我目前正在努力研究如何在其他时间聚合中汇总我的每日数据（周，月，季等）。
以下是我的原始数据类型的样子：
| date | traffic_type | visits |
|----------|--------------|---------|
| 20180101 | 1 | 1221650 |
| 20180101 | 2 | 411424 |
| 20180101 | 4 | 108407 |
| 20180101 | 5 | 298117 |
| 20180101 | 6 | 26806 |
| 20180101 | 7 | 12033 |
| 20180101 | 8 | 80368 |
| 20180101 | 9 | 69544 |
| 20180101 | 10 | 39919 |
| 20180101 | 11 | 26291 |
| 20180102 | 1 | 1218490 |
| 20180102 | 2 | 410965 |
| 20180102 | 4 | 108037 |
| 20180102 | 5 | 297727 |
| 20180102 | 6 | 26719 |
| 20180102 | 7 | 12019 |
| 20180102 | 8 | 80074 |
首先，无论traffic_type如何，我都想查看访问次数：
SELECT date, SUM(visits) as visits_per_day
FROM visits_tbl
GROUP BY date
结果如下：
| ymd | visits_per_day |
|:--------:|:--------------:|
| 20180101 | 2294563 |
| 20180102 | 2289145 |
| 20180103 | 2300367 |
| 20180104 | 2310256 |
| 20180105 | 2368098 |
| 20180106 | 2372257 |
| 20180107 | 2373863 |
| 20180108 | 2364236 |
但是，如果我想检查 visits_per_day 在每个时间聚合中最高的特定日期（例如：月），我正在努力检索正确的输出。
以下是我的所作所为：
SELECT
(date div 100) as y_month, MAX(visits_per_day) as max_visit_per_day
FROM
(SELECT date, SUM(visits) as visits_per_day
FROM visits_tbl
GROUP BY date) as t1
GROUP BY
y_month
这是我的查询的输出：
| y_month | max_visit_per_day |
|:-------:|:-----------------:|
| 201801 | 2435845 |
| 201802 | 2519000 |
| 201803 | 2528097 |
| 201804 | 2550645 |
但是，我不知道visits_per_day最高的确切日期是什么。
期望的输出：
| y_month | max_visit_per_day | ymd |
|:-------:|:-----------------:|:--------:|
| 201801 | 2435845 | 20180130 |
| 201802 | 2519000 | 20180220 |
| 201803 | 2528097 | 20180325 |
| 201804 | 2550645 | 20180406 |
ymd 代表visits_per_day最高的日期。 在编程的帮助下，该逻辑将用于仪表板中，以便自动选择时间聚合。 有人可以帮助我吗？
答案 0 :(得分：0)
这是结构化查询语言的结构化部分的工作。也就是说，您将编写一些子查询并将它们视为表。
您已经知道如何查找每天的访问次数。我们将每个月的月份添加到该查询（http://sqlfiddle.com/#!9/a8455e/13/0）。
SELECT date DIV 100 as month, date,
SUM(visits) as visits
FROM visits_tbl
GROUP BY date
接下来，您需要找到每月最多的每日访问次数。 （http://sqlfiddle.com/#!9/a8455e/12/0）
SELECT month, MAX(visits) max_daily_visits
FROM (
SELECT date DIV 100 as month, date,
SUM(visits) as visits
FROM visits_tbl
GROUP BY date
) dayvisits
GROUP BY month
然后，诀窍是检索每个月发生最大值的日期。这需要加入。没有common table expressions（MySQL缺少），你需要重复第一个子查询。 （http://sqlfiddle.com/#!9/a8455e/11/0）
SELECT detail.*
FROM (
SELECT month, MAX(visits) max_daily_visits
FROM (
SELECT date DIV 100 as month, date,
SUM(visits) as visits
FROM visits_tbl
GROUP BY date
) dayvisits
GROUP BY month
) maxvisits
JOIN (
SELECT date DIV 100 as month, date,
SUM(visits) as visits
FROM visits_tbl
GROUP BY date
) detail ON detail.visits = maxvisits.max_daily_visits
AND detail.month = maxvisits.month
这个相当复杂的查询的大纲有助于解释它。我们将使用名为dayvisits的虚构表来代替该子查询。
SELECT detail.*
FROM (
SELECT month, MAX(visits) max_daily_visits
FROM dayvisits
GROUP BY date DIV 100
) maxvisits
JOIN dayvisits detail ON detail.visits = maxvisits.max_daily_visits
AND detail.month = maxvisits.month
您正在为子查询中的每个month寻找极值。 （这是一种相当标准的SQL操作。）为此，您可以使用MAX() ... GROUP BY查询找到该值。然后将其连接到子查询本身，以查找与极值对应的其他值。
如果您确实有公用表表达式，则查询将如下所示。你可能会考虑采用名为MariaDB的MySQL分支，它有CTE。
WITH dayvisits AS (
SELECT date DIV 100 as month, date,
SUM(visits) as visits
FROM visits_tbl
GROUP BY date
)
SELECT dayvisits.*
FROM (
SELECT month, MAX(visits) max_daily_visits
FROM dayvisits
GROUP BY month
) maxvisits
JOIN dayvisits ON dayvisits.visits = maxvisits.max_daily_visits
AND dayvisits.month = maxvisits.month
答案 1 :(得分：0)
[查询MSSQL]快速高效。
select visit_sum_day_wise.date
, visit_sum_day_wise.Max_Visits
, visit_sum_day_wise.traffic_type
, LAST_VALUE(visit_sum_day_wise.visits) OVER(PARTITION BY
visit_sum_day_wise.date ORDER BY visit_sum_day_wise.date ROWS BETWEEN
UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ) AS max_visit_per_day
from (
select visits_tbl.date , visits_tbl.visits , visits_tbl.traffic_type
,max(visits_tbl.visits ) OVER ( PARTITION BY visits_tbl.date ORDER
BY visits_tbl.date ROWS BETWEEN UNBOUNDED PRECEDING AND 0
PRECEDING) Max_visits
from visits_tbl
) as visit_sum_day_wise
where visit_sum_day_wise.visits = (select max(visits_B.visits ) from
visits_tbl visits_B where visits_B.Date = visit_sum_day_wise.date )
