gpt4 book ai didi

mysql - 基于 ORDER 字段而不是首次出现的累积字段

转载 作者:行者123 更新时间:2023-11-30 00:07:52 29 4
gpt4 key购买 nike

我目前有这个事件表:

id | user | country | when
---+------+---------+-----------------------
5 | 1 | my | "2011-07-21 00:00:00"
19 | 1 | my | "2014-04-02 18:00:00"
20 | 1 | my | "2014-04-09 18:00:00"
29 | 1 | my | "2011-09-27 15:00:00"
30 | 1 | my | "2011-02-01 15:00:00"
31 | 1 | my | "2012-05-12 19:00:00"
75 | 1 | my | "2014-06-20 16:00:00"

在输出中,我需要获取用户出现国家/地区的次数以及累积的次数。为此,我使用以下查询:

SELECT f.id, f.user, f.country, f.`when`, (
SELECT COUNT(*) FROM events AS f2
WHERE f2.country = f.country AND f2.user = f.user
) AS countrytimes, (
SELECT count(*) FROM events AS f3
WHERE f3.country = f.country AND f3.id <= f.id AND f3.user = f.user
ORDER BY f.`when` DESC
) AS accumulated
FROM events AS f
WHERE f.user = 1
AND f.country = 'my'
ORDER BY f.`when` DESC

该查询的输出基于字段when,直到这里为止。问题在于,字段累积是基于所请求的用户国家/地区的首次出现,而不是基于when字段。

id | user | country | when                  | countrytimes | accumulated
---+------+---------+-----------------------+--------------+-------------
75 | 1 | my | "2014-06-20 16:00:00" | 7 | 7
20 | 1 | my | "2014-04-09 18:00:00" | 7 | 3
19 | 1 | my | "2014-04-02 18:00:00" | 7 | 2
31 | 1 | my | "2012-05-12 19:00:00" | 7 | 6
29 | 1 | my | "2011-09-27 15:00:00" | 7 | 4
5 | 1 | my | "2011-07-21 00:00:00" | 7 | 1
30 | 1 | my | "2011-02-01 15:00:00" | 7 | 5

表和查询在此 SQLFiddle

如何对其进行排序,以便 when = 2011 年 2 月 1 日的行获得 accumulated = 1?

最佳答案

您只需将累积标准从 f3.id <= f.id 更改即可。至f3.when <= f.when按日期而不是id获取累积值;

SELECT f.id, f.user, f.country, f.when, (
SELECT COUNT(*) FROM events AS f2
WHERE f2.country = f.country AND f2.user = f.user
) AS countrytimes, (
SELECT count(*) FROM events AS f3
WHERE f3.country = f.country AND f3.when <= f.when AND f3.user = f.user
ORDER BY f.`when` DESC
) AS accumulated
FROM events AS f
WHERE f.user = 1
AND f.country = 'my'
ORDER BY f.`when` DESC

An SQLfiddle to test with .

关于mysql - 基于 ORDER 字段而不是首次出现的累积字段,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24362232/

29 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com