在 crontab 中使用“%”

在 crontab 中使用“%”

我正在尝试添加一个 cron 作业来执行从远程服务器到(Ubuntu12)本地计算机的 rsync 并创建日志文件。

下面是crontab-l

00 18 * * * rsync -a -v --delete -e ssh user@centosvm:/home/user/rsync-test ~/backup > ~/rsync$(date +%Y%m%d_%H%M%S).log 2>&1

我不断收到这封邮件,告知工作中存在语法错误。

Received: by work-virtual-machine (Postfix, from userid 1002)

        id 697ADA24A0; Thu, 30 Apr 2015 16:21:01 -0700 (PDT)

From: root@work-virtual-machine (Cron Daemon)

To: user@work-virtual-machine

Subject: Cron <user@work-virtual-machine> "rsync -a -v --delete -e ssh user@centosvm:/home/user/rsync-test ~/backup > /home/user/rsync$(date +

Content-Type: text/plain; charset=ANSI_X3.4-1968

X-Cron-Env: <SHELL=/bin/sh>

X-Cron-Env: <HOME=/home/user>

X-Cron-Env: <PATH=/usr/bin:/bin>

X-Cron-Env: <LOGNAME=user>

Message-Id: <20150430232101.697ADA24A0@work-virtual-machine>

Date: Thu, 30 Apr 2015 16:21:01 -0700 (PDT)

/bin/sh: 1: Syntax error: end of file unexpected (expecting ")")

我还安装了 unix2dos 软件包。

答案1

好吧——想通了。发回来以防万一有人在某个时候遇到它。 The % sign has a special meaning in crontab. it's changed to newline and any string after the first % will be sent to the command as standard input. To force cron to interpret it literally, you have to escape it:

00 18 * * * rsync -a -v --delete -e ssh user@centosvm:/home/user/rsync-test ~/backup > ~/rsync$(date +\%Y\%m\%d_\%H\%M\%S).log 2>&1

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