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Accepted
Jedi6
Asked: 2024-09-04 01:24:19 +0800 CST

Ansible：更新字典列表中的值

  • 772

我有以下词典列表：

"errorlist": [
    {
        "error": "Not found",
        "path": "/tmp/working/directory1/file1"
    },
    {
        "error": "Not found",
        "path": "/tmp/working/directory2/file1"
    },
    {
        "error": "Not found",
        "path": "/tmp/working/directory1/file2"
    },
    {
        "error": "Not found",
        "path": "/tmp/working/directory2/file2"
    }
]

我想删除路径值中的“/tmp/working/”部分，即将绝对路径转换为相对路径。

我在网上搜索时也遇到过类似的问题，但没有找到解决方案。应该很简单，但我就是想不明白。

预期结果应该是具有更新值的原始列表或具有相同结构的新列表。

ansible

2 个回答

  1. Ansible 并不是一个出色的数据处理工具。对于您想要执行的操作，最简单的解决方案可能是用 Python 编写的自定义过滤器。将以下内容放入filter_plugins/filters.py（filter_plugins目录与您的剧本位于同一目录中）：

    def replaceInAttribute(data, attr, find, replace):
    for d in data:
        if attr in d:
            d[attr] = d[attr].replace(find, replace)

    return data


class FilterModule:
    def filters(self):
        return {
            "replaceInAttribute": replaceInAttribute,
        }

    现在你可以编写如下剧本：

    - hosts: localhost
  gather_facts: false
  vars:
    errorlist: [
        {
            "error": "Not found",
            "path": "/tmp/working/directory1/file1"
        },
        {
            "error": "Not found",
            "path": "/tmp/working/directory2/file1"
        },
        {
            "error": "Not found",
            "path": "/tmp/working/directory1/file2"
        },
        {
            "error": "Not found",
            "path": "/tmp/working/directory2/file2"
        }
    ]
  tasks:
  - debug:
      msg: "{{ errorlist | replaceInAttribute('path', '/tmp/working/', '') }}"

    并得到以下输出：

    TASK [debug] *************************************************************************************************************************************************************
ok: [localhost] => {
    "msg": [
        {
            "error": "Not found",
            "path": "directory1/file1"
        },
        {
            "error": "Not found",
            "path": "directory2/file1"
        },
        {
            "error": "Not found",
            "path": "directory1/file2"
        },
        {
            "error": "Not found",
            "path": "directory2/file2"
        }
    ]
}

    这里介绍的过滤器比它需要的稍微复杂一些，因为它支持对任意属性的查找/替换。

    • 1
  2. Best Answer

    过滤基本名称并创建字典列表

      paths: "{{ errorlist | map(attribute='path') | map('basename') |
                         map('community.general.dict_kv', 'path') }}"

    给出

      paths:
    - {path: file1}
    - {path: file1}
    - {path: file2}
    - {path: file2}

    压缩列表并合并项目

      result: "{{ errorlist | zip(paths) | map('combine') }}"

    给出

      result:
    - {error: Not found, path: file1}
    - {error: Not found, path: file1}
    - {error: Not found, path: file2}
    - {error: Not found, path: file2}

    根据您的需要安装管道。

    • 完整测试剧本的示例
    - hosts: localhost

  vars:

    errorlist:
      - {error: Not found, path: /tmp/working/directory1/file1}
      - {error: Not found, path: /tmp/working/directory2/file1}
      - {error: Not found, path: /tmp/working/directory1/file2}
      - {error: Not found, path: /tmp/working/directory2/file2}

    paths: "{{ errorlist | map(attribute='path') | map('basename') |
                           map('community.general.dict_kv', 'path') }}"
    result: "{{ errorlist | zip(paths) | map('combine') }}"

  tasks:

    - debug:
        var: paths | to_yaml
    - debug:
        var: result | to_yaml
    • 根据您的需要调整过滤器。例如，要删除字符串的任何部分，请使用过滤器regex_replace
      paths: "{{ errorlist | map(attribute='path') |
                         map('regex_replace', '/tmp/working/(.*)', '\\1') |
                         map('community.general.dict_kv', 'path') }}"

    给出

      paths:
    - {path: directory1/file1}
    - {path: directory2/file1}
    - {path: directory1/file2}
    - {path: directory2/file2}

    这将导致

      result:
    - {error: Not found, path: directory1/file1}
    - {error: Not found, path: directory2/file1}
    - {error: Not found, path: directory1/file2}
    - {error: Not found, path: directory2/file2}
    • 1

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