Cron作业未获取.bashrc中设置的环境变量 [英] Cron job does NOT get the environment variables set in .bashrc

问题描述

这是我的日常工作:

plee@dragon:~$ crontab -l
* * * * * /bin/bash -l -c 'source ~/.bashrc; echo $EDITOR > /tmp/cronjob.test'

在~/.bashrc文件中，我有export EDITOR=vim，但是在最后一个/tmp/cronjob.test文件中，它仍然是空的吗?

and inside ~/.bashrc file, I have export EDITOR=vim, but in the final /tmp/cronjob.test file, it's still empty?

那么如何获取环境变量(在.bashrc文件中设置)并在cron作业中使用它?

So how can I get the environment variables (set in .bashrc file) and use it in my cron job?

plee@dragon:~$ lsb_release -a
No LSB modules are available.
Distributor ID: Ubuntu
Description:    Ubuntu 12.04 LTS
Release:        12.04
Codename:       precise
plee@dragon:~$ uname -a
Linux dragon 3.2.0-26-generic-pae #41-Ubuntu SMP Thu Jun 14 16:45:14 UTC 2012 i686 i686 i386 GNU/Linux

如果使用此:

* * * * * /bin/bash -l -c -x 'source ~/.bashrc; echo $EDITOR > /tmp/cronjob.test' 2> /tmp/cron.debug.res

在/tmp/cron.debug.res中:

...
++ return 0
+ source /home/plee/.bashrc
++ '[' -z '' ']'
++ return
+ echo

顺便说一句，.bashrc文件是Ubuntu 12.04随附的默认文件，但我添加了一行export EDITOR=vim.

BTW, the .bashrc file is the default one came with Ubuntu 12.04, with the exception that I added one line export EDITOR=vim.

如果我不使用cron作业，请直接在命令行上执行此操作:

If I don't use the cron job, instead, just directly do this on the command line:

source .bashrc; echo $EDITOR # Output: vim

推荐答案

source ~/.bashrc不起作用的原因是~/.bashrc上的内容(Ubuntu 12.04中的默认设置).如果您查看它，您将在第5行和第6行看到以下内容:

The reason for source ~/.bashrc not working is the contents on your ~/.bashrc (default one from Ubuntu 12.04). If you look in it you will see on lines 5 and 6 the following:

# If not running interactively, don't do anything
[ -z "$PS1" ] && return

PS1变量是为交互式外壳程序设置的，因此即使您将其作为登录外壳程序执行，通过cron运行时也不存在.这由/bin/bash -l -c -x 'source ~/.bashrc; echo $EDITOR > /tmp/cronjob.test'生成的文件的内容确认:

PS1 variable is set for an interactive shell, so it's absent when run via cron, even though you are executing it as a login shell. This is confirmed by contents of the file produced by /bin/bash -l -c -x 'source ~/.bashrc; echo $EDITOR > /tmp/cronjob.test':

+ source /home/plee/.bashrc
++ '[' -z '' ']'
++ return

要使source ~/.bashrc工作，请注释掉检查~/.bashrc中PS1变量是否存在的行:

To make source ~/.bashrc work, comment out the line that checks for presence of the PS1 variable in ~/.bashrc:

#[ -z "$PS1" ] && return

这将使bash通过cron执行~/.bashrc的全部内容

This will make bash execute the entire contents of ~/.bashrc via cron

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