Question

我想计算包含一些客户标记的已查看页面的数量，我的代码目前返回所有3个计数的相同值，这些值太高，有人可以建议问题在这里吗？

SELECT
  Date,
  COUNT(CASE WHEN hits.page.pagepath LIKE "?aff=" THEN 1 ELSE 0 END) AS Affinity,
  COUNT(CASE WHEN hits.page.pagepath LIKE "?aff2=" THEN 1 ELSE 0 END) AS Addon_Affinity,
  COUNT(CASE WHEN hits.page.pagepath LIKE "?aff3=" THEN 1 ELSE 0 END) AS Style_Affinity

FROM `[tablename].ga_sessions_*` AS t
  CROSS JOIN UNNEST(hits) AS hits
WHERE (_TABLE_SUFFIX BETWEEN '20170703' AND '20170703')
group by date

Answer 1

如果输入不为空，COUNT会将其总和增加1。您应该使用SUM或COUNTIF：

SELECT
  Date,
  COUNTIF(hits.page.pagepath LIKE "?aff=") AS Affinity,
  COUNTIF(hits.page.pagepath LIKE "?aff2=") AS Addon_Affinity,
  COUNTIF(hits.page.pagepath LIKE "?aff3=") AS Style_Affinity
FROM `[tablename].ga_sessions_*` AS t
  CROSS JOIN UNNEST(hits) AS hits
WHERE (_TABLE_SUFFIX BETWEEN '20170703' AND '20170703')
group by date

Answer 2

其他可能性是解决此查询，如：

common

其中WITH data AS( select "20170701" as date, ARRAY<STRUCT<page STRUCT<pagePath STRING> >> [STRUCT(STRUCT('?aff=' as pagePath) as page), STRUCT(STRUCT('?aff=' as pagePath) as page), STRUCT(STRUCT('/test' as pagePath) as page), STRUCT(STRUCT('?aff2=' as pagePath) as page), STRUCT(STRUCT('?aff=' as pagePath) as page)] hits union all select "20170701" as date, ARRAY<STRUCT<page STRUCT<pagePath STRING> >> [STRUCT(STRUCT('?aff2=' as pagePath) as page), STRUCT(STRUCT('?aff2=' as pagePath) as page), STRUCT(STRUCT('?aff3=' as pagePath) as page), STRUCT(STRUCT('?aff3=' as pagePath) as page), STRUCT(STRUCT('?aff=' as pagePath) as page)] hits ) SELECT date, sum((select countif(hits.page.pagepath like "?aff=") from unnest(hits) hits)) Affinity, sum((select countif(hits.page.pagepath like "?aff2=") from unnest(hits) hits)) Addon_Affinity, sum((select countif(hits.page.pagepath like "?aff3=") from unnest(hits) hits)) Style_Affinity FROM data GROUP BY date是您data数据的模拟。

计算包含精确字符串的已查看页面的数量

2 个答案: