Select the next day occurance from the table - mysql

This is my db structure of the table game days. Every game ends at 8:00 pm . Here all the game id is same.So now i want to query by the game id like this way that if today is monday and it is before 8:00 pm so it will fetch only the game_days with value of day='Monday'.Once 8:00pm is over then it will show the row with the value of day='Wednesday' until wednesday 8:00pm and after 8:00 pm it will show the row with the value of day='Friday' until Friday 8:00pm and then again after Friday 8:00pm it will show Monday...
So what will be the query for this ?


I think you just want the date offset by 4 hours. One method is:
where date(updated_at + interval 4 hour) = curdate()
I am guessing that updated_at is the column you want to reference, but it can be any date/time column.
I prefer to express this as:
where updated_at >= curdate() - interval 4 hour and
updated_at < curdate() + interval 1 day - interval 4 hour
This version can make use of an index on the date/time column.


Please try this (updated based on comments, replace table_name with appropriate table name is your db):
SELECT *
FROM `table_name`
WHERE day = (
SELECT t3.day FROM `table_name` as t3
LEFT JOIN (
SELECT HOUR(NOW()),
CASE
WHEN HOUR(NOW()) <= 19 THEN DAYNAME(NOW() + INTERVAL t.p DAY)
ELSE DAYNAME(NOW() + INTERVAL (t.p + 1) DAY)
END AS day,
CASE
WHEN HOUR(NOW()) <= 19 THEN t.p
ELSE t.p + 1
END as p
FROM (
SELECT 0 as p
UNION SELECT 1 as p
UNION SELECT 2 as p
UNION SELECT 3 as p
UNION SELECT 4 as p
UNION SELECT 5 as p
UNION SELECT 6 as p
) t
) t2 on t3.day = t2.day
ORDER BY t2.p ASC LIMIT 1
)

Related

How to skip sundays (MySQL)

I don't have any idea how to do it... I need to skip sundays. My query is like
SELECT * FROM `table_1` WHERE `date`=(CURRENT_DATE + INTERVAL 1 DAY)
but if next day is sunday i need rows from query like this one
SELECT * FROM `table_1` WHERE `date`=(CURRENT_DATE + INTERVAL 2 DAY)
Any ideas will be helpful.

You can check if CURRENT_DATE is Saturday and produce the value accordingly.
For example:
SELECT *
FROM `table_1`
WHERE `date`=
case when dayofweek(current_date) = 7 -- saturday
then CURRENT_DATE + INTERVAL 2 DAY
else CURRENT_DATE + INTERVAL 1 DAY
end

You could put case inside the interval statement, like:
SELECT * FROM `table_1`
WHERE `date` = (
CURRENT_DATE + INTERVAL (
case when dayofweek(current_date + interval 1 day) = 1 then 2 else 1 end
) DAY
);

mysql LAST_DAY() only reads 1 subquery result, how to process all results? using joins?

I have an insurance policies table like this:
+-------------------------------------------------------------+
| id | cancellation_val | cancellation_interval | expire_date |
+-------------------------------------------------------------+
| 1 | 30 | day | 2019-06-09 |
| 2 | 2 | month | 2019-12-01 |
+-------------------------------------------------------------+
I need to get the ids of the policies that are going to expire based on cancellation, from today and within 4 months, calculating the last day of the month, like this pseudo-code:
'today' <= LAST_DAY( expire_date - cancellation_val/interval ) < 'today + 4 months'
Being not a pro I think I should use JOINs but I don't know how, after days of trying the only thing I achieved was this:
SELECT LAST_DAY(
DATE_FORMAT(
STR_TO_DATE(
(SELECT CASE cancellation_interval
WHEN "day" THEN date_sub(expire_date, INTERVAL cancellation_val DAY)
WHEN "month" THEN date_sub(data_scadenzaexpire_date, INTERVAL cancellation_val MONTH)
END
AS newDate
FROM insurance WHERE id=2
), '%Y-%m-%d'
), '%Y-%m-%d'
)
)
This is working but I don't need the "WHERE id=2" clause (because I need to process ALL rows of the table), and if I remove it I got error "subquery returns more than 1 row".
So how I can proceed? And using the result to stay between 'today' AND 'today + 4 months' ?
I think with some kind of JOIN I could do it in a easier way but I don't know how.
Thank you all

The problem is the structure of the query, not the LAST_DAY function.
We want to return the id values of rows that meet some condition. So the query would be of the form:
SELECT t.id
, ...
FROM insurance t
WHERE ...
HAVING ...
Introducing another SELECT keyword basically introduces a subquery. There are restrictions on subqueries... in the SELECT list, a subquery can return a single column and (at most) a single row.
So let's ditch that extra SELECT keyword.
We can derive the newdate as an expression of the SELECT list, and then we can reference that derived column in the HAVING clause. The spec said we wanted to return the id value, so we include that in the SELECT list. We don't have to return any other columns, but for testing/debugging, it can be useful to return the values that were used to derive the newdate column.
Something like this:
SELECT t.id
, LAST_DAY(
CASE t.cancellation_interval
WHEN 'day' THEN t.expire_date - INTERVAL t.cancellation_val DAY
WHEN 'month' THEN t.expire_date - INTERVAL t.cancellation_val MONTH
ELSE t.expire_date
END
) AS newdate
, t.expire_date
, t.cancellation_interval
, t.cancellation_val
FROM insurance t
HAVING newdate >= DATE(NOW())
AND newdate <= DATE(NOW()) + INTERVAL 4 MONTH
ORDER
BY newdate ASC
We don't have to include the newdate in the SELECT list; we could just replace occurrences of newdate in the HAVING clause with the expression.
We could also use an inline view to "hide" the derivation of the newdate column
SELECT v.id
, v.newdate
FROM ( SELECT t.id
, LAST_DAY(
CASE t.cancellation_interval
WHEN 'day' THEN t.expire_date - INTERVAL t.cancellation_val DAY
WHEN 'month' THEN t.expire_date - INTERVAL t.cancellation_val MONTH
ELSE t.expire_date
END
) AS newdate
FROM insurance t
) v
WHERE v.newdate >= DATE(NOW())
AND v.newdate <= DATE(NOW()) + INTERVAL 4 MONTH
ORDER
BY v.newdate ASC

check this query: remove the HAVING Line to see all rows
SELECT
IF(cancellation_interval = 'day',
i.expire_date - INTERVAL i.`cancellation_val` DAY,
i.expire_date - INTERVAL i.`cancellation_val` MONTH
) as cancellation_day,
i.*
FROM `insurance` i
HAVING cancellation_day < NOW() + INTERVAL 4 MONTH;
SAMPLES
MariaDB [test]> SELECT IF(cancellation_interval = 'day', i.expire_date - INTERVAL i.`cancellation_val` DAY, i.expire_date - INTERVAL i.`cancellation_val` MONTH ) as cancellation_day, i.* FROM `insurance` i HAVING cancellation_day < NOW() + INTERVAL 4 MONTH;
+------------------+----+------------------+-----------------------+-------------+
| cancellation_day | id | cancellation_val | cancellation_interval | expire_date |
+------------------+----+------------------+-----------------------+-------------+
| 2019-05-10 | 1 | 30 | day | 2019-06-09 |
+------------------+----+------------------+-----------------------+-------------+
1 row in set (0.001 sec)

When you use a SELECT query as an expression, it can only return one row.
If you want to process all the rows, you need to call LAST_DAY() inside the query, not on the result.
SELECT *
FROM insurance
WHERE CURDATE() <= LAST_DAY(
expire_date - IF(cancellation_interval = 'day',
INTERVAL cancellation_val DAY,
INTERVAL cancellation_val MONTH))
AND LAST_DAY(expire_date - IF(cancellation_interval = 'day',
INTERVAL cancellation_val DAY,
INTERVAL cancellation_val MONTH)) < CURDATE + INTERVAL 4 MONTH

between date inteval mysql not showing today

my query:
SELECT p.idprd,
p.nmprd,
pe.idprd,
pe.stockjual,
pe.stockkeluar,
pe.tothrgjual,
pe.tgljual
FROM tbproduk AS p
INNER JOIN (
SELECT idprd,
Sum(stockjual) AS 'stockjual',
Sum(stockkeluar) AS 'stockkeluar',
Sum(tothrgajual) AS 'tothrgjual',
tgljual
FROM tbpenjualan
WHERE '2019-01-06' >= '2019-01-06' - INTERVAL 7 day
AND '2019-01-06' < '2019-01-06' + INTERVAL 7 day
GROUP BY idprd
) AS pe ON p.idprd = pe.idprd
my result
my data:
date '2019-01-06' not showing. how to my currentdate to showing

GROUP BY has missing column - which would prevent you to even execute SQL statement successfully. But after adjusting SQL statement, results include correct sets of data.
SELECT p.idprd,
p.nmprd,
pe.idprd,
pe.stockjual,
pe.stockkeluar,
pe.tothrgjual,
pe.tgljual
FROM tbproduk AS p
INNER JOIN (
SELECT idprd,
Sum(stockjual) AS 'stockjual',
Sum(stockkeluar) AS 'stockkeluar',
Sum(tothrgajual) AS 'tothrgjual',
tgljual
FROM tbpenjualan
WHERE '2019-01-06' >= '2019-01-06' - INTERVAL 7 day
AND '2019-01-06' < '2019-01-06' + INTERVAL 7 day
GROUP BY idprd, tgljual
) AS pe ON p.idprd = pe.idprd
ORDER BY pe.tgljual desc;
Query Result:

Elapsed Time Between Two Dates for specified time range

I have a MYSQL table with a TIMESTAMP column 'Start' and a TIMESTAMP column 'End'. I want to return the number of minutes between the start and the end (End is always after than Start). Usually I'd just use 'TIMESTAMPDIFF()' but this time I need to get the minutes from 9am until 22pm, of each day in that date range.
If a row has a Start '2017-01-01 07:15:00' and an End of '2017-01-02 11:30:00' - the elapsed time should be 15.5 hours (930 minutes).
I'm having trouble coming up with a decent way of doing this and my searching online hasn't found quite what I'm looking for. Can someone help me along?
Edit:
CREATE TABLE date_ranges (
Start TIMESTAMP,
End TIMESTAMP
);
INSERT INTO date_ranges VALUES('2017-01-01 07:15:00','2017-01-02 11:30:00');
I came up with this:
SELECT Start, End, TIMESTAMPDIFF(MINUTE, Start, End) AS MinutesElapsed
FROM date_ranges;
I'm missing the part where the time in minutes is calculated only in the specified time range (9am until 22pm). Any ideas?

Here you go:
SELECT t1, t2, (TIMESTAMPDIFF(MINUTE, t1, t2) - TIMESTAMPDIFF(DAY, t1, t2)*660) FROM
(SELECT CASE WHEN t1 < STR_TO_DATE(concat(date_format(t1, '%Y-%m-%d'), ' 09:00:00'), '%Y-%m-%d %h:%i:%s')
THEN STR_TO_DATE(concat(date_format(t1, '%Y-%m-%d'), ' 09:00:00'), '%Y-%m-%d %h:%i:%s')
ELSE t1
END AS t1 FROM test) test1,
(SELECT CASE WHEN t2 > STR_TO_DATE(concat(date_format(t2, '%Y-%m-%d'), ' 22:00:00'), '%Y-%m-%d %h:%i:%s')
THEN STR_TO_DATE(concat(date_format(t2, '%Y-%m-%d'), ' 22:00:00'), '%Y-%m-%d %h:%i:%s')
ELSE t2
END AS t2 FROM test) test2;
660 = number of minutes between 22:00 and 09:00 (11 hours)
Here's the SQL Fiddle.

It's not very concise, but this should give you the results you want:
select started_at,ended_at,
(case
when date(ended_at) = date(started_at)
then
timestampdiff(
minute,
greatest(started_at,concat(date(started_at),' 09:00:00')),
least(ended_at,concat(date(ended_at),' 22:00:00'))
)
else
timestampdiff(
minute,
least(greatest(started_at,concat(date(started_at),' 09:00:00')),concat(date(started_at),' 22:00:00')),
concat(date(started_at),' 22:00:00')
)
+
timestampdiff(
minute,
concat(date(ended_at),' 09:00:00'),
greatest(least(ended_at,concat(date(ended_at),' 22:00:00')),concat(date(ended_at),' 09:00:00'))
)
+ ((datediff(ended_at,started_at)-1)*780)
end) as total_minutes
from your_table;

--Generating all dates in 2017.
CREATE TABLE CALENDAR AS --Use a different table name if CALENDAR already exists
SELECT '2017-12-31 09:00:00' - INTERVAL c.number DAY AS start_datetime,'2017-12-31 22:00:00' - INTERVAL c.number DAY AS end_datetime
FROM (SELECT singles + tens + hundreds number FROM
(SELECT 0 singles
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
) singles JOIN
(SELECT 0 tens
UNION ALL SELECT 10 UNION ALL SELECT 20 UNION ALL SELECT 30
UNION ALL SELECT 40 UNION ALL SELECT 50 UNION ALL SELECT 60
UNION ALL SELECT 70 UNION ALL SELECT 80 UNION ALL SELECT 90
) tens JOIN
(SELECT 0 hundreds
UNION ALL SELECT 100 UNION ALL SELECT 200 UNION ALL SELECT 300
UNION ALL SELECT 400 UNION ALL SELECT 500 UNION ALL SELECT 600
UNION ALL SELECT 700 UNION ALL SELECT 800 UNION ALL SELECT 900
) hundreds
ORDER BY number DESC) c
WHERE c.number BETWEEN 0 and 364
;
--End of table creation
--Actual query begins here
SELECT D.`START`,
D.`END`,
SUM(TIMESTAMPDIFF(MINUTE,GREATEST(D.`START`,C.START_DATETIME), LEAST(D.`END`,C.END_DATETIME))) AS TOTAL_TIME
FROM CALENDAR C
LEFT JOIN DATE_RANGES D ON DATE(C.START_DATETIME) >= DATE(D.`START`)
AND DATE(C.START_DATETIME) <= DATE(D.`END`)
WHERE D.`START` IS NOT NULL
AND D.`END` IS NOT NULL
GROUP BY D.`START`,
D.`END`
;
Construct a calendar table with a dates for a specified number of years. Each date having a start time of 09:00 and an end time of 22:00.
Left join on this table to get one row per date from the date ranges table.
Sum up the differences each day to get the total time worked.
Sample Demo

Day 1 Day 2 Day 3
|--********--|--********--|--********--|
|__________________________|
The question, IMHO is to know how many minutes the first day, and how many minutes the last day, the intermediate days have 780 minutes.
I've used a subquery just to help in the intermediate calculations.
select
if(hour(t1) < 9, date(t1) + interval 9 hour , t1) as tIni1,
date(t1) + interval 22 hour as tFin1,
date(t2) + interval 9 hour as tIni2,
if(hour(t2) > 22, date(t2) + interval 22 hour, t2) as tFin2,
TIMESTAMPDIFF(day, date(t1), date(t2)) numDays
from
tdt
tIni1 and tFin1 is the period of the first day, and tIni2, tFin2 the period of the last day, obviously first and last day can be the same.
Then calculate minutes of first day + minutes of second day + 780 minutes for every intermediate day.
select numDays, tIni1, tFin1, tIni2, tFin2,
if (numDays = 0,
TIMESTAMPDIFF(minute, tIni1, tFin2),
TIMESTAMPDIFF(minute, tIni1, tFin1)
+ TIMESTAMPDIFF(minute, tIni2, tFin2)
+ (numDays - 1) * 780
) as Minutes
from (
select
if(hour(t1) < 9, date(t1) + interval 9 hour , t1) as tIni1,
date(t1) + interval 22 hour as tFin1,
date(t2) + interval 9 hour as tIni2,
if(hour(t2) > 22, date(t2) + interval 22 hour, t2) as tFin2,
TIMESTAMPDIFF(day, date(t1), date(t2)) numDays
from
tdt
) ti
;
Try it here: http://rextester.com/GDHAB78973

how to get next three business days by mysql (current data and next three days)

I want to get the three working days from the current date as excluding Saturday and Sunday. can any one help me out here.
I have tried the interval method and DayOfWeek(day) <> 1 AND DayOfWeek(day) <> 7 but it is not giving me the proper result

Not very elegant but
select d
from
(
select curdate() as d
union all
select curdate() + interval 1 day
union all
select curdate() + interval 2 day
union all
select curdate() + interval 3 day
union all
select curdate() + interval 4 day
) tmp
where dayofweek(d) not in (1,7)
order by d
limit 3