Question

刚开始使用Jolt，我尝试转换以下输入：

{
    "task_name": "Data Full Load",
    "table": "enterprise",
    "schema": "dbo",
    "op": "F",
    "data": [
        {
            "enterprise_id": "00001"
        },
        {
            "enterprise_name": "FOO"
        },
        {
            "note": ""
        },
        {
            "delete_ind": "N"
        },
        {
            "create_timestamp": "2004-11-03 22:28:08.180"
        },
        {
            "modify_timestamp": "2016-05-02 13:02:53.437"
        },
        {
            "row_timestamp": "0x000000001CDCAAC3"
        }
    ]
}

进入此输出：

{
    "enterprise_id": "00001",
    "enterprise_name": "FOO",
    "note": "",
    "delete_ind": "N",
    "create_timestamp": "2004-11-03 22:28:08.180",
    "modify_timestamp": "2016-05-02 13:02:53.437",
    "row_timestamp": "0x000000001CDCAAC3"
}

我的数据将包含各种未知元素，因此我无法按名称对其进行转换。我能够使用此规范来提取数据数组，但我不确定如何组合元素

[
  {
    "operation": "shift",
    "spec": { "data": "" }
}
]

当前输出：

[ {
  "enterprise_id" : "00001"
}, {
  "enterprise_name" : "FOO"
}, {
  "note" : ""
}, {
  "delete_ind" : "N"
}, {
  "create_timestamp" : "2004-11-03 22:28:08.180"
}, {
  "modify_timestamp" : "2016-05-02 13:02:53.437"
}, {
  "row_timestamp" : "0x000000001CDCAAC3"
} ]

Answer 1

在jolt-demo.appspot.com上发生一点争吵之后，我明白了。对于那些遇到相同挑战的人来说，这是规范：

<a>a> //take a> away

如何使用Jolt展平n个对象的json数组？

1 个答案: