linux cron作业未运行
[英]linux cron job isn't running
问题描述
I have a Ubuntu Server 10. 
我有一个Ubuntu Server 10。
At the root cron I have a script to run every day at the same hour. 在root cron，我有一个脚本可以每天在同一时间运行。
Running the script manually it's run fine, but in the cron it doesn't run. 手动运行脚本可以正常运行，但在cron中则无法运行。
At the syslog I got this: 在系统日志中，我得到了：
Aug 23 09:22:01 database CRON[6884]: (root) CMD (./bkp.sh >> /tmp/bkp.log)
Aug 23 09:22:01 database CRON[6883]: (CRON) error (grandchild #6884 failed with exit status 2)
I found the possible source of the problem, I added some LOGs (echo) in the script... 我找到了问题的可能根源，我在脚本中添加了一些日志（回声）...
During script execution, it exits on a "function" definition... 在脚本执行期间，它以“功能”定义退出。
It seems that I have no access to command "function" 看来我无权访问命令“功能”
How can I solve this? 我该如何解决？
3 个解决方案
解决方案1
0 2012-08-23 12:43:43
Shell scripts launched by cron must carefully specify the environment that they're running in. Shells run within cron will be non-interactive, meaning that they will not read /etc/profile , where $PATH is typically set. 通过推出Shell脚本cron必须认真指定他们在正在运行的环境。中运行壳cron将非交互式的，这意味着他们将无法读取/etc/profile ，其中$PATH通常设置。 This means that you will need to specify the paths to all executables called in your script. 这意味着您将需要指定脚本中调用的所有可执行文件的路径。
If you run a lot of cron scripts, it may be worth your time to write a small shell script which will define and export environment variables to be used inside your script, starting with $PATH . 如果您运行许多cron脚本，则可能值得花时间编写一个小的Shell脚本，该脚本将定义并导出要在脚本中使用的环境变量，以$PATH开头。
解决方案2
0 已采纳 2012-08-23 12:55:29
解决方案3
0 2012-08-23 14:27:17
将脚本放在以下目录中，然后它将每天运行。
/etc/cron.daily/
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