I have tried looking at some similar examples like group by date range and weekdays etc but I couldnt fix it on my query.
as per my sample data screenshot, I need to only return
sum(salesamount)/sum(salescount) for week 1
and
sum(salesamount)/sum(salescount) for week 2.
Each of the week contain 5 days (in this example is wednesday - sunday).
My Attempt:
select salesstartdate, date_add(salesstartdate, interval 5 day) as gdate,
salesamount, salescount, sum(salesamount)/sum(salescount) as ATV
from testing
group by gdate;
My desired output is:
Week 1 15.34173913
Week 2 15.80365088
Calculation to get week 1 is (3507.1+3639.97+5258.77+8417.04+5994.48)/(285+273+344+478+368)
Calculation to get week 2 is the same as above except the date would now be from 8 to 12 of June.
You can do it with a subquery. In order to first group your result set properly and then execute aggregation on it:
SELECT
concat('WEEK', ' ', weekno) as `Week #`,
MIN(salesstartdate) as startDate,
MAX(salesstartdate) as endDate,
sum(salesamount)/sum(salescount) as ATV
FROM
(
SELECT
salesstartdate,
salesamount,
salescount,
WEEKOFYEAR(salesstartdate) as weekno -- get the week number of the current year
FROM
weekno
WHERE
WEEKDAY(salesstartdate) BETWEEN 2 AND 6 -- get index of week day
) as weeks
GROUP BY
weekno
I have used 2 MySQL functions here:
WEEKOFYEAR()
WEEKDAY()
Output:
WEEK 23 | 2016-06-08 | 2016-06-12 | 15.8040
WEEK 24 | 2016-06-16 | 2016-06-19 | 15.9323
and without subquery as well:
SELECT
concat('WEEK', ' ', WEEKOFYEAR(salesstartdate)) as `Week #`,
MIN(salesstartdate) as startDate,
MAX(salesstartdate) as endDate,
sum(salesamount)/sum(salescount) as ATV
FROM
weekno
WHERE
WEEKDAY(salesstartdate) BETWEEN 2 AND 6 -- get index of week day
GROUP BY
WEEKOFYEAR(salesstartdate)
You can do this way
select SUBDATE(salesstartdate, WEEKDAY(salesstartdate)) as week_range
, sum(salesamount)/sum(salescount)
from testing
where salesstartdate between SUBDATE(salesstartdate, WEEKDAY(salesstartdate))
and date_add(SUBDATE(salesstartdate, WEEKDAY(salesstartdate)), interval 5 day))
Group by week_range
Related
I have a MySQL requirement to select data from a table based on a start date and end date and group it by weekly also selecting the data in reverse order by date. Assume that, I have chosen the start date as 1st November and the end date as 04 December. Now, I would like to fetch the data as 04 December to 28 November, 27 November to 20 November, 19 November to 12 November and so on and sum the value count for that week.
Given an example table,
id
value
created_at
1
10
2021-10-11
2
13
2021-10-17
3
11
2021-10-25
4
8
2021-11-01
5
1
2021-11-10
6
4
2021-11-18
7
34
2021-11-25
8
17
2021-12-04
Now the result should be like 2021-12-04 to 2021-11-28 as one week, following the same in reverse order and summing the column value data for that week. I have tried in the query to add the interval of 7 days after the end date but it didn't work.
SELECT count(value) AS total, MIN(R.created_at)
FROM data_table AS D
WHERE D.created_at BETWEEN '2021-11-01' AND '2021-12-04' - INTERVAL 7 DAY ORDER BY D.created_at;
And it's also possible to have the last week may have lesser than 7 days.
Expected output:
end_interval
start_interval
total
2021-12-04
2021-11-27
17
2021-11-27
2021-11-20
34
2021-11-20
2021-11-13
4
2021-11-13
2021-11-06
1
2021-11-06
2021-10-30
8
2021-10-30
2021-10-25
11
Note that the last week is only 5 days depending upon the selected from and end dates.
One option to address this problem is to
generate a calendar of all your intervals, beginning from last date till first date, with a split of your choice, using a recursive query
joining back the calendar with the original table
capping start_interval at your start_date value
aggregating values for each interval
You can have three variables to be set, to customize your date intervals and position:
SET #start_date = DATE('2021-10-25');
SET #end_date = DATE('2021-12-04');
SET #interval_days = 7;
Then use the following query, as already described:
WITH RECURSIVE cte AS (
SELECT #end_date AS end_interval,
DATE_SUB(#end_date, INTERVAL #interval_days DAY) AS start_interval
UNION ALL
SELECT start_interval AS end_interval,
GREATEST(DATE(#start_date), DATE_SUB(start_interval, INTERVAL #interval_days DAY)) AS start_interval
FROM cte
WHERE start_interval > #start_date
)
SELECT end_interval, start_interval, SUM(_value) AS total
FROM cte
LEFT JOIN tab
ON tab.created_at BETWEEN start_interval AND end_interval
GROUP BY end_interval, start_interval
Check the demo here.
I have 2 dates to compare so I can count years between both date. If days is less than 365 days then I use interval to make it as 1 year. However I have problem if both date is similar in day for example 2019-10-15 to 2020-10-15 will be displayed as 2 years when I want it to be displayed as 1 year. Here is my code with not the result I want:
SELECT
TIMESTAMPDIFF(
YEAR,
'2019-10-15',
'2020-10-15' + INTERVAL 1 YEAR
) AS `year`
FROM
dob
Result:
What I want is:
2019-10-15 to 2020-10-14 = 1 year
2019-10-15 to 2020-10-15 = 1 year
2019-10-15 to 2020-10-16 = 2 year
How can I achieved that? Thanks in advance.
Does this do what you want? It seems you're only one day out from the expected output so I just subtracted a day.
SELECT
TIMESTAMPDIFF
( YEAR, '2019-10-15',
'2020-10-15'
+ INTERVAL 1 YEAR
- INTERVAL 1 DAY )
AS `year`
FROM dob
I've got a table with multiple columns and two of the columns are start_date and end_date.
I need to calculate the number of days in each month. Let's assume I have following data in my table
id | start_date | end_date
1 04.01.2016 15.02.2016
2 07.01.2016 22.01.2016
3 16.05.2016 11.07.2016
I want an output as follows
Month | numberOfTravelDays
January 51
February 15
May 15
June 31
July 11
This output I am expecting is the number of total travel days each month has been utilized. I am having trouble constructing the sql query for this. Can someone assist me on this?
This is what I have for now. And it's not doing the job. The below query also filters only this year's records(but ignore that).
select MONTH(start_date) as month,
COUNT(DATEDIFF(start_date, end_date)) as numberOfTravelDays
from travel
where YEAR(start_date) = YEAR(CURDATE())
group by MONTH(start_date),
MONTH(end_date)
Use a derived table:
select monstart,
sum(datediff(least(m.monend, t.end_date) + interval 1 day,
greatest(m.monstart, t.start_date)
)
) as days_worked
from travel t join
(select date('2016-01-01') as monstart, date('2016-01-31') as monend union all
select date('2016-02-01') as monstart, date('2016-02-29') as monend union all
. . .
) m
on t.end_date >= m.monstart and t.start_date <= m.monend
group by monstart;
I have a table like this:
id | created_on
1 2013-09-03 20:05:09
2 2013-09-05 17:03:13
...
How do I write a query to return a result of record counts that was created from Date X to Date Y in 7-day intervals?
So the result would look like this:
count | created_on
4 2013-09-17 00:00:00
2 2013-09-24 00:00:00
1 2013-09-31 00:00:00
10 2013-10-07 00:00:00
...
You can go to the beginning of the week by subtracting the day of the week. Here is one way to do that:
select date(created_on - interval dayofweek(created_on) day), count(*)
from t
group by date(created_on - interval dayofweek(created_on) day);
If this is not the day you want the week to start, then you can add an offset day.
Group by the date field, floored to the week:
SELECT
count(*),
YEARWEEK(created_on) as week
FROM
yourtable
GROUP BY week
This assumes that created_on is a type that can be interpreted as a date:
http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_yearweek
This will get you weekly groupings, but you may want to then convert that field (which will look like YYYYWW) back to something more readable.
You can try this
SELECT created_on, count( id ) AS count
FROM `test_table`
WHERE created_on
BETWEEN '2013-09-01'
AND '2013-10-10'
GROUP BY WEEK( created_on )
I am having an issue with a SELECT command in MySQL. I have a database of securities exchanged daily with maturity from 1 to 1000 days (>1 mio rows). I would like to get the outstanding amount per day (and possibly per category). To give an example, suppose this is my initial dataset:
DATE VALUE MATURITY
1 10 3
1 15 2
2 10 1
3 5 1
I would like to get the following output
DATE OUTSTANDING_AMOUNT
1 25
2 35
3 15
Outstanding amount is calculated as the total of securities exchanged still 'alive'. That means, in day 2 there is a new exchange for 10 and two old exchanges (10 and 15) still outstanding as their maturity is longer than one day, for a total outstanding amount of 35 on day 2. In day 3 instead there is a new exchange for 5 and an old exchange from day 1 of 10. That is, 15 of outstanding amount.
Here's a more visual explanation:
Monday Tuesday Wednesday
10 10 10 (Day 1, Value 10, matures in 3 days)
15 15 (Day 1, 15, 2 days)
10 (Day 2, 10, 1 day)
5 (Day 3, 5, 3 days with remainder not shown)
-------------------------------------
25 35 15 (Outstanding amount on each day)
Is there a simple way to get this result?
First of all in the main subquery we find SUM of all Values for current date. Then add to them values from previous dates according their MATURITY (the second subquery).
SQLFiddle demo
select T1.Date,T1.SumValue+
IFNULL((select SUM(VALUE)
from T
where
T1.Date between
T.Date+1 and T.Date+Maturity-1 )
,0)
FROM
(
select Date,
sum(Value) as SumValue
from T
group by Date
) T1
order by DATE
I'm not sure if this is what you are looking for, perhaps if you give more detail
select
DATE
,sum(VALUE) as OUTSTANDING_AMOUNT
from
NameOfYourTable
group by
DATE
Order by
DATE
I hope this helps
Each date considers each row for inclusion in the summation of value
SELECT d.DATE, SUM(m.VALUE) AS OUTSTANDING_AMOUNT
FROM yourTable AS d JOIN yourtable AS m ON d.DATE >= m.MATURITY
GROUP BY d.DATE
ORDER BY d.DATE
A possible solution with a tally (numbers) table
SELECT date, SUM(value) outstanding_amount
FROM
(
SELECT date + maturity - n.n date, value, maturity
FROM table1 t JOIN
(
SELECT 1 n UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5
) n ON n.n <= maturity
) q
GROUP BY date
Output:
| DATE | OUTSTANDING_AMOUNT |
-----------------------------
| 1 | 25 |
| 2 | 35 |
| 3 | 15 |
Here is SQLFiddle demo