If i have a database with 2 columns, date and account and i want to first count account per day and then group by week. How wrong is my code and how to do it?
I edited my code a little bit, i was not thinking right from the beginning. I want the sum to be 9 for week 48.
SELECT date, account,
(SELECT date, COUNT(DISTINCT account)
FROM t1
GROUP BY date
) AS sum
FROM t1
GROUP BY YEARWEEK(date)
You seem to be looking for a simple aggregate query with count(distinct ...):
select yearweek(date) year_week, count(distinct account) cnt_account
from t1
group by yearweek(date)
order by year_week
Note: yearweek() gives you the year and week; this is better than week(), if your data spreads over several years.
EDIT
From the comments, you need two levels of aggregation:
select yearweek(dy) year_week, sum(cnt) cnt_account
from (
select date(t1.date) dy, count(distinct t1.account) cnt
from t1
group by date(t1.date)
) t
group by yearweek(dy)
order by year_week
Related
I currently am trying to track the number of messages sent by month as well as the volume's percent change in comparison to one year prior.
Here is my current query:
Select
a.mo,
a.ye,
a.Messages,
((a.Messages - b.Messages) / b.Messages) as "% Change"
from(
select
MONTH(post_date) as mo,
count(*) as "Messages",
YEAR(post_date) as ye
from
pm_messages
WHERE
post_date > "2018-01-01 00:00:00"
group by
year(post_date),
month(post_date)
) a
left join (
select
MONTH(post_date) as mo,
YEAR(post_date) as ye,
count(*) as "Messages"
from
pm_messages
group by
year(post_date),
month(post_date)
) b on a.mo = b.mo
and a.ye -1 = b.ye
This works great, however, it places month and year in separate columns, which has been messing up the graphs I am working with. However, when I try to pull month and year into one columns as I've done in other queries from the same table, i.e. using:
SELECT DATE_FORMAT(`post_date`,'%M %Y')
My query does not work.
Does anyone know how I can combine my current query to still calculate the return from a year prior but have month and date come up as one column, as opposed to (Month | Year | Messages | % Change)
Thanks!!
you can use extract instead of separate year() and month() functions :
EXTRACT(YEAR_MONTH from post_date)
of course you have to group by this instead of year, month . for example :
select
EXTRACT(YEAR_MONTH from post_date) yearmonth,
count(*) as "Messages"
from
pm_messages
group by
EXTRACT(YEAR_MONTH from post_date)
If you have data for every month, you can use lag():
select year(post_date) as ye, month(post_date) as mo,
count(*) as Messages,
lag(count(*)) over (partition by month(post_date) order by year(post_date)) as prev_year
from pm_messages
where post_date >= '2018-01-01'
group by year(post_date), month(post_date)
I expect this query to give me the avg value from daily active users up to date and grouped by month (from Oct to December). But the result is 164K aprox when it should be 128K. Why avg is not working? Avg should be SUM of values / number of current month days up to today.
SELECT sq.month_year AS 'month_year', AVG(number)
FROM
(
SELECT CONCAT(MONTHNAME(date), "-", YEAR(DATE)) AS 'month_year', count(distinct id_user) AS number
FROM table1
WHERE date between '2020-10-01' and '2020-12-31 23:59:59'
GROUP BY EXTRACT(year_month FROM date)
) sq
GROUP BY 1
Ok guys thanks for your help. The problem was that on the subquery I was pulling the info by month and not by day. So I should pull the info by day there and group by month in the outer query. This finally worked:
SELECT sq.day_month, AVG(number)
FROM (SELECT date(date) AS day_month,
count(distinct id_user) AS number
FROM table_1
WHERE date >= '2020-10-01' AND
date < '2021-01-01'
GROUP BY 1
) sq
GROUP BY EXTRACT(year_month FROM day_month)
Do not use single quotes for column aliases!
SELECT sq.month_year, AVG(number)
FROM (SELECT CONCAT(MONTHNAME(date), '-', YEAR(DATE)) AS month_year,
count(distinct id_user) AS number
FROM table1
WHERE date >= '2020-10-01' AND
date < '2021-01-01'
GROUP BY month_year
) sq
GROUP BY 1;
Note the fixes to the query:
The GROUP BY uses the same columns as the SELECT. Your query should return an error (although it works in older versions of MySQL).
The date comparisons have been simplified.
No single quotes on column aliases.
Note that the outer query is not needed. I assume it is there just to illustrate the issue you are having.
I need to find the average time in days between a customer's second order and third order
I know that I need to use the timestampdiff but am quite at a loss for how to select the second and third dates and need some sort of nest.
SELECT CustomerID,
OrderDate,
diff,
avg(timestampdiff(day, start_date, end_date)) AS average_days
FROM () o3
WHERE date3, date2
ORDER BY CustomerID, OrderDate;
Table
To achieve your desired result, you first need to calculate ROW_NUMBER from your data PARTITION BY CustmerId. Then keep rows only with RowNumber IN (2,3) and then get the DateDiff between two days. The following query will help getting your desired results-
SELECT CustomerID,datediff(MAX(OrderDate),MIN(OrderDate))
FROM
(
SELECT *,
#row_num :=IF(#prev_value = concat_ws('',CsutomerID),#row_num+1,1)AS RowNumber
, #prev_value := concat_ws('',CsutomerID)
FROM your_table A
ORDER BY CustomerID,OrderDate
)B
WHERE B.RowNumber IN (2,3)
GROUP BY CustomerID;
Basically I have a table like this:
Table Time:
ID.......Date
1......08/26/2016
1......08/26/2016
2......05/29/2016
3......06/22/2016
4......08/26/2015
5......05/23/2015
5......05/23/2015
6......08/26/2014
7......04/26/2014
8......08/26/2013
9......03/26/2013
The query should return like this
Year........CountNum
2016........4
2015........3
To find out which year does its value tend to increase in. I notice that I want to display the years that have more values (number of row in this case) than the previous year.
What I've done so far
SELECT Year, count(*) as CountNum
FROM Time
GROUP BY Year
ORDER BY CountNum DESC;
I don't know how to get the year from date format. I tried year(Date) function, but I got Null data.
Please help!
It should works fine.
select year(date), count(*) as countNum
from time
group by year(date)
order by countNum
Join the grouped data to itself with 1 year offset:
select
a.*
from
(
select year(`Date`) as _year, count(*) as _n
from time group by 1
) a
left join
(
select year(`Date`) as _year, count(*) as _n
from time group by 1
) b
on a._year = b._year-1
where a._n > b._n
order by 1
I am trying to calculate how much each customer spends in their first year with us but I'm struggling with how to format the query. So far I have:
select customerId, sum(total)
from cms2_orders
where customerId = 254063
and
(dtc between
(select min(dtc)
from cms2_orders
group by customerId
)
AND
(select (date_add(min(dtc), interval 1 year))
from cms2_orders))
group by customerId
I know the problem is with the 'group by' on line 8 but if that is not present, it is interpreted as the first order ever placed by anyone. Is there a different way of doing this?
One cleaner way of doing this is to use a subquery to calculate the start date for each customer, and then join this to your main query to remove records from the total which did not occur in the first year.
SELECT t1.customerid,
SUM(t1.total)
FROM cms2_orders t1
INNER JOIN
(
SELECT customerid,
MIN(dtc) AS dtc_start
FROM cms2_orders
GROUP BY customerid
) t2
ON t1.customerid = t2.customerid
WHERE t1.dtc BETWEEN t2.dtc_start AND DATEADD(t2.dtc_start, INTERVAL 1 YEAR)
GROUP BY t1.customerid
here's the different way, hope it works
select customerid, sum(total)
from cms2_orders
where customerId = 254063
and year(dtc) = year(min(dtc))