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;
Related
I have table like this below :
And i want the result like : sum the amount group by date and find the difference between the result
If you are using MySQL 8.0 then you can use lag() to achieve expected output.
select
date,
amount,
coalesce((amount - last_val), amount) as diff
from
(
select
date,
amount,
lag(amount) over (order by date) as last_val
from
(
select
date,
sum(amount) as amount
from myTable
group by
date
) subq
) subo
order by
date
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
I currently have this that shows me the total of how much the amount is per month. What i want to get from this, is to show the month that made the highest amount per each year.
SELECT year(paymentDate), month(paymentDate) , SUM(amount)
FROM classicmodels.payments
GROUP BY year(paymentDate), month(paymentDate)
ORDER BY paymentDate ASC;
Here is the table that i want to only show the month that got the highest amount for each year
You can use ROW_NUMBER for this:
SELECT y, m, amount
FROM (
SELECT YEAR(paymentDate) AS y, MONTH(paymentDate) AS m,
amount,
ROW_NUMBER() OVER (PARTITION BY YEAR(paymentDate) ORDER BY amount DESC) AS rn
FROM classicmodels.payments) AS t
WHERE t.rn = 1
Note: In case of ties the query return an arbitrary month. To return all months having the biggest amount use RANK instead of ROW_NUMBER.
Using ROW_NUMBER() you can assign sequence number for each records based on your criteria, and select the record with sequence 1.
Sample below;
SELECT *
FROM (
SELECT year(paymentDate) year_val, month(paymentDate) month_val, SUM(amount) amt_val,
ROW_NUMBER() OVER(PARTITION BY year(paymentDate) ORDER BY SUM(amount) DESC) AS ROW_ORDER
FROM classicmodels.payments
GROUP BY year(paymentDate), month(paymentDate)
) AS D
WHERE D.ROW_ORDER = 1
There are lot of other ways to achieve the same.
and WHERE D.ROW_ORDER = n --you will return the n'th highest for each year.
Give SUM(Amount) a column alias, ORDER BY <that alias> DESC and add TOP(1) to the SELECT
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 have a lookup table that relates dates and people associated with those dates:
id, user_id,date
1,1,2014-11-01
2,2,2014-11-01
3,1,2014-11-02
4,3,2014-11-02
5,1,2014-11-03
I can group these by date(day):
SELECT DATE_FORMAT(
MIN(date),
'%Y/%m/%d 00:00:00 GMT-0'
) AS date,
COUNT(*) as count
FROM user_x_date
GROUP BY ROUND(UNIX_TIMESTAMP(created_at) / 43200)
But, how can get the number of unique users, that have now shown up previously? For instance this would be a valid result:
unique, non-unique, date
2,0,2014-11-01
1,1,2014-11-02
0,1,2014-11-03
Is this possibly without having to rely on a scripting language to keep track of this data?
I think this query will do what you want, at least it seems to work for your limited sample data.
The idea is to use a correlated sub-query to check if the user_id has occurred on a date before the date of the current row and then do some basic arithmetic to determine number of unique/non-unique users for each date.
Please give it a try.
select
sum(u) - sum(n) as "unique",
sum(n) as "non-unique",
date
from (
select
date,
count(user_id) u,
case when exists (
select 1
from Table1 i
where i.user_id = o.user_id
and i.date < o.date
) then 1 else 0
end n
from Table1 o
group by date, user_id
) q
group by date
order by date;
Sample SQL Fiddle
I didn't include the id column in the sample fiddle as it's not needed (or used) to produce the result and won't change anything.
This is the relevant question: "But, how can get the number of unique users, that have now shown up previously?"
Calculate the first time a person shows up, and then use that for the aggregation:
SELECT date, count(*) as FirstVisit
FROM (SELECT user_id, MIN(date) as date
FROM user_x_date
GROUP BY user_id
) x
GROUP BY date;
I would then use this as a subquery for another aggregation:
SELECT v.date, v.NumVisits, COALESCE(fv.FirstVisit, 0) as NumFirstVisit
FROM (SELECT date, count(*) as NumVisits
FROM user_x_date
GROUP BY date
) v LEFT JOIN
(SELECT date, count(*) as FirstVisit
FROM (SELECT user_id, MIN(date) as date
FROM user_x_date
GROUP BY user_id
) x
GROUP BY date
) fv
ON v.date = fv.date;