Ansible cron 作业不会转义百分号 `%`

Question

Trying to create a cron job to collect dpkg.logs and ship them to an s3 bucket. The task flows as below:

- 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)"

But, ss described in man crontab, percent-signs must be escaped. That is why you see a backslash before percent sign.

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.

Problem is, Ansible fails to execute the task:

--> 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 }}"

I have tried to ignore the escape of percent and just leave it as $(date +%F) , but cron job is created wrongly. Any ideas?!

Answer 1

尝试逃避反斜杠

\\%F