将多个对象合并到一个对象下，并在一个命令中添加同级对象

Question

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"
    }
  ]
}

我想将它们合并到一个“动物”对象下。我发现这会合并文件：

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"
  ]
}

我想合并到同一个文件中

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

我可以只用一个jq 命令完成这两项操作吗？

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

结果如下所示：

{
  "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"
  ]
}

Answer 1

假设您有一个（先验的）不确定或大量类型的动物，并且只有一个 owners 文件。在这种情况下，最好不要使用-s 选项（以节省内存），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，例如大致如下：

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

其中 program.jq 包含一个程序，例如：

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

（注意{"owners": $owners} 是如何缩写的。）

Answer 2

不确定这是否是可行的方法，但它通过以下方式获得所需的输出：

• 使用--slurp:
• 将前 3 个文件作为单个数组变量捕获
  [ .[0] * .[1] * .[2] ] as $all
• 将owners 对象作为单个变量捕获
  .[3].owners as $owners
• 根据需要创建对象
  { "animals": $all, "owners": $owners }

jq \
    --slurp \
    '[ .[0] * .[1] * .[2] ] as $all | .[3].owners as $owners | { "animals": $all, "owners": $owners }' cats.json dogs.json snakes.json owners.json

将产生：

{
  "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"
  ]
}