如何使用Jolt将JSON数组规范化为属性键值结构？

示例源JSON #

[
  {
    "itemNo": "ITEM001",
    "name": "笔记本电脑",
    "price": 5999,
    "brand": "XX品牌"
  },
  {
    "itemNo": "ITEM002",
    "category": "办公耗材",
    "stock": 100
  }
]

期望输出JSON #

[
  {
    "id": "ITEM001",
    "attribute": "itemNo",
    "value": "ITEM001"
  },
  {
    "id": "ITEM001",
    "attribute": "name",
    "value": "笔记本电脑"
  },
  {
    "id": "ITEM001",
    "attribute": "price",
    "value": 5999
  },
  {
    "id": "ITEM001",
    "attribute": "brand",
    "value": "XX品牌"
  },
  {
    "id": "ITEM002",
    "attribute": "itemNo",
    "value": "ITEM002"
  },
  {
    "id": "ITEM002",
    "attribute": "category",
    "value": "办公耗材"
  },
  {
    "id": "ITEM002",
    "attribute": "stock",
    "value": 100
  }
]

正确的Jolt Shift规范 #

[
  {
    "operation": "shift",
    "spec": {
      "*": {
        "*": {
          "@(2,itemNo)": "[#2].id",
          "$": "[#2].attribute",
          "@": "[#2].value"
        }
      }
    }
  },
  {
    "operation": "shift",
    "spec": {
      "*": "[]"
    }
  }
]

规范说明 #

1. 第一个Shift操作：

   • 外层*遍历源数组里的每个物品对象
   • 内层*遍历当前物品的所有属性
   • @(2,itemNo)：向上回溯2层拿到当前物品的itemNo值，作为新对象的id
   • $：取当前遍历的属性名，赋值给attribute字段
   • @：取当前属性的对应值，赋值给value字段
   • [#2]：用外层数组的索引做临时分组键，保证同一物品的属性对象归为一组

2. 第二个Shift操作：

   • 把分组后的多维数组平铺成一维数组，得到最终需要的结构

内容的提问来源于stack exchange，提问作者Katja Bürger