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社区首页 >问答首页 >如何将JSON数组中的值解析为BigQuery中的列

如何将JSON数组中的值解析为BigQuery中的列
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Stack Overflow用户
提问于 2020-09-04 11:28:35
回答 1查看 534关注 0票数 1

我有一个类似于下面的JSON数组

代码语言:javascript
运行
复制
{"key":"Email","slug":"customer-email","value":"abc@gmail.com"}
{"key":"Phone Number","slug":"mobile-phone-number","value":"123456789"}
{"key":"First Name","slug":"first-name","value":"abc"}
{"key":"Last Name","slug":"last-name","value":"xyz"}
{"key":"Date of birth","slug":"date-of-birth","value":"01/01/1990"}

我希望将数组变成如下所示的列

代码语言:javascript
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email|          phoneNumber |  firstName |  lastName |  dob
abc@gmail.com   123456789      abc          xyz         01/01/1990

我们将非常感谢您提供的任何指南或意见。

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2020-09-04 12:09:57

下面是针对BigQuery标准SQL的说明

代码语言:javascript
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#standardSQL
SELECT id, 
  MAX(IF(key = 'Email', value, NULL)) AS Email,
  MAX(IF(key = 'PhoneNumber', value, NULL)) AS PhoneNumber,
  MAX(IF(key = 'FirstName', value, NULL)) AS FirstName,
  MAX(IF(key = 'LastName', value, NULL)) AS LastName,
  MAX(IF(key = 'Dateofbirth', value, NULL)) AS Dateofbirth
FROM `project.dataset.table`,
UNNEST(ARRAY(
    SELECT AS STRUCT 
      REPLACE(JSON_EXTRACT_SCALAR(json, '$.key'), ' ', '') AS key,
      JSON_EXTRACT_SCALAR(json, '$.value') AS value
    FROM UNNEST(json_array) json
))
GROUP BY id   

您可以使用问题中的样本数据进行测试,如下面的示例所示

代码语言:javascript
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复制
#standardSQL
WITH `project.dataset.table` AS (
  SELECT 1 id, [
    '{"key":"Email","slug":"customer-email","value":"abc@gmail.com"}',
    '{"key":"Phone Number","slug":"mobile-phone-number","value":"123456789"}',
    '{"key":"First Name","slug":"first-name","value":"abc"}',
    '{"key":"Last Name","slug":"last-name","value":"xyz"}',
    '{"key":"Date of birth","slug":"date-of-birth","value":"01/01/1990"}'
  ] json_array
)
SELECT id, 
  MAX(IF(key = 'Email', value, NULL)) AS Email,
  MAX(IF(key = 'PhoneNumber', value, NULL)) AS PhoneNumber,
  MAX(IF(key = 'FirstName', value, NULL)) AS FirstName,
  MAX(IF(key = 'LastName', value, NULL)) AS LastName,
  MAX(IF(key = 'Dateofbirth', value, NULL)) AS Dateofbirth
FROM `project.dataset.table`,
UNNEST(ARRAY(
    SELECT AS STRUCT 
      REPLACE(JSON_EXTRACT_SCALAR(json, '$.key'), ' ', '') AS key,
      JSON_EXTRACT_SCALAR(json, '$.value') AS value
    FROM UNNEST(json_array) json
))
GROUP BY id   

带输出

代码语言:javascript
运行
复制
Row id  Email           PhoneNumber     FirstName   LastName    Dateofbirth  
1   1   abc@gmail.com   123456789       abc         xyz         01/01/1990  
票数 1
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/63734411

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