，否则请加入90年代早期并使用$（cmd ...）格式进行cmd替换，而不是使用反引号。 1995年出版的Ksh 93书评论说，命令替换的反引号已被弃用;-)

IHTH

#1