在一个命令中合并单个对象下的多个对象和添加同级对象 [英] Merge multiple objects under a single object and add sibling object in one command

问题描述

所以我有三个文件：

cats.json

{
  "cats": [
    {
      "name": "fluffles",
      "age": 10,
      "color": "white"
    }
  ]
}

dogs.json

{
  "dogs": [
    {
      "name": "sam",
      "age": 5,
      "color": "black and white"
    },
    {
      "name": "rover",
      "age": 2,
      "color": "brown and white"
    }
  ]
}

Snakes.json

{
  "snakes": [
    {
      "name": "noodles",
      "age": 10,
      "color": "green"
    }
  ]
}

我想将它们合并在一起，放在一个动物对象下。我发现这将合并文件：

jq -s '{"animals": .} ' cats.json dogs.json snakes.json > animals.json

{
  "animals": [
    {
      "cats": [
        {
          "name": "fluffles",
          "age": 10,
          "color": "white"
        }
      ]
    },
    {
      "dogs": [
        {
          "name": "sam",
          "age": 5,
          "color": "black and white"
        },
        {
          "name": "rover",
          "age": 2,
          "color": "brown and white"
        }
      ]
    },
    {
      "snakes": [
        {
          "name": "noodles",
          "age": 10,
          "color": "green"
        }
      ]
    }
  ]
}

现在我有一个额外的对象：

owners.json

{
  "owners": [
    "peter",
    "william",
    "sally"
  ]
}

我要使用

将其合并到同一文件中

jq -s '.[0] + .[1]' animals.json owners.json

我可以用一个jq命令同时执行这两个操作吗？

jq -s '{"animals": .} ' cats.json dogs.json snakes.json > animals.json
jq -s '.[0] + .[1]' animals.json owners.json

结果如下：

{
  "animals": [
    {
      "cats": [
        {
          "name": "fluffles",
          "age": 10,
          "color": "white"
        }
      ]
    },
    {
      "dogs": [
        {
          "name": "sam",
          "age": 5,
          "color": "black and white"
        },
        {
          "name": "rover",
          "age": 2,
          "color": "brown and white"
        }
      ]
    },
    {
      "snakes": [
        {
          "name": "noodles",
          "age": 10,
          "color": "green"
        }
      ]
    }
  ],
  "owners": [
    "peter",
    "william",
    "sally"
  ]
}

推荐答案

假设您有一个(先验的)不确定或大量的动物类型，并且只有一个owners文件。在这种情况下，不使用-s选项会更好(节省内存)，并且使用所有者文件作为第一个数据文件来调用JQ会更容易，例如

jq -n -f program.jq owners.json $(ls *.json | grep -v owners.json)

其中Program.jq包含程序，例如：

input as $owners | {$owners, animals: [inputs]}

(请注意{"owners": $owners}如何缩写。)

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