Question

我有一张包含跟踪数据的表格。除其他值外，该表还包含traffic_medium，traffic_source和traffic_campaign列。列有时包含（none）或null作为值。

我想使用左边的连接，中间，scource和广告系列作为匹配条件来匹配来自其他桌子的访问者总数。

如果所有列都包含数据，则此方法正常。如果一列有（无）或null为值，则它不起作用。

我使用BigQuery和旧版SQL。

SELECT  
A.id,
A.trafficSource_medium,
A.trafficSource_source,
A.trafficSource_campaign,
B.sum_visitor AS sum_visitor

FROM [table] AS A
left outer join (Select 
count(distinct fullvisitorID) as sum_visitor,
trafficSource_medium,
trafficSource_source,
trafficSource_campaign
FROM [table2] 
GROUP BY trafficSource_medium,
trafficSource_source,
trafficSource_campaign)
AS B
on A.trafficSource_medium=B.trafficSource_medium AND     
A.trafficSource_source=B.trafficSource_source AND 
A.trafficSource_campaign=B.trafficSource_campaign

感谢您的帮助！

Answer 1

尝试下面的内容假设各个字段是STRING类型。如果它们是INT - 将'n / a'替换为让我们说-999 - 重要的是选择不用作相应字段值的常量

#legacySQL
SELECT  
  A.id,
  CASE WHEN A.trafficSource_medium = 'n/a' THEN NULL ELSE A.trafficSource_medium END AS trafficSource_medium,
  CASE WHEN A.trafficSource_source = 'n/a' THEN NULL ELSE A.trafficSource_source END AS trafficSource_source,
  CASE WHEN A.trafficSource_campaign = 'n/a' THEN NULL ELSE A.trafficSource_campaign END AS trafficSource_campaign,
  B.sum_visitor AS sum_visitor
FROM (
  SELECT 
    id,
    IFNULL(trafficSource_medium, 'n/a') AS trafficSource_medium,
    IFNULL(trafficSource_source, 'n/a') AS trafficSource_source,
    IFNULL(trafficSource_campaign 'n/a') AS trafficSource_campaign
  FROM [table] 
) AS A
LEFT OUTER JOIN (
  SELECT 
    COUNT(DISTINCT fullvisitorID) AS sum_visitor,
    IFNULL(trafficSource_medium, 'n/a') AS trafficSource_medium,
    IFNULL(trafficSource_source, 'n/a') AS trafficSource_source,
    IFNULL(trafficSource_campaign 'n/a') AS trafficSource_campaign
  FROM [table2] 
  GROUP BY 
    trafficSource_medium,
    trafficSource_source,
    trafficSource_campaign
) AS B
ON A.trafficSource_medium = B.trafficSource_medium 
AND A.trafficSource_source = B.trafficSource_source 
AND A.trafficSource_campaign = B.trafficSource_campaign

这里的想法是将NULL“转换”为某个值，因此它们是JOIN'able - 然后在最终的SELECT中将其“转换”为NULL

如果你可以迁移到标准SQL - 你可以尝试下面的代码 - 它做的改动较少 - 主要是在ON子句中

#standardSQL
SELECT  
  A.id,
  A.trafficSource_medium,
  A.trafficSource_source,
  A.trafficSource_campaign,
  B.sum_visitor AS sum_visitor
FROM `table` AS A
LEFT OUTER JOIN (
  SELECT 
    COUNT(DISTINCT fullvisitorID) AS sum_visitor,
    trafficSource_medium,
    trafficSource_source,
    trafficSource_campaign
  FROM `table2`
  GROUP BY 
    trafficSource_medium,
    trafficSource_source,
    trafficSource_campaign
) AS B
ON IFNULL(A.trafficSource_medium, 'n/a') = IFNULL(B.trafficSource_medium, 'n/a') 
AND IFNULL(A.trafficSource_source, 'n/a') = IFNULL(B.trafficSource_source, 'n/a') 
AND IFNULL(A.trafficSource_campaign, 'n/a') = IFNULL(B.trafficSource_campaign, 'n/a')

左连接有多个标准，这些标准部分为空或为空

1 个答案: