使用Jolt转换复杂的Json数组

Question

当调用外部API时，我收到了一个复杂的json结构，我需要根据我们的需要将其转换为简单的json。我发现jolt具有转换json的能力，但无法提出jolt规范。

我的输入Json数组——

{
  "attribute": [
    "teamalloc",
    "prodAlloc"
  ],
  "item": {
    "id": "abcde",
    "name": "Champak Kumar",
    "allocDetails": {
      "updatedAt": "2020-08-10T14:26:48-07:00",
      "token": 1134,
      "items": [
        {
          "allocation": 0.2,
          "team": {
            "id": 90,
            "name": "Some Team Name 1",
            "createdAt": "2010-01-19T10:52:52-07:00"
          }
        },
        {
          "allocation": 0.9,
          "team": {
            "id": 80,
            "name": "Some Team Name 2",
            "createdAt": "2010-01-19T10:52:52-07:00",
            "product": {
              "id": 20,
              "name": "Some Product Name 1",
              "otherDetails": {
                "key": "Id",
                "value": "GEC"
              }
            }
          }
        },
        {
          "allocation": 0.1,
          "team": {
            "id": 10,
            "name": "Some Team Name 3",
            "createdAt": "2010-01-19T10:52:52-07:00",
            "product": {
              "id": 22,
              "name": "Some Product Name 2",
              "otherDetails": {
                "key": "Id1",
                "value": "GEC1"
              }
            }
          }
        }
      ]
    }
  }
}

我的输出 Json 结构应该看起来像 -

{
  "name": "Champak Kumar",
  "allocDetails": [
    {
      "allocation": 0.2,
      "team": {
        "id": 90,
        "name": "Some Team Name 1"
      }
    },
    {
      "allocation": 0.9,
      "team": {
        "id": 80,
        "name": "Some Team Name 2",
        "product": {
          "id": 20,
          "name": "Some Product Name 1"
        }
      }
    },
    {
      "allocation": 0.1,
      "team": {
        "id": 10,
        "name": "Some Team Name 3",
        "product": {
          "id": 22,
          "name": "Some Product Name 2"
        }
      }
    }
  ]
}

我尝试了多种jolt规格，但无法得到想要的输出。这种情况下最理想的jolt规格应该是什么？

Answer 1

工作震动规格——

  {
    "operation": "shift",
    "spec": {
      "item": {
        "name": "name",
        "allocDetails": {
          "items": {
            "*": {
              "allocation": "allocDetails[&1].allocation",
              "team": {
                "id": "allocDetails[&2].team.id",
                "name": "allocDetails[&2].team.name",
                "product": {
                  "id": "allocDetails[&3].team.product.id",
                  "name": "allocDetails[&3].team.product.name"
                }
              }
            }
          }
        }
      }
    }
  }
]```

Answer 2

这个规范应该使用移位操作应该起作用:

[
  {
    "operation": "shift",
    "spec": {
      "item": {
        "name": "name",
        "allocDetails": {
          "items": {
            "*": {
              "allocation": "allocDetails[&1].allocation",
              "team": {
                "id": "allocDetails[&2].team.id",
                "name": "allocDetails[&2].team.name",
                "product": "allocDetails[&2].team.product"
              }
            }
          }
        }
      }
    }
  }
]

编辑#1：

解释：shift 操作规范定义了我们希望将输入 json 中的值放在结果 json 中的位置。

1. 规范中的每个键值对都定义了源-目标关系。例如（让我们从最简单的开始）：来自["项目"]["名称"]的值将位于输出JSON中的["名称"]键下。
2. 的

 "items": {
    "*": {
      ...
    }
 }

部分说:“对于‘items’键下的数组的每个元素，执行< code >... "

...
"name": "allocDetails[&2].team.name"
...

说：“将”名称“键下的值（即在 item[”items“][56][”team“][”name“] 中放置在 ”allocDetails“ 键下的数组的第 57 个元素中。

“*”与第57个元素匹配

看看< code > shift operation javadocs ，尤其是在< code > "

编辑#2：考虑其他详细信息注释：您也可以像这样处理它：

...
    "team": {
        "id": "allocDetails[&2].team.id",
        "name": "allocDetails[&2].team.name",
        "product": {
            "otherDetails": null,
            "*": "allocDetails[&3].team.product.&"
        }
    }
...

以上：将产品的所有部件键（与“*”匹配）放入具有相同名称的键（”