从 Ansible 中的字符串中删除最后 n 个字符

Question

下面是我构造字符串的剧本。

   - 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，除了我在末尾留下了尾随 &&，即 "allcmd": " ls -lrt /tmp/scripts | tail -57 && ls -lrt /tmp/MON | tail -57 &&"。

我想从 allcmd 变量中删除最后 3 个字符，即 &&。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函数来删除 ansible 中的最后 n 个字符。

你能推荐一下吗？

Answer 1

- hosts: localhost
  vars:
    foo:
      - {"name": "SCRIPT", "path": "/tmp/scripts"}
      - {"name": "MON", "path": "/tmp/MON"}
  tasks:
    - debug:
        var: foo

    - set_fact:
        cmds: "{{ [ 'ls -lrt ' + item.path + ' | tail -57' ] + cmds | default([]) }}"
      loop: "{{ foo }}"

    - set_fact:
        allcmd: "{{ cmds | join(' && ')}}"

    - debug:
        var: allcmd

输出：

ok: [localhost] => {
    "allcmd": "ls -lrt /tmp/MON | tail -57 && ls -lrt /tmp/scripts | tail -57"
}

Answer 2

link

中概述了一种更简单的方法

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

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

从 Ansible 中的字符串中删除最后 n 个字符

Remove last n charecters from a string in Ansible

formatting

trim

filter

ansible

ansible-facts