Json 合并yaml文件并按键值列表的值重新组织

我有一个yaml文件列表，每个文件描述一个项目，其中包含一个代表可持续发展目标的数字列表

我希望合并所有文件并将它们转换为不同格式的json，sdg索引作为键，相关项目作为列表值

输入：

---
# gnu_health.yaml
description: >
  GNU Health is a Free/Libre project for health practitioners, health
  institutions and governments. It provides the functionality of Electronic
  Medical Record (EMR), Hospital Management (HMIS) and Health Information
  System (HIS).
sdgs: [3]
name: GNU Health 

---
# a11y.yaml
description: >
  This Accessibility Project is a community-driven effort to make web
  accessibility easier by leveraging a worldwide community of developer
  knowledge.
sdgs: [10]
name: A11Y

---
# bahmni.yaml
description: >
  Bahmni is an Open Source hospital Management System focusing
  on poor/underserved and public hospitals in the developing
  world.
  It's aimed to being a generic system which can be used for
  multiple diseases and hospitals in different countries.
sdgs: [1, 3]
name: Bahmni

预期产出：

{
  "1": [
    {
      "name": "Bahmni",
      "description: "..."
    }
  ],
  "3": [
    {
      "name": "GNU Health",
      "description: "..."
    },
    {
      "name": "Bahmni",
      "description: "..."
    }
  ],
  "10": [
    {
      "name: "A11Y",
      "description: "..."
    }
  ]
}

{
  "1": [
    {
      "name": "Bahmni",
      "description: "..."
    }
  ],
  "3": [
    {
      "name": "GNU Health",
      "description: "..."
    }
  ],
  "3": [
    {
      "name": "Bahmni",
      "description: "..."
    }
  ],
  "10": [
    {
      "name: "A11Y",
      "description: "..."
    }
  ]
}

我发现使用jq的过滤系统很难解决这个问题，即使在阅读了文章和其他资源之后

有人能给我指一下正确的方向吗

目前的最大努力：

# use as follow: yq -f $binDir/concat_sdgs.jq $srcDir/*.y*ml

# concat_sdgs.jq
{
  (.sdgs[]|tostring): [.]
}

不幸的是，这不会将来自同一sdg的项目合并在一起

当前输出不正确：

{
  "1": [
    {
      "name": "Bahmni",
      "description: "..."
    }
  ],
  "3": [
    {
      "name": "GNU Health",
      "description: "..."
    },
    {
      "name": "Bahmni",
      "description: "..."
    }
  ],
  "10": [
    {
      "name: "A11Y",
      "description: "..."
    }
  ]
}

{
  "1": [
    {
      "name": "Bahmni",
      "description: "..."
    }
  ],
  "3": [
    {
      "name": "GNU Health",
      "description: "..."
    }
  ],
  "3": [
    {
      "name": "Bahmni",
      "description: "..."
    }
  ],
  "10": [
    {
      "name: "A11Y",
      "description: "..."
    }
  ]
}

---

好消息是你已经接近了

为了简单起见，我将假设.yaml到.json的转换已经完成。稍微调整一下过滤器，很容易看出：

jq '{ (.sdgs[]|tostring): del(.sdgs) }' a11y.json gnu_health.json bahmni.json

生成一个由四个单关键点对象组成的流，这些对象与所需对象密切对应

将它们组合成单个对象有点棘手。为了简单起见，让我们首先定义一个helper函数，该函数可用于按键对单个键对象进行分组：

  def group_by_keys: reduce .[] as $o ({}; 
     reduce ($o | to_entries[]) as $kv (.; .[$kv.key]

接下来，我们将使用带有-n命令行选项的

输入

：

jq -n '
  def group_by_keys: reduce .[] as $o ({}; 
     reduce ($o | to_entries[]) as $kv (.; .[$kv.key] += [$kv.value]));
  [inputs | {(.sdgs[]|tostring): del(.sdgs) }] | group_by_keys

' a11y.json gnu_health.json bahmni.json

（不要忘记

-n

）

如果键的顺序很重要，则只需使用此过滤器：

def sort_by_keys:
  to_entries
  | sort_by(.key|tonumber)
  | from_entries;

---

谢谢你的回复。不幸的是，这似乎没有产生预期的结果。我没有为每个sdg键创建一个项目列表，而是将一个项目视为sdg的对象值。如果我可以为每个sdg连接项目，这将是完美的，我正在做-n。它仍然只在输出中列出每个sdg的一个项目。如果您尝试使用3个文件运行此操作，其中2个文件具有

sdg:[3]

，则只会返回一个项目抱歉，我已使用更好的示例更新了此问题：）让我们开始调试。

debug

是开始调试的好地方。我想你会发现花时间掌握

reduce

和

to_条目

过滤器家族是值得的。