转换JSON-JSON JOLT

我是JOLT的新手，我需要将我的JSON文件转换为所需的模式。这是我的输入转换JSON-JSON JOLT

[ 
    { 
     "PK": 12345, 
     "FULL_NAME":"Amit Prakash", 
     "BIRTHDATE":"1987-05-25", 
     "SEX":"M", 
     "EMAIL": "[email protected]", 
     "PHONE": "809386731", 
     "TS":"2015-11-19 14:36:34.0" 
    }, 
    { 
     "PK": 12654, 
     "FULL_NAME": "Rohit Dhand", 
     "BIRTHDATE":"1979-02-01", 
     "SEX":"M", 
     "EMAIL": "[email protected]", 
     "PHONE": "937013861", 
     "TS":"2015-11-20 11:03:02.6" 
    }, 
    ... 
]

，这是我想要的输出：

{ 
    "records": [ 
     { 
      "attribs": [{ 
       "type": "customer", 
       "reference": "CUST" 
      }], 
      "name": "Amit Prakash", 
      "personal_email": "[email protected]", 
      "mobile": "809386731", 
      "id": 12345 
     }, 
     { 
      "attribs": [{ 
       "type": "customer", 
       "reference": "CUST" 
      }], 
      "name": "Rohit Dhand", 
      "personal_email": "[email protected]", 
      "mobile": "937013861", 
      "id": 12654 
     }, 
     ... 
    ] 
}

到目前为止，我只设法到这一点：

[ 
    { 
    "operation": "remove", 
    "spec": { 
     "*": { 
     "BIRTHDATE": "", 
     "SEX": "", 
     "TS": "" 
     } 
    } 
    }, 
    { 
    "operation": "shift", 
    "spec": { 
     "*": "records" 
    } 
    } 
]

但我不能去从这里开始。我不知道如何重命名输出中的键。

另外，有什么可以替代删除操作？删除如果您排除的键数少于包含的键，那么操作是很好的，但反过来如何（包含几个键，多于在JSON对象内排除）？

来源

2017-05-30 menorah84

规格

[ 
    { 
    "operation": "shift", 
    "spec": { 
     "*": { 
     "PK": "records[&1].id", 
     "PHONE": "records[&1].mobile", 
     "EMAIL": "records[&1].personal_email", 
     "FULL_NAME": "records[&1].name" 
     } 
    } 
    }, 
    { 
    "operation": "default", 
    "spec": { 
     "records[]": { 
     "*": { 
      "attribs[]": { 
      "0": { 
       "type": "customer", 
       "reference": "CUST" 
      } 
      } 
     } 
     } 
    } 
    } 
]

移，使您的数据副本，而其他业务则没有。因此，删除东西的一种方法就是不要将它复制到您的换档规格中。

一般来说，删除是用来摆脱那些会“搅乱”这种转变的东西。

来源

2017-05-31 13:46:44

转换JSON-JSON JOLT

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