新列的多个 BigQuery 子选择

Question

我有一张表，其中包含企业名称与邮政编码的一对多关系，因为多个行业代码与给定邮政编码的该企业相匹配。一个单独的表格包含按邮政编码的家庭。为了将邮政编码作为行和给定企业作为一列的家庭相加，同时对多行中匹配相同邮政编码的家庭进行重复数据删除（以避免过度计算家庭），我查询

SELECT ZIPCode, SUM(SumHouseholds1) AS Company1  
FROM (  
    SELECT ZIPCode, SUM(Households) OVER (PARTITION BY ZIPCode, DBAName) AS SumHouseholds1  
    FROM Business  
    JOIN Location  
    ON Location.ZIPCode = Business.ZIPCode  
    WHERE DBAName='Company1'  
GROUP BY DBAName, ZIPCode, Households)  
GROUP BY ZIPCode  

这样的输出：

---

邮政编码公司1
10001 17007
10003 54084

当我尝试向原始 SELECT 语句添加其他列（Company2、Company3 等）时：

SELECT ZIPCode, SUM(SumHouseholds1) AS Company1  
FROM (  
    SELECT ZIPCode, SUM(Households) OVER (PARTITION BY ZIPCode, DBAName) AS SumHouseholds1  
    FROM Business  
    JOIN Location  
    ON Location.ZIPCode = Business.ZIPCode  
    WHERE DBAName='Company1'  
GROUP BY DBAName, ZIPCode, Households),  
SUM(SumHouseholds2) AS Company2  
FROM (  
    SELECT ZIPCode, SUM(Households) OVER (PARTITION BY ZIPCode, DBAName) AS SumHouseholds2  
    FROM Business  
    JOIN Location  
    ON Location.ZIPCode = Business.ZIPCode  
    WHERE DBAName='Company2'  
GROUP BY DBAName, ZIPCode, Households)
GROUP BY ZIPCode 

我遇到了一个“遇到”“FROM”“FROM”“错误。

Answer 1

好的，假设您的初始代码确实适合您 - 下面将解决第二个查询的问题

SELECT 
  c1.ZIPCode AS ZIPCode, 
  c1.Company1 AS Company1, 
  c2.Company2 AS Company2, 
  c3.Company3 AS Company3
FROM (
  SELECT ZIPCode, SUM(SumHouseholds1) AS Company1  
  FROM (  
      SELECT ZIPCode, SUM(Households) OVER (PARTITION BY ZIPCode, DBAName) AS SumHouseholds1  
      FROM Business  
      JOIN Location  
      ON Location.ZIPCode = Business.ZIPCode  
      WHERE DBAName='Company1'  
  GROUP BY DBAName, ZIPCode, Households)
) AS c1
JOIN (
  SELECT ZIPCode, SUM(SumHouseholds2) AS Company2  
  FROM (  
      SELECT ZIPCode, SUM(Households) OVER (PARTITION BY ZIPCode, DBAName) AS SumHouseholds2  
      FROM Business  
      JOIN Location  
      ON Location.ZIPCode = Business.ZIPCode  
      WHERE DBAName='Company2'  
  GROUP BY DBAName, ZIPCode, Households)
) AS c2
ON c1.ZIPCode = c2.ZIPCode
JOIN (
  SELECT ZIPCode, SUM(SumHouseholds3) AS Company3  
  FROM (  
      SELECT ZIPCode, SUM(Households) OVER (PARTITION BY ZIPCode, DBAName) AS SumHouseholds3  
      FROM Business  
      JOIN Location  
      ON Location.ZIPCode = Business.ZIPCode  
      WHERE DBAName='Company3'  
  GROUP BY DBAName, ZIPCode, Households)
) AS c3
ON c1.ZIPCode = c3.ZIPCode

但是即使现在它可以工作（我希望因为我根本没有测试过它）它太重且难以管理
下面解决了这个问题（仍然没有经过测试，但应该可以工作，至少应该给你一个想法）

SELECT
  ZIPCode,
  SUM(CASE WHEN DBAName='Company1' THEN Company ELSE 0 END) AS Company1,
  SUM(CASE WHEN DBAName='Company2' THEN Company ELSE 0 END) AS Company2,
  SUM(CASE WHEN DBAName='Company3' THEN Company ELSE 0 END) AS Company3
FROM (
  SELECT ZIPCode, DBAName, SUM(SumHouseholds1) AS Company
  FROM (  
      SELECT ZIPCode, SUM(Households) OVER (PARTITION BY ZIPCode, DBAName) AS SumHouseholds  
      FROM Business  
      JOIN Location  
      ON Location.ZIPCode = Business.ZIPCode  
  GROUP BY DBAName, ZIPCode, Households)
)
GROUP BY ZIPCode

2016 年 7 月 12 日更新，基于 cmets 中的更多信息

SELECT
 ZIPCode,
 SUM(CASE WHEN DBAName='Company1' THEN Company ELSE 0 END) AS Company1,
 SUM(CASE WHEN DBAName='Company2' THEN Company ELSE 0 END) AS Company2,
 SUM(CASE WHEN DBAName='Company3' THEN Company ELSE 0 END) AS Company3
FROM (
 SELECT ZIPCode, DBAName, SUM(SumHouseholds) AS Company
 FROM (  
    SELECT ZIPCode, DBAName, SUM(Households) OVER (PARTITION BY ZIPCode, DBAName) AS SumHouseholds  
    FROM Business  
    JOIN Location  
    ON Location.ZIPCode = Business.Market  
    GROUP BY DBAName, ZIPCode, Households
 )
 GROUP BY DBAName, ZIPCode
)
GROUP BY ZIPCode

输出是

ZIPCode Company1    Company2    Company3     
10001    5           5           5   
10016    8           8           8   
12345   17          17          17   
16420   10           0           0   

进一步的想法

上面的“修复”仍然完全依赖于假设你的逻辑是正确的。

我确实有填充它不是：

我认为以下调整使其更正：

首先 - 按家庭分组看起来非常可疑，在查看您的笔记后，我认为您需要在下面

SELECT
  ZIPCode,
  SUM(CASE WHEN DBAName='Company1' THEN Company ELSE 0 END) AS Company1,
  SUM(CASE WHEN DBAName='Company2' THEN Company ELSE 0 END) AS Company2,
  SUM(CASE WHEN DBAName='Company3' THEN Company ELSE 0 END) AS Company3
FROM (  
  SELECT ZIPCode, DBAName, SUM(Households) AS Company  
  FROM (
    SELECT Market, DBAName 
    FROM AS Business 
    GROUP BY Market, DBAName
  ) AS Business
  JOIN Location  
  ON Location.ZIPCode = Business.Market  
  GROUP BY DBAName, ZIPCode
)
GROUP BY ZIPCode

这反过来 - 可以进一步简化为

SELECT
  ZIPCode,
  SUM(CASE WHEN DBAName='Company1' THEN Households ELSE 0 END) AS Company1,
  SUM(CASE WHEN DBAName='Company2' THEN Households ELSE 0 END) AS Company2,
  SUM(CASE WHEN DBAName='Company3' THEN Households ELSE 0 END) AS Company3
FROM (  
  SELECT Market, DBAName 
  FROM Business 
  GROUP BY Market, DBAName
) AS Business
JOIN Location  
ON Location.ZIPCode = Business.Market  
GROUP BY ZIPCode

不知何故，我的感觉 - 最后一个查询就是您要查找的内容！

但它仍然是一个选项，我只是不知道您真实的一些细节 - 很可能更复杂 - 用例，所以在这种情况下，您的原始逻辑可能是正确的