MySQL 查询一天中房协访问量最高的时间

MySQL query to find hour of the day when the number of housing society visits is highest

房协到访数据

Id Contact Entry_time Exit_time Duration of Stay
1 8080808080 26/07/2021 08:00:05 26/07/2021 08:23:06 181
2 9692596925 26/07/2021 08:12:49 26/07/2021 08:14:44 115
3 7099270992 26/07/2021 11:02:49 26/07/2021 11:14:44 715
4 8900289002 26/07/2021 16:12:49 26/07/2021 16:14:44 115
5 9089590895 26/07/2021 15:12:49 26/07/2021 15:14:44 115
6 8765087650 26/07/2021 19:12:49 26/07/2021 19:14:44 115
7 7862178621 26/07/2021 18:12:49 26/07/2021 18:14:44 115

访问数据是多年可用的,可以包含数百万行,因此该解决方案的时间复杂度应该很低。
预期输出:8-9 AM(因为最高访问次数 (2) 是在该小时内进行的)。

当我阅读这个问题时,这将对您有很大帮助。 您可以做几件事来改善这一点:

  • 用小时数创建单独的 table 以避免联合(或用数字 table 来做);
  • 单独存储时间,TIME() 方法会减慢查询速度。

.

SELECT 
    hours.`hour`,
    COUNT(occurrence.perfId) AS occurrences
FROM (
    SELECT '00:00' AS `hour`
    UNION SELECT '01:00'
    UNION SELECT '02:00'
    UNION SELECT '03:00'
    UNION SELECT '04:00'
    UNION SELECT '05:00'
    UNION SELECT '06:00'
    UNION SELECT '07:00'
    UNION SELECT '08:00'
    UNION SELECT '09:00'
    UNION SELECT '10:00'
    UNION SELECT '11:00'
    UNION SELECT '12:00'
    UNION SELECT '13:00'
    UNION SELECT '14:00'
    UNION SELECT '15:00'
    UNION SELECT '16:00'
    UNION SELECT '17:00'
    UNION SELECT '18:00'
    UNION SELECT '19:00'
    UNION SELECT '20:00'
    UNION SELECT '21:00'
    UNION SELECT '22:00'
    UNION SELECT '23:00'    
) hours
LEFT JOIN YOUR_TABLE occurrence ON hours.`hour` BETWEEN TIME(occurrence.Entry_time) AND TIME(occurrence.Exit_time)
GROUP BY hours.`hour`
ORDER BY occurrences DESC

将产生(来自测试数据库的数据):

[hour] [occurrences]
10:00   73554
11:00   67492
09:00   65679
08:00   63886
13:00   63565
12:00   62525
07:00   61500
14:00   53095
15:00   49017
16:00   41955
17:00   31991
18:00   21251
06:00   17591
19:00   13717
20:00   8532
21:00   4421
22:00   2050
23:00   818
05:00   796
04:00   561
01:00   175
03:00   123
02:00   120
00:00   23

要获取小时,只需用查询和 select 第一个包裹:

SELECT hour FROM (
    # Paste query from above here...
) hourData
LIMIT 1;

我能够使用以下查询获得所需的输出:
SELECT HOUR(Entry_time) as hr FROM table_name GROUP BY HOUR(Entry_time) ORDER BY COUNT(*) DESC LIMIT 1;
感谢@nicholascarey 的解决方案:SQL to determine peak volume and hour for all days