Create a row for every day in a date range? - mysql

Question

I have a table like this:
+----+---------+------------+
| id | price | date |
+----+---------+------------+
| 1 | 340 | 2018-09-02 |
| 2 | 325 | 2018-09-05 |
| 3 | 358 | 2018-09-08 |
+----+---------+------------+
And I need to make a view which has a row for every day. Something like this:
+----+---------+------------+
| id | price | date |
+----+---------+------------+
| 1 | 340 | 2018-09-02 |
| 1 | 340 | 2018-09-03 |
| 1 | 340 | 2018-09-04 |
| 2 | 325 | 2018-09-05 |
| 2 | 325 | 2018-09-06 |
| 2 | 325 | 2018-09-07 |
| 3 | 358 | 2018-09-08 |
+----+---------+------------+
I can do that using PHP with a loop (foreach) and making a temp variable which holds the previous price til there is a new date.
But I need to make a view ... So I should do that using pure-SQL .. Any idea how can I do that?

Answer 1

You could use a recursive CTE to generate the records in the "gaps". To avoid that an infinite gap after the last date is "filled", first get the maximum date in the source data and make sure not to bypass that date in the recursion.
I have called your table tbl:
with recursive cte as (
select id,
price,
date,
(select max(date) date from tbl) mx
from tbl
union all
select cte.id,
cte.price,
date_add(cte.date, interval 1 day),
cte.mx
from cte
left join tbl
on tbl.date = date_add(cte.date, interval 1 day)
where tbl.id is null
and cte.date <> cte.mx
)
select id,
price,
date
from cte
order by 3;
demo with mysql 8

Answer 2

Here is an approach which should work without analytic functions. This answer uses a calendar table join approach. The first CTE below is the base table on which the rest of the query is based. We use a correlated subquery to find the most recent date earlier than the current date in the CTE which has a non NULL price. This is the basis for finding out what the id and price values should be for those dates coming in from the calendar table which do not appear in the original data set.
WITH cte AS (
SELECT cal.date, t.price, t.id
FROM
(
SELECT '2018-09-02' AS date UNION ALL
SELECT '2018-09-03' UNION ALL
SELECT '2018-09-04' UNION ALL
SELECT '2018-09-05' UNION ALL
SELECT '2018-09-06' UNION ALL
SELECT '2018-09-07' UNION ALL
SELECT '2018-09-08'
) cal
LEFT JOIN yourTable t
ON cal.date = t.date
),
cte2 AS (
SELECT
t1.date,
t1.price,
t1.id,
(SELECT MAX(t2.date) FROM cte t2
WHERE t2.date <= t1.date AND t2.price IS NOT NULL) AS nearest_date
FROM cte t1
)
SELECT
(SELECT t2.id FROM yourTable t2 WHERE t2.date = t1.nearest_date) id,
(SELECT t2.price FROM yourTable t2 WHERE t2.date = t1.nearest_date) price,
t1.date
FROM cte2 t1
ORDER BY
t1.date;
Demo
Note: To make this work on MySQL versions earlier than 8+, you would need to inline the CTEs above. It would result in verbose code, but, it should still work.

Answer 3

Since you are using MariaDB, it is rather trivial:
MariaDB [test]> SELECT '2019-01-01' + INTERVAL seq-1 DAY FROM seq_1_to_31;
+-----------------------------------+
| '2019-01-01' + INTERVAL seq-1 DAY |
+-----------------------------------+
| 2019-01-01 |
| 2019-01-02 |
| 2019-01-03 |
| 2019-01-04 |
| 2019-01-05 |
| 2019-01-06 |
(etc)
There are variations on this wherein you generate a large range of dates, but then use a WHERE to chop to what you need. And use LEFT JOIN with the sequence 'derived table' on the 'left'.
Use something like the above as a derived table in your query.