如何刪除空的 object 標簽

Question

給定以下 JSON：

{
      "identifers": {
        "_id": "f073895e207c3157",
        "attrsDigest": "-706912700",
        "entityTypes": [
          {
            "_id": "bacf9e903a6c5def",
            "attrsDigest": "1537152651"
          },
          {
            "_id": "08aac63dacd579c8",
            "attrsDigest": "-1461027456"
          }
        ],
        "statuses": [
          {
            "_id": "50912252d4781056",
            "attrsDigest": "154590869"
          }
        ]
      }
    }

我需要制作以下內容：

{
  "identifers": [
    {
      "_id": "f073895e207c3157",
      "attrsDigest": "-706912700"
    },
    {
        "_id": "bacf9e903a6c5def",
        "attrsDigest": "1537152651"
    },
    {
      "_id": "08aac63dacd579c8",
      "attrsDigest": "-1461027456"
    },
    {
      "_id": "50912252d4781056",
      "attrsDigest": "154590869"
    }
    ]
  }

我有以下內容：

[**].${ "identifers": _id, "digest": attrsDigest }

產生這個：

{
  "identifers": [
    "f073895e207c3157",
    "bacf9e903a6c5def",
    "08aac63dacd579c8",
    "50912252d4781056"
  ],
  "digest": [
    "-706912700",
    "1537152651",
    "-1461027456",
    "154590869"
  ]
}

Answer 1

您可以使用“reduce”運算符按 ID 對嵌套數據進行分組，然后將其格式化為所需形狀的數組：

**{ _id: attrsDigest }
  ~> $each(function($attrsDigest, $id) {
    { "_id": $id, "attrsDigest": $attrsDigest }
  })

直播游樂場： https://stedi.link/A1yo9Pz