雪花 SQL 計數與 window function 不同

Question

我有一個包含每周客戶和商店信息的數據集，

問題-我必須計算截至本周為止的最后 1、2、3、4、5、6 ..so 周內的獨特客戶總數等特征。

在 window function 上使用不同客戶列的計數時出現錯誤-

我嘗試使用 concat function 創建數組，但也沒有成功 - 感謝您的幫助！

SELECT STORE,WEEK,

count(distinct Customers) over (partition by STORE order by WEEK rows between 1 preceding and 1 preceding) as last_1_week_customers,
count(distinct Customers) over (partition by STORE order by WEEK rows between 2 preceding and 2 preceding) as last_2_week_customers
    from TEST_TABLE
    group by STORE,WEEK

錯誤 - SQL 編譯錯誤： distinct 不能與 window 框架或訂單一起使用。

我該如何解決這個錯誤？

輸入

CREATE TABLE TEST_TABLE (STORE STRING,WEEK STRING,Customers STRING);


INSERT INTO TEST_TABLE VALUES

('A','1','AA'),
('A','1','DD'),
('A','2','AA'),
('A','2','BB'),
('A','2','CC'),
('A','3','AA'); 

Output

Answer 1

嗯...我認為您根本不需要 window 函數...

首先，我們可以從一個簡單的分組開始：

select
    store,
    week,
    count(distinct customers) as cnt
from
    test_table
where
    week >= [this week's number minus 5]
group by
   store, week

這將產生一個簡單的表：

店鋪	星期	cnt
一個	1	2
一個	2	3
一個	3	1

在這一點上，我會要求您考慮這是否已經足夠了。 可能您已經可以將這種格式的數據用於您需要的任何目的。 但如果不是，那么我們可以進一步修改它以獲得“樞軸”output。

在此查詢中，將${w}替換為本周的數字：

select
    store,
    count(distinct case when week=${w} then customers else null end) as cnt_now,
    count(distinct case when week=${w-1} then customers else null end) as cnt_minus_1,
    count(distinct case when week=${w-2} then customers else null end) as cnt_minus_2,
    count(distinct case when week=${w-3} then customers else null end) as cnt_minus_3,
    count(distinct case when week=${w-4} then customers else null end) as cnt_minus_4,
    count(distinct case when week=${w-5} then customers else null end) as cnt_minus_5
from
    test_table
where
    week >= {$w-5}
group by
   store

請記住 - COUNT()和COUNT(DISTINCT)只計算NON-NULL值。

店鋪	cnt_now	cnt_minus_1	cnt_minus_2	cnt_minus_3	cnt_minus_4	cnt_minus_5
一個	1	2	3	4	5	6
乙	9	8	7	6	5	4