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我想选择多个列并将电子邮件分组GROUP BY

#standardSQL
SELECT
      customers.orderCustomerEmail AS email,      
      customers.orderCustomerNumber AS customerNumber,
      customers.billingFirstname AS billingFirstname,
      customers.billingLastname AS billingLastname
FROM dim_customers AS customers
GROUP BY customers.orderCustomerEmail

失败：

Error: SELECT list expression references customers.orderCustomerNumber
       which is neither grouped nor aggregated at [4:7]

这类似于这个问题Bigquery select distinct values

但这并不能解决我的问题，因为将所有列添加到GROUP BY相同的结果不同SELECT DISTINCT

dim_customer 架构：

orderCustomerEmail:STRING,
billingFirstname:STRING,
billingLastname:STRING,
orderCustomerNumber:STRING,
OrderNumber:STRING

虚拟数据：https ://docs.google.com/spreadsheets/d/1T1JZRWni18hhU4tO-9kQqq5Y3hVWgpP-aE7o6ij9bDE/edit?usp=sharing

4

1 回答 1

3

当您按某些列分组时，您需要确保将一些聚合函数应用于其余列。否则你会得到你在问题中显示的错误

试试下面的 BigQuery 标准 SQL 示例 

#standardSQL
SELECT 
  customers.orderCustomerEmail AS email,      
  ARRAY_AGG(STRUCT(customers.orderCustomerNumber AS customerNumber,
  customers.billingFirstname AS billingFirstname,
  customers.billingLastname AS billingLastname)) AS info
FROM `dim_customers`, UNNEST(customers) AS customers
GROUP BY email

或者只是简单的 DISTINCT 

#standardSQL
SELECT DISTINCT 
  customers.orderCustomerEmail AS email,      
  customers.orderCustomerNumber AS customerNumber,
  customers.billingFirstname AS billingFirstname,
  customers.billingLastname AS billingLastname
FROM `dim_customers`, UNNEST(customers) AS customers

请注意：就您期望的确切输出而言，您的问题不够具体，因此上面很可能需要对您的特定需求进行一些调整

更新

我基本上需要每个客户一行（电子邮件是唯一标识符，因此是组）详细信息（号码，名字，姓氏）可以从最后一个条目中获取，例如 

#standardSQL
WITH `dim_customers` AS (
  SELECT [
    STRUCT('a' AS orderCustomerEmail, 1 AS orderCustomerNumber, 'af' AS billingFirstname, 'al' AS billingLastname),
    STRUCT('a' AS orderCustomerEmail, 4 AS orderCustomerNumber, 'af1' AS billingFirstname, 'al2' AS billingLastname),
    STRUCT('b' AS orderCustomerEmail, 2 AS orderCustomerNumber, 'bf' AS billingFirstname, 'bl' AS billingLastname),
    STRUCT('c' AS orderCustomerEmail, 3 AS orderCustomerNumber, 'cf' AS billingFirstname, 'cl' AS billingLastname)
    ] AS customers UNION ALL
  SELECT [
    STRUCT('a' AS orderCustomerEmail, 1 AS orderCustomerNumber, 'af' AS billingFirstname, 'al' AS billingLastname),
    STRUCT('a' AS orderCustomerEmail, 4 AS orderCustomerNumber, 'af1' AS billingFirstname, 'al2' AS billingLastname),
    STRUCT('b' AS orderCustomerEmail, 2 AS orderCustomerNumber, 'bf' AS billingFirstname, 'bl' AS billingLastname),
    STRUCT('c' AS orderCustomerEmail, 3 AS orderCustomerNumber, 'cf' AS billingFirstname, 'cl' AS billingLastname)
    ] AS customers
)
SELECT
  customers.orderCustomerEmail AS email,      
  ARRAY_AGG(STRUCT(customers.orderCustomerNumber AS customerNumber,
    customers.billingFirstname AS billingFirstname,
    customers.billingLastname AS billingLastname))[OFFSET(0)] AS info
FROM `dim_customers`, UNNEST(customers) AS customers
GROUP BY email

更新

以下是更新的架构！

dim_customer 架构：

orderCustomerEmail:STRING,
billingFirstname:STRING,
billingLastname:STRING,
orderCustomerNumber:STRING,
OrderNumber:STRING 

#standardSQL
WITH `dim_customers` AS (
  SELECT 10201 AS orderCustomerNumber, 'a@email.com' AS orderCustomerEmail, 'Alex' AS billingFirstname, 'Miller' AS billingLastname UNION ALL
  SELECT 10202, 'b@email.com', 'Ben', 'Williams' UNION ALL
  SELECT 10203, 'c@email.com', 'Chris', 'Collins' UNION ALL
  SELECT 10204, 'd@email.com', 'David', 'Hems' UNION ALL
  SELECT 10201, 'a@email.com', 'A.', 'Miller' UNION ALL
  SELECT 10201, 'a@email.com', 'A.', 'Miller' UNION ALL
  SELECT 10202, 'b@email.com', 'Ben', 'Williams' UNION ALL
  SELECT 10202, 'b@email.com', 'Bens Father', 'Williams' UNION ALL
  SELECT 10205, 'a@email.com', 'A.', 'Miller' UNION ALL
  SELECT 10206, 'e@email.com', 'Ed', 'Winchell'
)
SELECT info.* FROM (
  SELECT
    orderCustomerEmail AS email, 
    ARRAY_AGG(STRUCT(
      orderCustomerEmail AS email, 
      orderCustomerNumber AS customerNumber,
      billingFirstname AS billingFirstname,
      billingLastname AS billingLastname))[OFFSET(0)] AS info
  FROM `dim_customers`
  GROUP BY email
)
-- ORDER BY email
于 2017-07-26T19:00:01.483 回答