mysql - SQL:在同一个表中查找重复计数、添加的新值和删除的值(动态)

Question

我希望使用 SQL 完成以下目标:

1) 查找重复记录数
根据“快照日期”列提取重复值的数量，并将其与上一个日期进行比较
2) 查找添加的记录数
3) 查找删除的记录数

请参阅下面的示例表格:

当前表

snapshot_date | unique ID
 2018-08-15        1
 2018-08-15        2
 2018-08-15        3
 2018-08-15        4
 2018-08-15        5

 2018-08-16        1
 2018-08-16        3
 2018-08-16        4
 2018-08-16        6
 2018-08-16        7
 2018-08-16        8
 2018-08-16        9

 2018-08-17        3
 2018-08-17        8
 2018-08-17        10
 2018-08-17        11
 2018-08-17        12
 2018-08-17        13

需要的表

snapshot date | count | # of dupe from previous date | sum of ID added | sum of ID removed
 2018-08-15       5                 N/A                     N/A                  N/A 
 2018-08-16       7                  3                       4                    2
 2018-08-17       6                  2                       4                    5

如果有人知道到达所需餐 table 的脚本，我将不胜感激!提前谢谢你们!

Answer 1

这是我的看法。基于SQL Server

SELECT  snapshot_date       = COALESCE(c.snapshot_date, DATEADD(day, 1, p.snapshot_date)),
        [count]             = COUNT(c.snapshot_date),
        dup_from_prev_day   = SUM(CASE WHEN c.snapshot_date is not null 
                                       AND  p.snapshot_date is not null 
                                       THEN 1 END),
        sum_of_id_added     = SUM(CASE WHEN c.snapshot_date is not null 
                                       AND  p.snapshot_date is null 
                                       THEN 1 END),
        sum_of_id_removed   = SUM(CASE WHEN c.snapshot_date is null 
                                       AND  p.snapshot_date is not null 
                                       THEN 1 END)
FROM    yourTable c         -- current
        FULL OUTER JOIN yourTable p -- previous
        ON  c.snapshot_date     = DATEADD(DAY, 1, p.snapshot_date)
        AND c.uniqueID          = p.uniqueID
GROUP BY COALESCE(c.snapshot_date, DATEADD(DAY, 1, p.snapshot_date))
HAVING COUNT(c.snapshot_date) > 0

/* RESULT : 
snapshot_date  count  dup_from_prev_day  sum_of_id_added  sum_of_id_removed
2018-08-15     5      NULL               5                NULL
2018-08-16     7      3                  4                2
2018-08-17     6      2                  4                5
*/