I need to calculate the hours between two dates, one of them being NOW(), excluding weekends.
I have found a few solutions that use stored procedures or mysql functions, but i'm looking to do this within a single query.
I can get the general hour difference by using this:
SELECT HOUR(TIMEDIFF(NOW(), created_at))
But I need this to only return hours of weekdays, not weekends. So from Friday at 23:00 to Monday at 1:00 there should only be a 2 hour difference.
Any help would be appreciated.
Thank you!
Mysql 8
In a programming language you'd write a loop and sum up the hours by iterating through the days. In SQL you can do the same with a recursive query:
with recursive cte (id, dt, minutes) as
(
select
id,
date(created_at),
case
when dayofweek(created_at) in (1, 7) then 0
else timestampdiff(minute, created_at, least(now(), date(created_at) + interval 1 day))
end
from mytable
union all
select
id,
dt + interval 1 day,
case
when dayofweek(dt + interval 1 day) in (1, 7) then 0
else timestampdiff(minute, dt + interval 1 day, least(now(), dt + interval 2 day))
end
from cte
where dt + interval 1 day <= curdate()
)
select id, round(sum(minutes) / 60, 1) as hours
from cte
group by id
order by id;
Demo: https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=a08f96bc29fb8a51302da10679496e22
Related
I am trying to write a query which outputs weekly dates from a given date to the end of the current year.
I am stuck here:
Select DATE_ADD(input_date, INTERVAL 7 DAY)
as alarm_date from userInput;
How do I get all the dates?
WITH RECURSIVE
cte AS ( SELECT id, input_date + INTERVAL 7 DAY alarm_date FROM userInput
UNION ALL
SELECT id, alarm_date + INTERVAL 7 DAY FROM cte WHERE YEAR(alarm_date) = YEAR(NOW()) )
SELECT * FROM cte;
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
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
I need to select data from MySQL database between the 1st day of the current month and current day.
select*from table_name
where date between "1st day of current month" and "current day"
Can someone provide working example of this query?
select * from table_name
where (date between DATE_ADD(LAST_DAY(DATE_SUB(CURDATE(), interval 30 day), interval 1 day) AND CURDATE() )
Or better :
select * from table_name
where (date between DATE_FORMAT(NOW() ,'%Y-%m-01') AND NOW() )
I was looking for a similar query where I needed to use the first day of a month in my query.
The last_day function didn't work for me but DAYOFMONTH came in handy.
So if anyone is looking for the same issue, the following code returns the date for first day of the current month.
SELECT DATE_SUB(CURRENT_DATE, INTERVAL DAYOFMONTH(CURRENT_DATE)-1 DAY);
Comparing a date column with the first day of the month :
select * from table_name where date between
DATE_SUB(CURRENT_DATE, INTERVAL DAYOFMONTH(CURRENT_DATE)-1 DAY) and CURRENT_DATE
select * from table_name
where `date` between curdate() - dayofmonth(curdate()) + 1
and curdate()
SQLFiddle example
I have used the following query. It has worked great for me in the past.
select date(now()) - interval day(now()) day + interval 1 day
try this :
SET #StartDate = DATE_SUB(DATE(NOW()),INTERVAL (DAY(NOW())-1) DAY);
SET #EndDate = ADDDATE(CURDATE(),1);
select * from table where (date >= #StartDate and date < #EndDate);
Complete solution for mysql current month and current year, which makes use of indexing properly as well :)
-- Current month
SELECT id, timestampfield
FROM table1
WHERE timestampfield >= DATE_SUB(CURRENT_DATE, INTERVAL DAYOFMONTH(CURRENT_DATE)-1 DAY)
AND timestampfield <= LAST_DAY(CURRENT_DATE);
-- Current year
SELECT id, timestampfield
FROM table1
WHERE timestampfield >= DATE_SUB(CURRENT_DATE, INTERVAL DAYOFYEAR(CURRENT_DATE)-1 DAY)
AND timestampfield <= LAST_DAY(CURRENT_DATE);
select * from table
where date between
(date_add (CURRENT_DATE, INTERVAL(1 - DAYOFMonth(CURRENT_DATE)) day)) and current_date;
select * from <table>
where <dateValue> between last_day(curdate() - interval 1 month + interval 1 day)
and curdate();
I found myself here after needing this same query for some Business Intelligence Queries I'm running on an e-commerce store. I wanted to add my solution as it may be helpful to others.
set #firstOfLastLastMonth = DATE_SUB(LAST_DAY(DATE_ADD(NOW(), INTERVAL -2 MONTH)),INTERVAL DAY(LAST_DAY(DATE_ADD(NOW(), INTERVAL -2 MONTH)))-1 DAY);
set #lastOfLastLastMonth = LAST_DAY(DATE_ADD(NOW(), INTERVAL -2 MONTH));
set #firstOfLastMonth = DATE_SUB(LAST_DAY(DATE_ADD(NOW(), INTERVAL -1 MONTH)),INTERVAL DAY(LAST_DAY(DATE_ADD(NOW(), INTERVAL -1 MONTH)))-1 DAY);
set #lastOfLastMonth = LAST_DAY(DATE_ADD(NOW(), INTERVAL -1 MONTH));
set #firstOfMonth = DATE_ADD(#lastOfLastMonth, INTERVAL 1 DAY);
set #today = CURRENT_DATE;
Today is 2019-10-08 so the output looks like
#firstOfLastLastMonth = '2019-08-01'
#lastOfLastLastMonth = '2019-08-31'
#firstOfLastMonth = '2019-09-01'
#lastOfLastMonth = '2019-09-30'
#firstOfMonth = '2019-10-01'
#today = '2019-10-08'
A less orthodox approach might be
SELECT * FROM table_name
WHERE LEFT(table_name.date, 7) = LEFT(CURDATE(), 7)
AND table_name.date <= CURDATE();
as a date being between the first of a month and now is equivalent to a date being in this month, and before now. I do feel that this is a bit easier on the eyes than some other approaches, though.
SELECT date_sub(current_date(),interval dayofmonth(current_date())-1 day) as first_day_of_month;
I had some what similar requirement - to find first day of the month but based on year end month selected by user in their profile page.
Problem statement - find all the txns done by the user in his/her financial year. Financial year is determined using year end month value where month can be any valid month - 1 for Jan, 2 for Feb, 3 for Mar,....12 for Dec.
For some clients financial year ends on March and some observe it on December.
Scenarios - (Today is `08 Aug, 2018`)
1. If `financial year` ends on `July` then query should return `01 Aug 2018`.
2. If `financial year` ends on `December` then query should return `01 January 2018`.
3. If `financial year` ends on `March` then query should return `01 April 2018`.
4. If `financial year` ends on `September` then query should return `01 October 2017`.
And, finally below is the query. -
select #date := (case when ? >= month(now())
then date_format((subdate(subdate(now(), interval (12 - ? + month(now()) - 1) month), interval day(now()) - 2 day)) ,'%Y-%m-01')
else date_format((subdate(now(), interval month(now()) - ? - 1 month)), '%Y-%m-01') end)
where ? is year end month (values from 1 to 12).
The key here is to get the first day of the month. For that, there are several options. In terms of performance, our tests show that there isn't a significant difference between them - we wrote a whole blog article on the topic. Our findings show that what really matters is whether you need the result to be VARCHAR, DATETIME, or DATE.
The fastest solution to the real problem of getting the first day of the month returns VARCHAR:
SELECT CONCAT(LEFT(CURRENT_DATE, 7), '-01') AS first_day_of_month;
The second fastest solution gives a DATETIME result - this runs about 3x slower than the previous:
SELECT TIMESTAMP(CONCAT(LEFT(CURRENT_DATE, 7), '-01')) AS first_day_of_month;
The slowest solutions return DATE objects. Don't believe me? Run this SQL Fiddle and see for yourself 😊
In your case, since you need to compare the value with other DATE values in your table, it doesn't really matter what methodology you use because MySQL will do the conversion implicitly even if your formula doesn't return a DATE object.
So really, take your pick. Which is most readable for you? I'd pick the first since it's the shortest and arguably the simplest:
SELECT * FROM table_name
WHERE date BETWEEN CONCAT(LEFT(CURRENT_DATE, 7), '-01') AND CURDATE;
SELECT * FROM table_name
WHERE date BETWEEN DATE(CONCAT(LEFT(CURRENT_DATE, 7), '-01')) AND CURDATE;
SELECT * FROM table_name
WHERE date BETWEEN (LAST_DAY(CURRENT_DATE) + INTERVAL 1 DAY - INTERVAL 1 MONTH) AND CURDATE;
SELECT * FROM table_name
WHERE date BETWEEN (DATE(CURRENT_DATE) - INTERVAL (DAYOFMONTH(CURRENT_DATE) - 1) DAY) AND CURDATE;
SELECT * FROM table_name
WHERE date BETWEEN (DATE(CURRENT_DATE) - INTERVAL (DAYOFMONTH(CURRENT_DATE)) DAY + INTERVAL 1 DAY) AND CURDATE;
SELECT * FROM table_name
WHERE date BETWEEN DATE_FORMAT(CURRENT_DATE,'%Y-%m-01') AND CURDATE;
I used this one
select DATE_ADD(DATE_SUB(LAST_DAY(now()), INTERVAL 1 MONTH),INTERVAL 1 day) first_day
,LAST_DAY(now()) last_day, date(now()) today_day
All the responses here have been way too complex. You know that the first of the current month is the current date but with 01 as the date. You can just use YEAR() and MONTH() to build the month date by inputting the NOW() method.
Here's the solution:
select * from table_name
where date between CONCAT_WS('-', YEAR( NOW() ), MONTH( NOW() ), '01') and DATE( NOW() )
CONCAT_WS() joins a series of strings with a separator (a dash in this case).
So if today is 2020-08-28, YEAR( NOW() ) = '2020' and MONTH( NOW() ) = '08' and then you just need to append '01' at the end.
Voila!
Get first date and last date from month and year.
select LAST_DAY(CONCAT(year,'.',month,'.','01')) as registerDate from user;
select date_add(date_add(LAST_DAY(end_date),interval 1 DAY),interval -1 MONTH) AS closingDate from user;
SET #date:='2012-07-11';
SELECT date_add(date_add(LAST_DAY(#date),interval 1 DAY),
interval -1 MONTH) AS first_day
I am trying to select the sum of an integer field for the past 5 days, and I need to group it for each day.
I'm having a bit of issues figuring out the grouping. Here's my sql query so far:
select
sum(`amount_sale`) as total
from `sales`
where the_date >= unix_timestamp((CURDATE() - INTERVAL 5 DAY))
that works fine for generating the sum for all 5 days together, but I need to break this down so that it shows the sum for each of the past 5 days i.e:
day 1 - $200
day 2- $500
day 3 - $20
etc.
SELECT DATE(FROM_UNIXTIME(the_date)) AS dt, SUM(amount_sale) AS total
FROM sales
WHERE the_date >= UNIX_TIMESTAMP((CURDATE() - INTERVAL 5 DAY))
GROUP BY
dt
To returns 0 for missing dates:
SELECT dt, COALESCE(SUM(amount_sale), 0) AS total
FROM (
SELECT CURDATE() - INTERVAL 1 DAY AS dt
UNION ALL
SELECT CURDATE() - INTERVAL 2 DAY AS dt
UNION ALL
SELECT CURDATE() - INTERVAL 3 DAY AS dt
UNION ALL
SELECT CURDATE() - INTERVAL 4 DAY AS dt
UNION ALL
SELECT CURDATE() - INTERVAL 5 DAY AS dt
) d
LEFT JOIN
sales
ON the_date >= UNIX_TIMESTAMP(dt)
AND the_date < UNIX_TIMESTAMP(dt + INTERVAL 1 DAY)
GROUP BY
dt
This is not a very elegant solution, however, MySQL lacks a way to generate recordsets from scratch.
use the format function to return weekday nr: SELECT DATE_FORMAT(the_date, '%w');
use between
like select * from XXX where date between date(...) and date(...) group by date Limit 0,5
should do it