計算訪問者排名在一個頁面上高於另一個頁面的情況

Question

我想計算在/page_y中的排名高於在page_x中排名的數字fullvisitorID 。 所以在這種情況下，結果將是 1，只有111

全訪問者 ID	秩	頁
111	1	/page_x
111	2	/page_y
222	1	/page_x
222	2	/page_x
333	2	/page_x
333	1	/page_y

Answer 1

考慮以下方法

select count(*) from (
  select distinct fullvisitorID
  from your_table
  qualify max(if(page='/page_y',rank,null)) over win > max(if(page='/page_x',rank,null)) over win
  window win as (partition by fullvisitorID)
)

Answer 2

對於計數，您可以使用COUNT和GROUP BY

SELECT fullvisitorID, COUNT(fullvisitorID), Page FROM table t1
WHERE rank = (SELECT MAX(t2.rank) FROM table t2 WHERE t2.fullvisitorID = t1.fullvisitorID)
Group By fullvisitorID, Page 

Answer 3

您可以通過匹配“ fullvisitorID ”字段在兩個表之間應用SELF JOIN ，然后強制

• 第一個具有“ page_y ”值的表
• 第二個表有“ page_x ”值
• 第一個表的排名具有第二個表的更高排名

SELECT *
FROM       tab t1
INNER JOIN tab t2
        ON t1.fullvisitorID = t2.fullvisitorID
       AND t1.page = '/page_y'
       AND t2.page = '/page_x'
       AND t1.rank > t2.rank

Answer 4

表分離方法：

DECLARE @t1 TABLE ( fullvisitorID INT, [rank] INTEGER,[page] VARCHAR (max)) --here where page = x
DECLARE @t2 TABLE ( fullvisitorID INT, [rank] INTEGER,[page] VARCHAR (max)) --here where page = y 
INSERT INTO @t1 SELECT * FROM @test t WHERE t.[page] LIKE '/page_x'
 INSERT INTO @t2 SELECT * FROM @test t WHERE t.[page] LIKE '/page_y'
          
SELECT COUNT(*) FROM @t1 INNER JOIN @t2 ON [@t1].fullvisitorID = [@t2].fullvisitorID WHERE [@t1].rank < [@t2].rank

Answer 5

SELECT COUNTIF(page = '/page_y') cnt FROM ( SELECT * FROM sample_table WHERE page IN ('/page_x', '/page_y') QUALIFY ROW_NUMBER() OVER (PARTITION BY fullvisitorID ORDER BY rank DESC) = 1 );