如何将 JSON 数组中的值解析为 BigQuery 中的列

Posted 2023-03-25

【中文标题】如何将 JSON 数组中的值解析为 BigQuery 中的列

【英文标题】：How do I parse value from JSON array into columns in BigQuery

【发布时间】：2020-09-04 03:28:34

【问题描述】：

我有一个类似的 JSON 数组

"key":"Email","slug":"customer-email","value":"[e-mail]"
"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"

我希望将数组变成这样的列

email|          phoneNumber |  firstName |  lastName |  dob
[e-mail]   123456789      abc          xyz         01/01/1990

任何指南或意见将不胜感激。

【问题讨论】：

它在 BigQuery 表中吗？如果是这样 - 这是一个重复的字符串还是实际上是一个代表 json 数组的字符串？请澄清

对不起，不够清晰。是的，这是在 BigQuery 中，这是一个重复的字符串。

【参考方案1】：

以下是 BigQuery 标准 SQL

#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   

您可以使用您问题中的示例数据进行测试，如以下示例所示

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 1 id, [
    '"key":"Email","slug":"customer-email","value":"[e-mail]"',
    '"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   

有输出

Row id  Email           PhoneNumber     FirstName   LastName    Dateofbirth  
1   1   [e-mail]   123456789       abc         xyz         01/01/1990