get Mondays between two dates mysql - mysql

Question

I want to get the Mondays between two dates with a select statement, but I only have searched in sql server.
I have tried this, but it doesn´t work:
SELECT DATE_ADD('2017-01-01', INTERVAL ROW DAY) AS Date
FROM (
SELECT #row := #row + 1 AS row
FROM (
SELECT 0 UNION ALL
SELECT 1 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6
) t1,
(
SELECT 0 UNION ALL
SELECT 1 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6
) t2,
(
SELECT #row:=-1
) t3 LIMIT 31
) b
WHERE DATE_ADD('2017-01-01', INTERVAL ROW DAY) BETWEEN '2017-01-01' AND '2017-12-31'
AND DAYOFWEEK(DATE_ADD('2017-01-01', INTERVAL ROW DAY)) = 2
This query only gives me the Monday´s date for one month.
I hope that you can help me
Regards!

Answer 1

If you really want to solve this SQL only, you have to delve a bit deeper and generate a sequence of days, this is by far the most difficult side task to solve this. Besides this, as you have already guessed, you can use DAYOFWEEK or WEEKDAY to get your desired days.
SELECT *
FROM (
SELECT DATE_ADD('2013-01-01',
INTERVAL n4.num*1000+n3.num*100+n2.num*10+n1.num DAY ) AS DATE
FROM (
SELECT 0 AS num
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 n1,
(
SELECT 0 AS num
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 n2,
(
SELECT 0 AS num
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 n3,
(
SELECT 0 AS num
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 n4
) AS a
WHERE DATE >= '2017-01-01' AND DATE < NOW()
AND WEEKDAY(DATE) = 0
ORDER BY DATE
Some other RDBMS have inbuild functions for that.
If would be way easier if you have a calendar table where all dates are already present.

Answer 2

I think that this solution will work with better performance than previous
SET #date_start = date('2019-07-06');
SET #date_end = date('2019-09-18');
SELECT CASE WHEN MOD(DAYOFWEEK(#date_start), 7) <= 2 AND
MOD(DAYOFWEEK(#date_start), 7) + DATEDIFF(v.selected_date, #date_start) - 7 * FLOOR(DATEDIFF(v.selected_date, #date_start) / 7) >= 2 OR
MOD(DAYOFWEEK(#date_start), 7) > 2 AND
MOD(DAYOFWEEK(#date_start), 7) + DATEDIFF(v.selected_date, #date_start) - 7 * FLOOR(DATEDIFF(v.selected_date, #date_start) / 7) >= 9 THEN 1 ELSE 0 END
+ FLOOR(DATEDIFF(v.selected_date, #date_start) / 7) num_of_mo
, #date_start start_date,v.selected_date end_date
FROM (select adddate(#date_start,t4*10000 + t3*1000 + t2*100 + t1*10 + t0) selected_date from
(select 0 t0 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
(select 0 t1 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
(select 0 t2 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
(select 0 t3 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
(select 0 t4 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
WHERE v.selected_date <= #date_end;