[SQL]511+512+534+550+569

511. 游戏玩法分析 I

solution

SELECT player_id, MIN(event_date) AS first_login
FROM Activity
GROUP BY player_id;

512. 游戏玩法分析 II

SELECT player_id, device_id
FROM Activity 
WHERE (player_id, event_date) IN (SELECT player_id, MIN(event_date)
                                  FROM Activity
                                    GROUP BY player_id);

HAVING不行的原因

having子句执行在select 之后, 因此having中的字段必须在select子句中, event_date没有再select子句里,所以不行

534. 游戏玩法分析 III

SELECT a1.player_id, a1.event_date, SUM(a2.games_played) AS games_played_so_far
FROM Activity a1, Activity a2
WHERE a1.player_id = a2.player_id
AND a1.event_date >= a2.event_date
GROUP BY a1.player_id, a1.event_date --这一定要有a1.event_date，否则Result会根据player_id自动合并
ORDER BY a1.player_id, a1.event_date;

另一种方法：

SELECT player_id, event_date, 
    CASE WHEN @prev = player_id THEN @cnt := @cnt + games_played 
        WHEN @prev := player_id THEN @cnt := games_played 
    END 'games_played_so_far'
FROM (SELECT player_id, event_date, games_played 
    FROM activity 
    ORDER BY player_id, event_date) a, 
    (SELECT @cnt := 0, @prev := null) t;

550. 游戏玩法分析 IV

方法一

SELECT ROUND(COUNT(DISTINCT player_id)/(SELECT COUNT(DISTINCT player_id)
                                        FROM Activity), 2) AS fraction
FROM Activity
WHERE (player_id, event_date) IN(SELECT player_id, DATE(MIN(event_date)+1)
                                 FROM Activity
                                 GROUP BY player_id);

方法二

SELECT ROUND(SUM(CASE WHEN DATEDIFF(a.event_date, b.first_date)=1 THEN 1 ELSE 0 END) / (SELECT COUNT(DISTINCT player_id)
FROM Activity), 2) AS fraction
FROM Activity a, --千万不要漏掉这个逗号！！
(SELECT player_id, MIN(event_date) AS first_date
FROM Activity
GROUP BY player_id) b
WHERE a.player_id = b.player_id;

569. 员工薪水中位数

SELECT b.id,b.company,b.salary
-- 3. 连接结果
FROM (
    -- 1. 按 company 分组排序，记为 `rk`
    SELECT id,company,salary,
    CASE @com WHEN company THEN @rk:=@rk+1 ELSE @rk:=1 END rk,
    @com:=company
    FROM employee,(SELECT @rk:=0, @com:='') a
    ORDER BY company,salary) b
LEFT JOIN 
    (-- 2. 计算各 company 的记录数除以2，记为 `cnt`
    SELECT company,COUNT(1)/2 cnt FROM employee GROUP BY company) c
ON b.company=c.company
-- 4. 找出符合中位数要求的记录
WHERE b.rk in (cnt+0.5,cnt+1,cnt);