我通常有几个问题,如何cron执行脚本,因为他们通常没有我的环境设置。是否有一种方法可以调用bash(?)以同样的方式cron,这样我就可以在安装它们之前测试脚本?
当前回答
已接受的答案确实提供了一种使用cron将使用的环境运行脚本的方法。正如其他人指出的那样,这并不是调试cron作业所需的唯一标准。
事实上,cron也使用非交互式终端,没有附加的输入等等。
如果这有帮助的话,我已经编写了一个脚本,可以轻松地运行命令/脚本,就像由cron运行一样。使用命令/脚本作为第一个参数调用它,就没问题了。
这个脚本也托管(可能更新)在Github上。
#!/bin/bash
# Run as if it was called from cron, that is to say:
# * with a modified environment
# * with a specific shell, which may or may not be bash
# * without an attached input terminal
# * in a non-interactive shell
function usage(){
echo "$0 - Run a script or a command as it would be in a cron job, then display its output"
echo "Usage:"
echo " $0 [command | script]"
}
if [ "$1" == "-h" -o "$1" == "--help" ]; then
usage
exit 0
fi
if [ $(whoami) != "root" ]; then
echo "Only root is supported at the moment"
exit 1
fi
# This file should contain the cron environment.
cron_env="/root/cron-env"
if [ ! -f "$cron_env" ]; then
echo "Unable to find $cron_env"
echo "To generate it, run \"/usr/bin/env > /root/cron-env\" as a cron job"
exit 0
fi
# It will be a nightmare to expand "$@" inside a shell -c argument.
# Let's rather generate a string where we manually expand-and-quote the arguments
env_string="/usr/bin/env -i "
for envi in $(cat "$cron_env"); do
env_string="${env_string} $envi "
done
cmd_string=""
for arg in "$@"; do
cmd_string="${cmd_string} \"${arg}\" "
done
# Which shell should we use?
the_shell=$(grep -E "^SHELL=" /root/cron-env | sed 's/SHELL=//')
echo "Running with $the_shell the following command: $cmd_string"
# Let's route the output in a file
# and do not provide any input (so that the command is executed without an attached terminal)
so=$(mktemp "/tmp/fakecron.out.XXXX")
se=$(mktemp "/tmp/fakecron.err.XXXX")
"$the_shell" -c "$env_string $cmd_string" >"$so" 2>"$se" < /dev/null
echo -e "Done. Here is \033[1mstdout\033[0m:"
cat "$so"
echo -e "Done. Here is \033[1mstderr\033[0m:"
cat "$se"
rm "$so" "$se"
其他回答
我发现的另一种简单方法(但可能容易出错,我仍在测试中)是在命令之前获取用户的配置文件。
编辑/etc/cron.d /脚本:
* * * * * user1 comand-that-needs-env-vars
会变成:
* * * * * user1 source ~/.bash_profile; source ~/.bashrc; comand-that-needs-env-vars
很脏,但它帮我完成了任务。是否有方法模拟登录?只是一个你可以运行的命令?Bash—登录不起作用。听起来这是更好的方法。
编辑:这似乎是一个可靠的解决方案:http://www.epicserve.com/blog/2012/feb/7/my-notes-cron-directory-etccrond-ubuntu-1110/
* * * * * root su --session-command="comand-that-needs-env-vars" user1 -l
别忘了,因为cron的父进程是init,所以它运行的程序没有控制终端。你可以用这样的工具来模拟:
http://libslack.org/daemon/
将此添加到您的crontab(临时):
* * * * * env > ~/cronenv
运行后,执行以下操作:
env - `cat ~/cronenv` /bin/sh
这假设您的cron运行/bin/sh,不管用户的默认shell是什么,它都是默认的。
脚注:如果env包含更高级的配置,例如PS1=$(__git_ps1 " (%s)")$,它将神秘地错误env: ":没有这样的文件或目录。
一些方法:
Export cron env and source it: Add * * * * * env > ~/cronenv to your crontab, let it run once, turn it back off, then run env - `cat ~/cronenv` /bin/sh And you are now inside a sh session which has cron's environment Bring your environment to cron You could skip above exercise and just do a . ~/.profile in front of your cron job, e.g. * * * * * . ~/.profile; your_command Use screen Above two solutions still fail in that they provide an environment connected to a running X session, with access to dbus etc. For example, on Ubuntu, nmcli (Network Manager) will work in above two approaches, but still fail in cron. * * * * * /usr/bin/screen -dm Add above line to cron, let it run once, turn it back off. Connect to your screen session (screen -r). If you are checking the screen session has been created (with ps) be aware that they are sometimes in capitals (e.g. ps | grep SCREEN) Now even nmcli and similar will fail.
取决于shell的账户
sudo su
env -i /bin/sh
or
sudo su
env -i /bin/bash --noprofile --norc
从http://matthew.mceachen.us/blog/howto -模拟- - cron -环境- 1018. - html