Question

以下结果集源自具有少量联接和联合的SQL查询。 sql查询已经在日期和游戏上对行进行分组。我需要一个列来描述按日期列分区的游戏尝试次数。

Username   Game     ID   Date

johndoe1   Game_1   100  7/22/14 1:52 AM
johndoe1   Game_1   100  7/22/14 1:52 AM
johndoe1   Game_1   100  7/22/14 1:52 AM
johndoe1   Game_1   100  7/22/14 1:52 AM
johndoe1   Game_1   121  7/22/14 1:56 AM
johndoe1   Game_1   121  7/22/14 1:56 AM
johndoe1   Game_1   121  7/22/14 1:56 AM
johndoe1   Game_1   121  7/22/14 1:56 AM
johndoe1   Game_1   121  7/22/14 1:56 AM
johndoe1   Game_1   130  7/22/14 1:59 AM
johndoe1   Game_1   130  7/22/14 1:59 AM
johndoe1   Game_1   130  7/22/14 1:59 AM
johndoe1   Game_1   130  7/22/14 1:59 AM
johndoe1   Game_1   130  7/22/14 1:59 AM
johndoe1   Game_1   200  7/22/14 2:54 AM
johndoe1   Game_1   200  7/22/14 2:54 AM
johndoe1   Game_1   200  7/22/14 2:54 AM
johndoe1   Game_1   200  7/22/14 2:54 AM
johndoe1   Game_1   210  7/22/14 3:54 AM
johndoe1   Game_1   210  7/22/14 3:54 AM
johndoe1   Game_1   210  7/22/14 3:54 AM
johndoe1   Game_1   210  7/22/14 3:54 AM

我有以下sql查询枚举分区内的行但不完全正确，因为我想根据日期和游戏计算该游戏的实例。在这种情况下，johndoe1已在Game_1尝试五次按时间戳划分。

此查询返回

下面的结果集

select *
, row_number() over (partition by ct."date" order by ct."date") as "Attempts"
from csv_temp as ct

Username   Game     ID   Date             Attempts  (Desired Attempts col.)

johndoe1   Game_1   100  7/22/14 1:52 AM  1          1
johndoe1   Game_1   100  7/22/14 1:52 AM  2          1
johndoe1   Game_1   100  7/22/14 1:52 AM  3          1
johndoe1   Game_1   100  7/22/14 1:52 AM  4          1
johndoe1   Game_1   121  7/22/14 1:56 AM  1          2
johndoe1   Game_1   121  7/22/14 1:56 AM  2          2
johndoe1   Game_1   121  7/22/14 1:56 AM  3          2
johndoe1   Game_1   121  7/22/14 1:56 AM  4          2
johndoe1   Game_1   121  7/22/14 1:56 AM  5          2
johndoe1   Game_1   130  7/22/14 1:59 AM  1          3   
johndoe1   Game_1   130  7/22/14 1:59 AM  2          3
johndoe1   Game_1   130  7/22/14 1:59 AM  3          3
johndoe1   Game_1   130  7/22/14 1:59 AM  4          3
johndoe1   Game_1   130  7/22/14 1:59 AM  5          3
johndoe1   Game_1   200  7/22/14 2:54 AM  1          4
johndoe1   Game_1   200  7/22/14 2:54 AM  2          4
johndoe1   Game_1   200  7/22/14 2:54 AM  3          4
johndoe1   Game_1   200  7/22/14 2:54 AM  4          4
johndoe1   Game_1   210  7/22/14 3:54 AM  1          5
johndoe1   Game_1   210  7/22/14 3:54 AM  2          5
johndoe1   Game_1   210  7/22/14 3:54 AM  3          5
johndoe1   Game_1   210  7/22/14 3:54 AM  4          5

任何指针都会有很大的帮助。

Answer 1

考虑partition by与group by的字段相似，然后，当分区值发生变化时，窗口函数将重新开始1

EDIT 如a_horse_with_no_name所示，为此需要dense_rank() 与row_number() rank()或dense_rank()不同，重复它指定的数字。对于分区中的每一行，row_number()必须是不同的值。 rank()和dense_rank()之间的差异是后者不会“跳过”数字。

对于您的查询，请尝试：

dense_rank() over (partition by Username, Game order by ct."date") as "Attempts"

顺便说一下，你不要按相同的字段进行分区和排序;如果需要，只需订购即可。它不在这里。

PostgreSQL窗口函数：row_number（）over（分区col顺序col2）

1 个答案: