每天从MYSQL返回
我想在mysql中填写从今天到30天前的日期 例如： 以下是mysql：每天从MYSQL返回,mysql,Mysql,我想在mysql中填写从今天到30天前的日期 例如： 以下是mysql： SELECT IFNULL(SUM(units.price), 0) as price, DATE(units.solddate) as date, DAY(units.solddate) as day, ( select a.Date from ( select curdate() - INTERVAL (a.a + (10 * b
SELECT IFNULL(SUM(units.price), 0) as price, DATE(units.solddate) as date, DAY(units.solddate) as day,
(
select a.Date
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 >= DATE_SUB(NOW(), INTERVAL 30 day)
) as onemonth
FROM (tables)
GROUP BY date
请给我一些建议。谢谢。在更好地理解了您想要实现的目标后，我整理了这个
MySQLSELECT
`dates_table`.`date`
-- ,`dates_table`.`day`
,SUM( COALESCE( `units`.`price` ,0 ) ) AS `price`
FROM
-- Start of query for making dates_table
(
SELECT
DATE( DATE_SUB( CURDATE( ) ,INTERVAL `intervals_table`.`days` DAY ) ) AS `date`
,DAY( DATE_SUB( CURDATE( ) ,INTERVAL `intervals_table`.`days` DAY ) ) AS `day`
FROM
(
SELECT 0 AS `days`
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 UNION SELECT 10 UNION SELECT 11 UNION SELECT 12
UNION SELECT 13 UNION SELECT 14 UNION SELECT 15 UNION SELECT 16
UNION SELECT 17 UNION SELECT 18 UNION SELECT 19 UNION SELECT 20
UNION SELECT 21 UNION SELECT 22 UNION SELECT 23 UNION SELECT 24
UNION SELECT 25 UNION SELECT 26 UNION SELECT 27 UNION SELECT 28
UNION SELECT 29 UNION SELECT 30 UNION SELECT 31
) `intervals_table`
HAVING
`date` >= DATE_SUB( CURDATE( ) ,INTERVAL 30 DAY )
)
-- End of query for making dates_table
`dates_table`
LEFT OUTER JOIN `units` ON
`units`.`sold_date` >= CONCAT( `dates_table`.`date` ,' 00:00:00' )
AND `units`.`sold_date` <= CONCAT( `dates_table`.`date` ,' 23:59:59' )
WHERE
1
GROUP BY
`dates_table`.`date`
ORDER BY
`dates_table`.`date`
date price
---------- -----
2015-07-07 0
2015-07-08 0
2015-07-09 0
2015-07-10 0
2015-07-11 0
2015-07-12 0
2015-07-13 0
2015-07-14 0
2015-07-15 0
2015-07-16 0
2015-07-17 0
2015-07-18 0
2015-07-19 0
2015-07-20 0
2015-07-21 0
2015-07-22 0
2015-07-23 0
2015-07-24 0
2015-07-25 0
2015-07-26 0
2015-07-27 0
2015-07-28 0
2015-07-29 0
2015-07-30 0
2015-07-31 0
2015-08-01 0
2015-08-02 10
2015-08-03 25
2015-08-04 3
2015-08-05 0
这是通过纠正您的方法来解决的
SELECT
`dates`.`date` ,SUM( COALESCE( `units`.`price` ,0 ) ) AS `price`
FROM
(
-- Start of query for making dates
SELECT
`dates`.`date`
,CONCAT( `dates`.`date` ,' 00:00:00' ) AS `day_start_datetime`
,CONCAT( `dates`.`date` ,' 23:59:59' ) AS `day_end_datetime`
FROM
(
SELECT
DATE_SUB( CURDATE( ) ,INTERVAL `intervals`.`days` DAY ) AS `date`
FROM
(
SELECT
( `hundreds_place`.`num` * 100 )
+( `tens_place`.`num` * 10 )
+( `ones_place`.`num` * 1 ) AS `days`
FROM ( SELECT 0 AS `num` 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 ) AS `ones_place`
JOIN ( SELECT 0 AS `num` 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 ) AS `tens_place`
JOIN ( SELECT 0 AS `num` 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 ) AS `hundreds_place`
) `intervals`
HAVING
`date` >= DATE_SUB( CURDATE() ,INTERVAL 30 DAY )
) `dates`
-- End of query for making dates
) `dates`
LEFT OUTER JOIN `units`
ON `units`.`sold_date` BETWEEN `day_start_datetime` AND `day_end_datetime`
WHERE
1
GROUP BY
`dates`.`date`
ORDER BY
`dates`.`date`
为什么？这两种方法都可以达到相同的效果。这可以用于1000天的间隔，大约是33个月，并且可以很容易地延长到更大的间隔。
date price
---------- -----
2015-07-07 0
2015-07-08 0
2015-07-09 0
2015-07-10 0
2015-07-11 0
2015-07-12 0
2015-07-13 0
2015-07-14 0
2015-07-15 0
2015-07-16 0
2015-07-17 0
2015-07-18 0
2015-07-19 0
2015-07-20 0
2015-07-21 0
2015-07-22 0
2015-07-23 0
2015-07-24 0
2015-07-25 0
2015-07-26 0
2015-07-27 0
2015-07-28 0
2015-07-29 0
2015-07-30 0
2015-07-31 0
2015-08-01 0
2015-08-02 10
2015-08-03 25
2015-08-04 3
2015-08-05 0
SELECT
`dates`.`date` ,SUM( COALESCE( `units`.`price` ,0 ) ) AS `price`
FROM
(
-- Start of query for making dates
SELECT
`dates`.`date`
,CONCAT( `dates`.`date` ,' 00:00:00' ) AS `day_start_datetime`
,CONCAT( `dates`.`date` ,' 23:59:59' ) AS `day_end_datetime`
FROM
(
SELECT
DATE_SUB( CURDATE( ) ,INTERVAL `intervals`.`days` DAY ) AS `date`
FROM
(
SELECT
( `hundreds_place`.`num` * 100 )
+( `tens_place`.`num` * 10 )
+( `ones_place`.`num` * 1 ) AS `days`
FROM ( SELECT 0 AS `num` 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 ) AS `ones_place`
JOIN ( SELECT 0 AS `num` 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 ) AS `tens_place`
JOIN ( SELECT 0 AS `num` 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 ) AS `hundreds_place`
) `intervals`
HAVING
`date` >= DATE_SUB( CURDATE() ,INTERVAL 30 DAY )
) `dates`
-- End of query for making dates
) `dates`
LEFT OUTER JOIN `units`
ON `units`.`sold_date` BETWEEN `day_start_datetime` AND `day_end_datetime`
WHERE
1
GROUP BY
`dates`.`date`
ORDER BY
`dates`.`date`