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我有两个不同的结构化 json 文件从雪管中输入。唯一的区别是它有许多嵌套数组，而不是嵌套字典。我试图弄清楚如何将结构 1转换为一个最终表格。我已成功将结构 2转换为表格并包含以下代码。

我知道我需要使用横向展平，但没有成功。

**Structure 1: Nested Arrays (Need help on)**
This json lives within a table and in column **JSONTEXT**
[
  {
    "ID": "xxx-xxxx-xxxx xxx-xxx",
    "caseTypeID": "xx-xxxx-xxxx-xxxxx",
    "content": {
      "AccountID": "xx-xxxxx-xxxx-xxxx xxxx-xxxxx",
      "AccountName": "XXXX",
      "Address": {
        "pxObjClass": "Data-Address-Postal"
      },
      "Addresses": [],
      "AllKickoffsComplete": "true",
      "BillingContactList": [],
      "ClientCurrency": "USD",
      "ClientID": "XXXXXX",
      "ClientNSID": "XXXXXXXX-00",
      "ClientName": "XXXXX XXXX Inc.",
      "CompanyPhoneNumber": "XXX-XXX-XXXX",
      "CrmSearchOrg": "XXXX",
      "EEList": [
        {
          "AccountID": "xxx-xxxxx-xxxx-xxxxx xxxx-xxxxx",
          "AccountName": "XXXX",
          "AllowanceList": [
            {
              "AllowanceAmount": "327",
              "AllowanceName": "Car Allowance",
              "pxObjClass": "xxxxx-xxxxx-xxxxx"
]

结构 2：嵌套字典 此 json 存在于表和列JSONTEXT中

[
  {
    "OppID": "xxxx-xxxxx",
    "pxObjClass": "xx-xxxxx-xxxx-xxxxxx",
    "pxPages": {
      "EEList": {
        "Country": "xxx",
        "CountryName": "xxx",
        "Currency": "xxx",
        "EstimatedICPCost": "xxxxxxxxxxx",
        "ICPCurrency": "xxxxx",
        "ICPID": "xxxxxxxxx.",
        "ICPNSID": "xxxx-xx",
        "ICPName": "xxx xx xx.",
        "LocalMonthlySalary": "xxxxxx",
        "MinFee": "xxxx",
        "MonthlyGrossCost": "xxxxx",
        "NewOrRepeatCustomer": "xxxxx",
        "OppCloseDate": "xxx-xxx-xx",
        "OppID": "xxx-xxxx",
        "OpportunityName": "xxx - xxx xxx - xxx - xxxx",
        "ReferralSource": "xxxxxx",
        "pxObjClass": "Index-xx-xxxx-xxxx-xxxxxx",
        "pxSubscript": "EEList"
      }
    },
    "pyID": "xxxxxx",
    "pzInsKey": "xxxx-xxxx-xxxx xxxxx-xxx"
  },
]

这是我的第二个有效结构的代码。

create or replace table xxxx
    as select 
    value:ID::varchar as ID,
    value:caseTypeID::varchar as caseTypeID,
    value:content:AccountID::varchar as AccountID,
    value:content:AccountName::varchar as AccountName,
    value:content:AllKickoffsComplete::boolean as AllKickoffsComplete,
    value:content:ClientCurrency::varchar as ClientCurrency,
    value:content:ClientID::varchar as ClientID,
    value:content:ClientNSID::varchar as ClientNSID,
    value:content:ClientName::varchar as ClientName,
    value:content:CompanyAddressCountryName::varchar as CompanyAddressCountryName,
    value:content:CompanyPhoneNumber::varchar as CompanyPhoneNumber,
    value:content:CreateNew::boolean as CreateNew,
    value:content:CrmSearchOrg::varchar as CrmSearchOrg,
    value:content:EEList:AccountID::varchar as EE_AccountID,
    value:content:EEList:AccountName::varchar as EE_AccountName
from new_raw_json, 
    lateral flatten (input =>jsontext);

这是我尝试过的代码，它仅在您放置 jsontext[Nth] 时才有效。

select
    value:ID::varchar as ID,
    value:EEListID::varchar as EEListID,
    value:caseTypeID::varchar as caseTypeID
    from new_raw_json,

    lateral flatten (input => jsontext[0]:content:EEList);

感谢任何帮助！

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1 回答 1

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您可以使用 FLATTEN 链接多个横向视图以继续分解为嵌套结构（数组中的数组）。

明确定义的方法可能会以这种方式出现（此处仅投影了一些列，以说明级别）：

SELECT
  outer_object.value:caseTypeID AS caseTypeID,
  outer_object.value:content.AccountID AS parentAccountID,
  eelist_object.value:AccountID AS eeListAccountID,
  allowance_object.value:AllowanceName
FROM
  new_raw_json,
  LATERAL FLATTEN (input => jsontext) outer_object,
  LATERAL FLATTEN (input => outer_object.value:content.EEList) eelist_object,
  LATERAL FLATTEN (input => eelist_object.value:AllowanceList) allowance_object;

请注意，这只会分解一个已识别的多值路径 ( List -> EEList -> AllowanceList)。从问题中不清楚是否必须分解所有路径（例如List -> EEList -> Addresses AND AllowanceList），或者是否可以将其中一些路径存储为VARIANT（或其他复杂的）类型在最终结果中。

例如，如果需要为underAllowanceList中列出的每个地址复制值，这可以通过执行来自两个爆炸查询结果（一个是连锁的，另一个是连锁的）来实现。AddressesEEListJOINList -> AddressesList -> EEList -> AllowanceList

于 2020-02-15T19:07:13.050 回答