如何使用Jolt将嵌套数据转换为线性数据

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英文:

How to convert nested data into linear data using jolt

问题

如何使用jolt将这个嵌套数据转换为如下所示的线性格式数据？需要为所有嵌套数据创建单独的条目。每条记录应包含以下5个数据：practice_loc，prac_num，topId，S1和S2。

我必须编写一个jolt来进行转换，下面是数据的所有可能情况：

数据

第1种情况：

输入：

  1. {
  2.   "practice_loc": 120,
  3.   "prac_num": 234,
  4.   "topId": "t1",
  5.   "subList": [
  6.     {
  7.       "S1": "A1",
  8.       "S2": "B1"
  9.     },
  10.     {
  11.       "S1": "A2"
  12.     }
  13.   ]
  14. }

转换后：

  1. {
  2.     "practice_loc": 120,
  3.     "prac_num": 234,
  4.     "topId": "t1",
  5.     "S1": "A1",
  6.     "S2": "B1"
  7.   },
  8.   {
  9.     "practice_loc": 120,
  10.     "prac_num": 234,
  11.     "topId": "t1",
  12.     "S1": "A2"
  13.  }

第2种情况可以是：

输入：

  1. {
  2.   "practice_loc": 987,
  3.   "prac_num": 232,
  4.   "topId": "artica",
  5.   "subList": [
  6.     {
  7.       "S1": "A5",
  8.       "S2": "B7"
  9.     }
  10.   ]
  11. }

转换后：

  1. {
  2.   "practice_loc": 987,
  3.   "prac_num": 232,
  4.   "topId": "artica",
  5.   "S1": "A5",
  6.   "S2": "B7"
  7. }

第3种情况可以是：

输入：

  1. {
  2.   "practice_loc": 334,
  3.   "prac_num": 233,
  4.   "topId": "plumcherry",
  5.   "subList": [
  6.     {
  7.       "S1": "A3"
  8.     }
  9.   ]
  10. }

转换后：

  1. {
  2.   "practice_loc": 334,
  3.   "prac_num": 233,
  4.   "topId": "plumcherry",
  5.   "S1": "A3"
  6. }

第4种情况可以是：

输入：

  1. {
  2.   "practice_loc": 987,
  3.   "prac_num": 232,
  4.   "topId": "rose",
  5.   "subList": [
  6.     {}
  7.   ]
  8. }

转换后：

  1. {
  2.   "practice_loc": 987,
  3.   "prac_num": 232,
  4.   "topId": "rose"
  5. }

我该如何编写一个jolt，可以处理所有这些情况，并返回转换后的数据，无论输入是哪种类型的？

英文:

How can I convert this nested data to linear format data as mentioned below using jolt. Need to create separate entry for all the nested data. every record should have 5 data in it practice_loc,prac_num,topId,S1 and S2.

I have to write one jolt for transforming, data below are all possibilities of data

Data

1st case ;

Input :

  1. {
  2.   "practice_loc": 120,
  3.   "prac_num": 234,
  4.   "topId": "t1",
  5.   "subList": [
  6.     {
  7.       "S1": "A1",
  8.       "S2": "B1"
  9.     },
  10.     {
  11.       "S1": "A2"
  12.     }
  13.   ]
  14. }

After transformation:

  1. {
  2.     "practice_loc": 120,
  3.     "prac_num": 234,
  4.     "topId": "t1",
  5.     "S1": "A1",
  6.     "S2": "B1"
  7.   },
  8.   {
  9.     "practice_loc": 120,
  10.     "prac_num": 234,
  11.     "topId": "t1",
  12.     "S1": "A2"
  13.  }

2nd case can be;

Input:

  1. {
  2.   "practice_loc": 987,
  3.   "prac_num": 232,
  4.   "topId": "artica",
  5.   "subList": [
  6.     {
  7.       "S1": "A5",
  8.       "S2": "B7"
  9.     }
  10.   ]
  11. }

After transformation :

  1. {
  2.   "practice_loc": 987,
  3.   "prac_num": 232,
  4.   "topId": "artica",
  5.   "S1": "A5",
  6.   "S2": "B7"
  7. }

3rd case can be;

Input :

  1. {
  2.   "practice_loc": 334,
  3.   "prac_num": 233,
  4.   "topId": "plumcherry",
  5.   "subList": [
  6.     {
  7.       "S1": "A3"
  8.     }
  9.   ]
  10. }

After transformation :

  1. {
  2.   "practice_loc": 334,
  3.   "prac_num": 233,
  4.   "topId": "plumcherry",
  5.   "S1": "A3"
  6. }

4th case can be ;

Input :

  1. {
  2.   "practice_loc": 987,
  3.   "prac_num": 232,
  4.   "topId": "rose",
  5.   "subList": [
  6.     {}
  7.   ]
  8. }

After transformation :

  1. {
  2.   "practice_loc": 987,
  3.   "prac_num": 232,
  4.   "topId": "rose"
  5. }

How can I write only one jolt which can cover all these cases and return the transformed data with whichever type of input it get

答案1

得分: 2

以下是翻译好的内容：

您可以使用以下转换

  1. [
  2.   { // 通过上一级对象包装节点对元素进行分组
  3.     "operation": "shift",
  4.     "spec": {
  5.       "subList": {
  6.         "*": {
  7.           "@2|@": "&3_&1" // 通过@2选择来自相同级别的“sublist”的外部元素，同时@返回该数组的值
  8.         }
  9.       }
  10.     }
  11.   },
  12.   { // 摆脱包装
  13.     "operation": "shift",
  14.     "spec": {
  15.       "*": {
  16.         "*": {
  17.           "*": "[#3].&"
  18.         }
  19.       }
  20.     }
  21.   },
  22.   { // 摆脱额外生成的数组，即“subList”
  23.     "operation": "remove",
  24.     "spec": {
  25.       "*": {
  26.         "subList": ""
  27.       }
  28.     }
  29.   }
  30. ]
英文:

You can use the following transformation

  1. [
  2.   { // group elements by upper level objects wrapper nodes
  3.     "operation": "shift",
  4.     "spec": {
  5.       "subList": {
  6.         "*": {
  7.           "@2|@": "&3_&1" // pick the outer elements as well from the same level of "sublist" through use of @2 while @ returns the value from this array
  8.         }
  9.       }
  10.     }
  11.   },
  12.   { // get rid of the wrappers
  13.     "operation": "shift",
  14.     "spec": {
  15.       "*": {
  16.         "*": {
  17.           "*": "[#3].&"
  18.         }
  19.       }
  20.     }
  21.   },
  22.   { // get rid of the extra generated array, namely "subList"
  23.     "operation": "remove",
  24.     "spec": {
  25.       "*": {
  26.         "subList": ""
  27.       }
  28.     }
  29.   }
  30. ]

huangapple
  • 本文由 发表于 2023年6月8日 16:16:10
  • 转载请务必保留本文链接：https://go.coder-hub.com/76429897.html
  • apache-nifi
  • jolt
  • json
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