从 Ansible 中的字符串中删除最后 n 个字符

Question

下面是我构建字符串的剧本。

   - name: Construct command for all paths on a particular IP
     set_fact:
       allcmd: "{{ allcmd | default('') + '\"ls -lrt ' + item.path + ' | tail -57 &&' }}"
     loop: "{{ user1[inventory_hostname] }}"

   - debug:
       msg: "allcmd is:{{ allcmd }}"

输出：

ok: [10.9.9.11] => (item={u'path': u'/tmp/scripts', u'name': u'SCRIPT'}) => {
    "ansible_facts": {
        "allcmd": "ls -lrt /tmp/scripts | tail -57 &&"
    },
    "ansible_loop_var": "item",
    "changed": false,
    "item": {
        "name": "SCRIPT",
        "path": "/tmp/scripts"
    }
}
ok: [10.9.9.11] => (item={u'path': u'/tmp/MON', u'name': u'MON'}) => {
    "ansible_facts": {
        "allcmd": " ls -lrt /tmp/scripts | tail -57 && ls -lrt /tmp/MON | tail -57 &&"
    },
    "ansible_loop_var": "item",
    "changed": false,
    "item": {
        "name": "MON",
        "path": "/tmp/MON"
    }
}

&&循环完成后，我得到了所需的字符串，除了最后留下了尾随的事实ie "allcmd": " ls -lrt /tmp/scripts | tail -57 && ls -lrt /tmp/MON | tail -57 &&"。

我希望&&从 allcmd 变量中删除最后 3 个字符。期望的输出：

"allcmd": " ls -lrt /tmp/scripts | tail -57 && ls -lrt /tmp/MON | tail -57"

找不到任何过滤器或功能来删除 ansible 中的最后 n 个字符。

你能建议一下吗？

Answer 1

there is a much easier way outlined in this link

http://www.freekb.net/Article?id=2884

  debug: 
    msg: "{{ 'Hello World'[:-3] }}"