PostgreSQL排名和相關

Question

我有一個表，它存儲不同類別的userids和他們的點。用戶數量將不斷變化。點也將不斷變化（如在計算器中的點）。因此，登錄用戶將看到一個儀表盤，其中說明了3個類別中的每一個 - 您有850點，在您之前有950個用戶。這是我現在的查詢 -

WITH USERS AS (
    SELECT COUNT(*) TOT 
    FROM user_pointS 
) 
SELECT ' You have ' || points_cat1 ||' points and there are '|| tot-rnk || ' ahead of you ' 
FROM (
    SELECT ID, 
      points_cat1, 
      rank() OVER (ORDER BY points_cat1 DESC) AS RNK 
    FROM user_pointS 
) AS RANKED, 
    USERS 
WHERE ID = 10 

有沒有更好的方法（性能方面）？我將不得不重複這3列？

Answer 1

好了，你可以做到這一點，而不CTE：

SELECT ' You have ' || points_cat1 ||' points and there are '|| tot-rnk || ' ahead of you ' 
FROM (SELECT ID, points_cat1, 
       rank() OVER (ORDER BY points_cat1 DESC) AS RNK , 
       count(*) over() as TOT 
     FROM user_pointS 
    ) RANKED 
WHERE ID = 10; 

您可以一次做到這一點對所有三個名作：

SELECT ' You have ' || points_cat1 ||' points and there are '|| tot-rnk1 || ' ahead of you ', 
     ' You have ' || points_cat2 ||' points and there are '|| tot-rnk2 || ' ahead of you ', 
     ' You have ' || points_cat3 ||' points and there are '|| tot-rnk3 || ' ahead of you ' 

FROM (SELECT ID, points_cat1, points_cat2, points_cat3, 
       rank() OVER (ORDER BY points_cat1 DESC) AS RNK , 
       rank() OVER (ORDER BY points_cat2 DESC) AS RNK1 , 
       rank() OVER (ORDER BY points_cat3 DESC) AS RNK2 , 
       count(*) over() as TOT 
     FROM user_pointS 
    ) RANKED 
WHERE ID = 10; 

你或許可以取代tot-rnk具有反向排序：

rank() OVER (ORDER BY points_cat1 ASC) AS RNK 

但你想測試一下，以確保它給你的結果你期望。