编写Jolt转换规格，将含数组字段的JSON转为多对象数组

输入示例 #

{
  "KUNNR": "123456789",
  "NAME1": "Milchwerke \"Mittelelbe\" GmbH",
  "NAME2": "Milchwerke \"Mittelelbe\" GmbH",
  "NAME3": "Milchwerke \"Mittelelbe\" GmbH",
  "NAME4": "Milchwerke \"Mittelelbe\" GmbH",
  "LAND1": "DE",
  "VKORG": "1255",
  "VTWEG": "00",
  "SPART": "00",
  "PARVW": [
    "WE",
    "WE"
  ],
  "PARZA": [
    "000",
    "000"
  ],
  "KUNN2": [
    "0000045959",
    "0000045959"
  ]
}

期望输出示例 #

[
  {
    "KUNNR": "123456789",
    "NAME1": "Milchwerke \"Mittelelbe\" GmbH",
    "NAME2": "Milchwerke \"Mittelelbe\" GmbH",
    "NAME3": "Milchwerke \"Mittelelbe\" GmbH",
    "NAME4": "Milchwerke \"Mittelelbe\" GmbH",
    "LAND1": "DE",
    "VKORG": "1255",
    "VTWEG": "00",
    "SPART": "00",
    "PARVW": "WE",
    "PARZA": "000",
    "KUNN2": "0000045959"
  },
  {
    "KUNNR": "123456789",
    "NAME1": "Milchwerke \"Mittelelbe\" GmbH",
    "NAME2": "Milchwerke \"Mittelelbe\" GmbH",
    "NAME3": "Milchwerke \"Mittelelbe\" GmbH",
    "NAME4": "Milchwerke \"Mittelelbe\" GmbH",
    "LAND1": "DE",
    "VKORG": "1255",
    "VTWEG": "00",
    "SPART": "00",
    "PARVW": "WE",
    "PARZA": "000",
    "KUNN2": "0000045959"
  }
]

正确的Jolt转换规格 #

[
  {
    "operation": "shift",
    "spec": {
      "KUNNR|NAME1|NAME2|NAME3|NAME4|LAND1|VKORG|VTWEG|SPART": "common.&",
      "PARVW": {
        "*": {
          "$": "&1.PARVW"
        }
      },
      "PARZA": {
        "*": {
          "$": "&1.PARZA"
        }
      },
      "KUNN2": {
        "*": {
          "$": "&1.KUNN2"
        }
      }
    }
  },
  {
    "operation": "shift",
    "spec": {
      "*": {
        "@(2,common)": "&",
        "*": "&"
      },
      "common": ""
    }
  },
  {
    "operation": "shift",
    "spec": {
      "*": "[]"
    }
  }
]

转换逻辑说明 #

1. 第一步Shift操作：

   • 将所有非数组字段归类到common临时节点，避免重复处理。
   • 按数组索引把PARVW、PARZA、KUNN2的元素分别存入对应下标的节点中（如0.PARVW、1.KUNN2）。

2. 第二步Shift操作：

   • 遍历每个索引节点，将common中的所有字段复制到当前节点，与该节点的数组元素字段合并。
   • 删除临时的common节点，清理冗余数据。

3. 第三步Shift操作：

   • 将所有索引节点转换为根数组的元素，生成最终的数组结构。

内容的提问来源于stack exchange，提问作者Thiyagaraj Narayanan