Question

尝试创建cron作业以收集dpkg.logs并将其发送到s3存储桶。任务流程如下：

- name: Configure cron job to export patch logs
  cron:
    name: export patch logs daily
    minute: 0
    hour: 0
    user: root
    cron_file: patch_logs
    job: "/usr/local/bin/aws s3 cp /var/log/dpkg.log s3://'{{ patch_logs_bucket }}'/dpkg.log.$(hostname).$(date +\%F)"

但是，在man crontab中描述的ss，必须转义百分号。这就是为什么在百分号前会看到一个反斜杠。

man (5) crontab:

Percent-signs (%) in the command, unless escaped with backslash (\), 
will be changed into newline characters, and all data after the 
first % will be sent to the command as standard input.

问题是，Ansible无法执行任务：

--> Action: 'converge'
ERROR! Syntax Error while loading YAML.
  found unknown escape character '%'

The error appears to have been in '/Users/<user>/<directory>/<to>/<ansible_project>/tasks/base.yml': line 132, column 115, but may
be elsewhere in the file depending on the exact syntax problem.

The offending line appears to be:

    cron_file: patch_logs
    job: "/usr/local/bin/aws s3 cp /var/log/dpkg.log s3://'{{ patch_logs_bucket }}'/dpkg.log.$(hostname).$(date +\%F)"
                                                                                                                  ^ here
We could be wrong, but this one looks like it might be an issue with
missing quotes.  Always quote template expression brackets when they
start a value. For instance:

    with_items:
      - {{ foo }}

Should be written as:

    with_items:
      - "{{ foo }}"

我试图忽略百分号的转义，只是将其保留为$(date +%F)，但是cron作业创建错误。有什么想法吗？！

Answer 1

尝试逃避反冲

\\%F

Ansible Cron作业无法逃脱百分号“％”

1 个答案: