MySQL Query To Produce Amount Per Month - mysql

Question

I have a table that contains:
id date user_id duration amount
1 2014-01-01 00:00:00 1 1 £10
2 2014-01-02 00:00:00 2 2 £10
3 2014-01-03 00:00:00 3 3 £10
I'm trying to display the amount per month. Any ideas how to do this in a query?

Answer 1

Working on the assumptions that you can extract the month from you datetime easily, so the real question is about the aggregation logic, and that you can create a numbers table.
Here is a simple example that shows the pattern.
sqlfiddle
CREATE TABLE Num (num int);
INSERT INTO Num VALUES (0),(1),(2),(3),(4);
CREATE TABLE Tbl (start int, run int);
INSERT INTO Tbl VALUES (1,2),(2,3);
SELECT start + num active_month
,count(*) * 10 income
FROM Tbl
INNER JOIN
Num ON num < run
GROUP BY start + num

Answer 2

Like Karl, I'm pretty sure some kind of numbers table is necessary here. Personally I like the approach given here, which defines a view (well, several of them) to generate numbers, instead of having to actually store a table full of numbers. Whether you use a table or a view, when you SELECT from it, it just looks like this:
n
---
0
1
2
3
…
With that you can construct a query like this:
SELECT
purchases.purchase_id,
purchases.date_purchased,
purchases.duration,
-- generator_16 is our numbers table
generator_16.n,
-- Below we calculate the year and month (year_mon) in the following way:
-- (1) Get the first day of the year, e.g. if date_purchased is 2012-12-28,
-- this gives us 2012-01-01.
-- (2) Get the month number, e.g. 12 for 2012-12-28) and add that many months
-- to the first day of the year, which gives us the first day of the
-- month, 2012-12-01.
-- (3) Add "n" months, where "n" is the number we get from the numbers table,
-- starting at 0.
DATE_ADD( -- (3)
DATE_ADD( -- (2)
MAKEDATE( YEAR(purchases.date_purchased), 1 ), -- (1)
INTERVAL MONTH(purchases.date_purchased) - 1 MONTH -- (2)
),
INTERVAL generator_16.n MONTH -- (3)
) AS year_mon,
purchases.amount_income / purchases.duration AS amount
FROM purchases
-- The below JOIN means that if `purchases.duration` is 3, we get three rows
-- that have 0, 1, and 2 in the `n` column, which we use as the number of dates
-- to add in (3) above.
JOIN generator_16
ON generator_16.n BETWEEN 0 AND purchases.duration - 1
ORDER BY purchases.purchase_id, year_mon;
This gives us a result like this (SQL Fiddle):
purchase_id date_purchased duration n year_mon amount
----------- -------------- -------- - ------------ ------
1 2013-12-28 … 2 0 2013-12-01 … 7.5
1 2013-12-28 … 2 1 2014-01-01 … 7.5
2 2014-01-04 … 1 0 2014-01-01 … 10
3 2014-02-04 … 6 0 2014-02-01 … 6.6667
3 2014-02-04 … 6 1 2014-03-01 … 6.6667
3 2014-02-04 … 6 2 2014-04-01 … 6.6667
3 2014-02-04 … 6 3 2014-05-01 … 6.6667
3 2014-02-04 … 6 4 2014-06-01 … 6.6667
3 2014-02-04 … 6 5 2014-07-01 … 6.6667
I inserted blank lines to separate the purchase_id groups so you can see how n increases from 0 to duration - 1 with each item in the group. As you can see, year_mon is equal to n months after the first day of the date_purchased month plus n months, and amount is equal to amount_income / duration.
We're almost done, but as you can see year_mon has repetition: 2014-01-01 is shown twice. This is great news, because we can then use GROUP BY to group by that column and SUM(amount) to get the total for that month:
SELECT
DATE_ADD(
DATE_ADD(
MAKEDATE( YEAR(purchases.date_purchased), 1 ),
INTERVAL MONTH(purchases.date_purchased) - 1 MONTH
),
INTERVAL generator_16.n MONTH
) AS year_mon,
SUM(purchases.amount_income / purchases.duration) AS total
FROM purchases
JOIN generator_16
ON generator_16.n BETWEEN 0 AND purchases.duration - 1
GROUP BY year_mon
ORDER BY year_mon;
The only difference between this query and the previous month is that we do GROUP BY year_mon and then SUM(amount_income / duration) to get total for the month, yielding this result (SQL Fiddle):
year_mon total
------------ ------
2013-12-01 … 7.5
2014-01-01 … 17.5
2014-02-01 … 6.6667
2014-03-01 … 6.6667
2014-04-01 … 6.6667
2014-05-01 … 6.6667
2014-06-01 … 6.6667
2014-07-01 … 6.6667
...and of course you can use DATE_FORMAT and CAST or ROUND to get nicely-formatted dates and amounts, or you can do that in your front-end code.

Answer 3

What about :
SELECT a.my_date, a.income, IFNULL(SUM(DISTINCT(a.income)) + sum( b.income ), a.income) as roll_up
FROM (
SELECT purchase_id, DATE_FORMAT( date_purchased, '%y-%m') AS my_date, SUM( amount_income / duration ) AS "income"
FROM incomes
GROUP BY my_date
) AS a
LEFT OUTER JOIN (
SELECT purchase_id, DATE_FORMAT( date_purchased, '%y-%m') AS my_date, SUM(amount_income / duration ) AS "income"
FROM incomes
GROUP BY my_date
) AS b ON ( a.purchase_id > b.purchase_id )
GROUP BY a.purchase_id
It's a bit tricky to do that in one shot - and it might be improved - but that gives the following results :
my_date income roll_up
13-12 8.5000 8.5000
14-01 10.0000 18.5000
14-02 16.6667 35.1667
My data set is :
1 2013-12-28 00:00:00 1 2 15
2 2014-01-04 00:00:00 2 1 10
3 2014-02-04 00:00:00 3 6 40
4 2013-12-29 00:00:00 4 1 1
5 2014-02-28 00:00:00 5 2 20