MySql Single Table, Select last 7 days and include empty rows - mysql

Question

I have searched similar problems here on stackoverflow but I could not understand how to make this work, what I'm trying to do...
So, I want to get last 7 days transactions from database and get total sales amount and also include empty rows if there is no data for some day.
What I have so far:
http://sqlfiddle.com/#!2/f4eda/6
This outputs:
| PURCHASE_DATE | AMOUNT |
|---------------|--------|
| 2014-04-25 | 19 |
| 2014-04-24 | 38 |
| 2014-04-22 | 19 |
| 2014-04-19 | 19 |
What I want:
| PURCHASE_DATE | AMOUNT |
|---------------|--------|
| 2014-04-25 | 19 |
| 2014-04-24 | 38 |
| 2014-04-23 | 0 |
| 2014-04-22 | 19 |
| 2014-04-21 | 0 |
| 2014-04-20 | 0 |
| 2014-04-19 | 19 |
Any help appreciated :)

Answer 1

This is not easy. I took help from this thread generate days from date range and combined it with your query.
So the idea was to get the list of dates from last 7 days then left join these dates with a static amount 0 to the query you have and then finally sum them. This could be used for any date range, just need to change them in both the queries
select
t1.purchase_date,
coalesce(SUM(t1.amount+t2.amount), 0) AS amount
from
(
select DATE_FORMAT(a.Date,'%Y-%m-%d') as purchase_date,
'0' as amount
from (
select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date
from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
) a
where a.Date BETWEEN NOW() - INTERVAL 7 DAY AND NOW()
)t1
left join
(
SELECT DATE_FORMAT(purchase_date, '%Y-%m-%d') as purchase_date,
coalesce(SUM(amount), 0) AS amount
FROM transactions
WHERE purchase_date BETWEEN NOW() - INTERVAL 7 DAY AND NOW()
AND vendor_id = 0
GROUP BY purchase_date
)t2
on t2.purchase_date = t1.purchase_date
group by t1.purchase_date
order by t1.purchase_date desc
DEMO

Answer 2

Simply put together a subquery with the dates you want and use left outer join:
select d.thedate, coalesce(SUM(amount), 0) AS amount
from (select date('2014-04-25') as thedate union all
select date('2014-04-24') union all
select date('2014-04-23') union all
select date('2014-04-22') union all
select date('2014-04-21') union all
select date('2014-04-20') union all
select date('2014-04-19')
) d left outer join
transactions t
on t.purchase_date = d.thedate and vendor_id = 0
GROUP BY d.thedate
ORDER BY d.thedate DESC;

Answer 3

This is for last 7 days;
select d.thedate, coalesce(SUM(amount), 0) AS amount
from (select DATE(NOW()) as thedate union all
select DATE(DATE_SUB( NOW(), INTERVAL 1 DAY)) union all
select DATE(DATE_SUB( NOW(), INTERVAL 2 DAY)) union all
select DATE(DATE_SUB( NOW(), INTERVAL 3 DAY)) union all
select DATE(DATE_SUB( NOW(), INTERVAL 4 DAY)) union all
select DATE(DATE_SUB( NOW(), INTERVAL 5 DAY)) union all
select DATE(DATE_SUB( NOW(), INTERVAL 6 DAY))) d left outer join
transactions t
on t.purchase_date = d.thedate and vendor_id = 0
GROUP BY d.thedate
ORDER BY d.thedate DESC;

Answer 4

with recursive all_dates(dt) as (
select '2014-04-19' as dt
union all
select dt + interval 1 day
from all_dates
where dt + interval 1 day <= '2014-04-25'
)
select d.dt as purchase_date, coalesce(m.amount, 0) as purchased
from all_dates as d
left join mytable m
on d.dt = m.purchase_date
order by purchase_date desc;