json - 使用jq将多个JSON的一个属性连接到一个文件中

Question

我有多个格式相似的 JSON 文件，这里有两个例子:

message_1.json

{
  "participants": [
    {
      "name": "Person One"
    },
    {
      "name": "Person Two"
    }
  ],

  "messages": [
    {
      "sender_name": "Person One",
      "timestamp_ms": 0002,
      "content": "Text2.",
      "type": "Generic"
    },
    {
      "sender_name": "Person Two",
      "timestamp_ms": 0001,
      "content": "Text1.",
      "type": "Generic"
    }
  ],
  "title": "Person One",
  "is_still_participant": true,
  "thread_type": "Regular",
  "thread_path": "inbox/SomeString"
}

message_2.json

{
  "participants": [
    {
      "name": "Person One"
    },
    {
      "name": "Person Two"
    }
  ],

  "messages": [
    {
      "sender_name": "Person Two",
      "timestamp_ms": 0004,
      "content": "Text4.",
      "type": "Generic"
    },
    {
      "sender_name": "Person One",
      "timestamp_ms": 0003,
      "content": "Text3.",
      "type": "Generic"
    }
  ],
  "title": "Person One",
  "is_still_participant": true,
  "thread_type": "Regular",
  "thread_path": "inbox/SomeString"
}

有没有一种方法可以使用 jq 来合并 JSON 文件，以便连接 messages 属性(顺序无关紧要)而其他的则单独保留？

合并 message_1.json 和 message_2.json 的结果如下所示:

消息.json

{
  "participants": [
    {
      "name": "Person One"
    },
    {
      "name": "Person Two"
    }
  ],

  "messages": [
    {
      "sender_name": "Person One",
      "timestamp_ms": 0002,
      "content": "Text2.",
      "type": "Generic"
    },
    {
      "sender_name": "Person Two",
      "timestamp_ms": 0001,
      "content": "Text1.",
      "type": "Generic"
    },
    {
      "sender_name": "Person Two",
      "timestamp_ms": 0004,
      "content": "Text4.",
      "type": "Generic"
    },
    {
      "sender_name": "Person One",
      "timestamp_ms": 0003,
      "content": "Text3.",
      "type": "Generic"
    }
  ],
  "title": "Person One",
  "is_still_participant": true,
  "thread_type": "Regular",
  "thread_path": "inbox/SomeString"
}

我有 11 个 JSON 文件，message_1.json，...，message_11.json。我想将它们全部合并到一个 messages.json 文件中，该文件包含 JSON 文件中的所有消息。我如何通过 bash 使用 jq 来做到这一点？

Answer 1

这是一种方法，它的优点是不需要 -s 选项，这会消耗比必要更多的内存:

jq 'reduce inputs as $in (.;
  .messages += $in.messages)
' $(for i in $(seq 1 11); do echo message_$i.json ; done)

请注意，在这种情况下，调用 jq 时应不带 -n 选项。