以下结果集源自具有一些连接和联合的 SQL 查询。 SQL 查询已经对 Date 和 game 上的行进行了分组。我需要一列来描述按日期列分区的游戏的尝试次数。
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,它枚举分区内的行,但并不完全正确,因为我希望根据日期和游戏计算该游戏的实例数。在本例中,johndoe1 已尝试 Game_1 按时间戳划分五次。
该查询返回以下结果集
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
任何指示都会有很大帮助。
考虑partition by
与您想要的字段相似group by
,然后,当分区值发生变化时,窗口函数从 1 重新开始
编辑
如 a_horse_with_no_name 所示,为此我们需要dense_rank()
unlike row_number()
rank()
or dense_rank()
重复它分配的数字。row_number()
分区中的每一行必须是不同的值。和...之间的不同rank()
and dense_rank()
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"
顺便说一句,您不按同一字段进行分区和排序;如果需要的话,只需按顺序就足够了。它不在这里。
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