How can I retrieve all the columns on a timerange aggregation? - mysql

Question

I am currently struggling on how to aggregate my daily data in other time aggregations (weeks, months, quarters etc).
Here is how my raw data type looks like:
| 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 |
First, I would like to check the sum of visits regardless of traffic_type:
SELECT date, SUM(visits) as visits_per_day
FROM visits_tbl
GROUP BY date
Here is the outcome:
| ymd | visits_per_day |
|:--------:|:--------------:|
| 20180101 | 2294563 |
| 20180102 | 2289145 |
| 20180103 | 2300367 |
| 20180104 | 2310256 |
| 20180105 | 2368098 |
| 20180106 | 2372257 |
| 20180107 | 2373863 |
| 20180108 | 2364236 |
However, if I want to check the specific day which the visits_per_day was the highest for each time aggregation (eg.: Month), I am struggling to retrieve the right output.
Here is what I did:
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
And here is the output of my query:
| y_month | max_visit_per_day |
|:-------:|:-----------------:|
| 201801 | 2435845 |
| 201802 | 2519000 |
| 201803 | 2528097 |
| 201804 | 2550645 |
However, I cannot know what was the exact day where the visits_per_day was the highest.
Desired output:
| y_month | max_visit_per_day | ymd |
|:-------:|:-----------------:|:--------:|
| 201801 | 2435845 | 20180130 |
| 201802 | 2519000 | 20180220 |
| 201803 | 2528097 | 20180325 |
| 201804 | 2550645 | 20180406 |
ymd would represent the day in which the visits_per_day was the highest.
This logic would be used in a dashboard with the help of programming in order to automatically select the time aggregation.
Can someone please help me?

Answer 1

This is a job for the structured part of structured query language. That is, you will write some subqueries and treat them as tables.
You already know how to find the number of visits per day. Let's add the month for each day to that query (http://sqlfiddle.com/#!9/a8455e/13/0).
SELECT date DIV 100 as month, date,
SUM(visits) as visits
FROM visits_tbl
GROUP BY date
Next you need to find the largest number of daily visits in each month. (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
Then, the trick is retrieving the date on which that maximum occurred in each month. That requires a join. Without common table expressions (which MySQL lacks) you need to repeat the first subquery. (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
The outline of this rather complex query helps explain it. Instead of that subquery, we'll use an imaginary table called dayvisits.
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
You're seeking an extreme value for each month in the subquery. (This is a fairly standard sort of SQL operation.) To do that you find that value with a MAX() ... GROUP BY query. Then you join that to the subquery itself to find the other values corresponding to the extreme value.
If you did have common table expressions, the query would look like this. YOu might consider adopting the MySQL fork called MariaDB, which has CTEs.
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

Answer 2

[Query Check on MSSQL] its quick and efficient.
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 )
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