SQL选择由

Question

周

分组顶部SUM我如何将需要更改以下SQLite3的查询来获取只是按星期分组最高总计行：

SELECT 
strftime('%W', date(p.match_date, 'unixepoch', 'localtime')) AS week_number, 
sum(red_cards + yellow_cards) AS cards, user_id 
FROM user_records 
GROUP BY week_number, user_id 
ORDER BY week_number, cards DESC 

上面的查询返回以下结果：

week_number - cards - user_id 
44   5  1 
44   1  2 
45   2  2 
45   1  1 

我想只显示顶行，而每星期：

week_number - cards - user_id 
44   5  1 
45   2  2 

有一些技巧来调整QUER ❖删除不必要的额外行？

Answer 1

尝试使用NOT EXISTS()：

SELECT s.* FROM ( 
    SELECT 
     strftime('%W', date(p.match_date, 'unixepoch', 'localtime')) AS week_number, 
     sum(red_cards + yellow_cards) AS cards, user_id 
    FROM user_records 
    GROUP BY week_number, user_id) s 
WHERE NOT EXISTS(SELECT 1 FROM user_records p 
       WHERE strftime('%W', date(p.match_date, 'unixepoch', 'localtime')) = s.week_number 
       GROUP BY p.user_id 
       HAVING SUM(red_cards + yellow_cards) > s.cards) 

Answer 2

首先，使用CTE。然后简单JOIN或WHERE条款会做：

WITH w as (
     SELECT strftime('%W', date(p.match_date, 'unixepoch', 'localtime')) AS week_number, 
      SUM(red_cards + yellow_cards) AS cards, user_id 
     FROM user_records 
     GROUP BY week_number, user_id 
    ) 
SELECT w.* 
FROM w 
WHERE w.cards = (SELECT MAX(w2.cards) 
       FROM w w2 
       WHERE w2.week_number = w.week_number 
       ) 
ORDER BY week_number; 

注意，这将返回多行如果有多个行有某一周相同最大值。

Answer 3

在SQLite的3.7.11或更高版本，你可以用MAX（）返回一组与最高值的行：

SELECT week_number, 
     max(cards) AS cards, 
     user_id 
FROM (SELECT strftime('%W', date(p.match_date, 'unixepoch', 'localtime')) AS week_number, 
      sum(red_cards + yellow_cards) AS cards, 
      user_id 
     FROM user_records 
     GROUP BY week_number, user_id) 
GROUP BY week_number 
ORDER BY week_number;