I have table in MySQL which contain data like,
Week No. Month Year
1 April 2016
4 April 2016
I need to find date of Monday for the particular week.
I think this should do the trick, maybe you should have to adjust type of y parameter.
DELIMITER $$
CREATE FUNCTION `calc_date`(y TEXT, m TEXT, w INT)
RETURNS date
BEGIN
DECLARE d DATE;
SET d = DATE_ADD(STR_TO_DATE(CONCAT(CONCAT(CONCAT(y,'-'),m),'-01'), '%Y-%M-%d'), INTERVAL w week);
RETURN DATE_SUB(d, INTERVAL (DAYOFWEEK(d) + 5)%7 DAY);
end;
$$
DELIMITER ;
then you just have to call
select calc_date(`Year`,`Month`,`Week No.`) from yourtable;
Related
I would like to change the interval in this SQL statement, based on a parameter in a stored procedure. I want to use three different intervals: 1 day, 8 hours, 1 hour
CREATE DEFINER= 'dbshizzle' PROCEDURE `getData`(in sD text(17), in sT text(8))
BEGIN
select stime, sval
from tblNumber
where sDix = 'allright'
and timestamp >= now() - interval 1 day
order by timestamp;
END
Should I use an IF statement with an integer parameter, or a text parameter?
How about just adjusting the parameter and passing in the value as hours?
CREATE DEFINER = 'dbshizzle' PROCEDURE `getData`(
in in_sD text(17), -- should change to varchar
in in_sT text(8), -- should change to varchar
in in_hours int
)
BEGIN
select stime, sval
from tblNumber
where sDix = 'allright'
and timestamp >= now() - interval in_hours hour
order by timestamp;
END;
I'm currently using DATE_ADD(date,INTERVAL expr type) to set a due date as a trigger in a mySQL Database.
What I'm wanting to know is if it is possible to skip weekends (Saturday, Sunday) as part of the trigger.
You'd have to create an own function for doing that. You can look how to do that in this answer, for example (just use function instead of procedure). As for how to write such a function, here's a working algorithm. The code is quite straightforward: it loops through days and skips weekends.
CREATE FUNCTION `DAYSADDNOWK`(addDate DATE, numDays INT) RETURNS date
BEGIN
IF (WEEKDAY(addDate)=5) THEN
SET addDate=DATE_ADD(addDate, INTERVAL 1 DAY);
END IF;
IF (WEEKDAY(addDate)=6) THEN
SET addDate=DATE_ADD(addDate, INTERVAL 1 DAY);
END IF;
WHILE numDays>0 DO
SET addDate=DATE_ADD(addDate, INTERVAL 1 DAY);
IF (WEEKDAY(addDate)=5) THEN
SET addDate=DATE_ADD(addDate, INTERVAL 1 DAY);
END IF;
IF (WEEKDAY(addDate)=6) THEN
SET addDate=DATE_ADD(addDate, INTERVAL 1 DAY);
END IF;
SET numDays=numDays-1;
END WHILE;
RETURN addDate;
END
Currently SELECT DAYSADDNOWK(CURDATE(), 5) yields 2016-03-07, which is correct.
Of course you only can use it with days, so no arbitrary interval, but your question mentioned date datatype, and I don't quite see how one could add a month not counting working days.
This function simply creates a list of dates starting at the date given in the arguments, and then figures out which date is x number of days (the interval) out while disregarding days 1 and 7 (which are Sunday and Saturday respectively on SQL Server).
CREATE FUNCTION [dbo].[udf_days_add_no_wknd]
(
#start_date date
, #interval int
)
RETURNS date
AS
BEGIN
declare #answer date
; with dates as
(
select #start_date as date_val
union all
select dateadd(d, 1, date_val) as date_val
from dates
where date_val < dateadd(d, #interval * 10, #start_date)
)
, final as
(
select top 1 lead(ld.date_val, #interval, NULL) over (order by ld.date_val asc) as new_date_val
from dates as ld
where 1=1
and datepart(dw, ld.date_val) not in (1,7) --eliminating weekends
)
select #answer = (select new_date_val from final)
return #answer
END
It is worth nothing that this solution is dependent on having SQL Server 2012 or later, considering the use of the lead() function.
I am developing a Java application using MySQL. I need to know which is the week of each month, of the stored dates. Is there any MySQL function for that ? Basically , if i was to use this for the current date (13.09) it would show me its in week number 2 and tomorrow it will be week number 3.
You can play with the WEEK() function, and see if it suits your needs. Here I'm using WEEK(date, 3) that will return the week of the year from 1 to 53, starting from Mondays:
set #my_date = '2015-09-13';
SELECT
WEEK(#my_date, 3) -
WEEK(#my_date - INTERVAL DAY(#my_date)-1 DAY, 3) + 1
AS week_number;
WEEK(date, 3) will return the week of the year of the selected date
WEEK(date - INTERVAL DAY(#my_date)-1 DAY, 3) will return the week of the year of the first day of the month of the selected date
It will return 1 for 01-March-2015 (because it's the first day of the month so it's week 1) and 2 for 02-March-2015 (because weeks starts from Mondays, so it's a new week). If this is not the desidered behaviour you should specify your requirements more precisely.
Please see a fiddle here.
Unfortunately, there isn't a "weekofmonth" function, but you could use dayofmonth, and manipulate the result a bit:
SELECT CURRENT_DATE(),
FLOOR((DAYOFMONTH(CURRENT_DATE()) - 1) / 7) + 1 AS week_of_month
Create a mysql function.
CREATE FUNCTION `WEEK_OF_MONTH`(
datee DATE
) RETURNS INT(11)
BEGIN
DECLARE DayNamee VARCHAR(20);
DECLARE StartDatee DATE;
DECLARE DayNumber INT DEFAULT 0;
SET DayNamee = (SELECT DAYNAME(datee));
SET StartDatee = (SELECT FIRST_DAY(datee));
WHILE StartDatee <= datee DO
IF DayNamee = DAYNAME(StartDatee) THEN
SET DayNumber = DayNumber + 1;
END IF;
SET StartDatee = DATE_ADD( StartDatee, INTERVAL 1 DAY);
END WHILE;
RETURN DayNumber;
END$$
DELIMITER ;
Call as --
SELECT `WEEK_OF_MONTH`('2018-12-31');
Result : 5
I created a function that returns a date in a different format from the original.
Basically, I am testing with this Select MonthSub('2014-04-10',2)# statement, and it should return
2014-02, instead of 2014-02-10.
Could someone check my code and see what I am doing is wrong?
If I do anything to format by using date_format(new_in_date, '%y-%m') it returns this error:
ERROR 1292 (22007): Incorrect date value: '2014-02' for column 'new_in_date' at row 1
The function I wrote:
Create function MonthSub (in_date date, in_mn_adjust int)
returns date
Begin
declare new_in_date date default in_date;
set new_in_date := date_sub(new_in_date, interval in_mn_adjust month);
return new_in_date;
end;
If I understand you correctly, what you want your function to do is:
Given a date d and a number n, you want to substract n months to d
Then you want to return only the year and month of the date
So, you can do the following:
SQL Fiddle
MySQL 5.5.32 Schema Setup:
create function MonthSub(d date, n int)
returns varchar(50) -- You're not returning a date, but a string
-- ('yyyy-mm' is not a date)
begin
declare t date;
declare ans varchar(50);
set t = date_add(d, interval -(day(d)-1) day); -- Substract the days
-- to avoid problems with things
-- like 'February 30th'
set t = date_add(t, interval -n month); -- Substract the months
set ans = date_format(t, '%Y-%m'); -- Format the date to 'yyyy-mm'
return ans; -- Return the date
end //
Query 1:
-- Test it!
select MonthSub('2014-05-02', 2)
Results:
| MONTHSUB('2014-05-02', 2) |
|---------------------------|
| 2014-03 |
How would I put together a query to display all of the hours in the next week as I want to compare a timetable against this for appointment purposes.
Thanks for any help!
edit--
the expected result would be great as between 9 to 5
| client_date | client_time |
10/01/2010 09:00:00
10/01/2010 10:00:00
10/01/2010 11:00:00
10/01/2010 12:00:00
10/01/2010 13:00:00
10/01/2010 14:00:00
10/01/2010 15:00:00
10/01/2010 16:00:00
10/01/2010 17:00:00
You will need to create a table to store the date and time values.
CREATE TABLE calendarhours (caldaytime DATETIME);
You will then need to create a stored procedure to loop through the two dates and insert the date time values for the time sheet times into the table.
DELIMITER $$
CREATE DEFINER=`root`#`localhost` PROCEDURE `timesheetdays`(startdate DATETIME, enddate DATETIME)
BEGIN
DECLARE tempdate DATETIME;
DELETE FROM `calendarhours`;
-- set the temp date to 9am of the start date
SET tempdate = DATE_ADD(DATE(startdate), INTERVAL '0 9' DAY_HOUR);
-- while the temp date is less than or equal to the end date, insert the date
-- into the temp table
WHILE ( tempdate <= enddate ) DO
BEGIN
-- insert temp date into temp table
INSERT INTO `calendarhours` (caldaytime) VALUES (tempdate);
-- increment temp date by an hour
SET tempdate = DATE_ADD(tempdate, INTERVAL '0 1' DAY_HOUR);
-- if the temp date is greater than 5 PM (17:00) then increment to the next day
IF TIMEDIFF(tempdate, DATE_ADD(DATE(tempdate), INTERVAL '0 17' DAY_HOUR)) > 0 THEN
BEGIN
-- increment to the next day
SET tempdate = DATE_ADD(DATE(tempdate), INTERVAL '1 9' DAY_HOUR);
-- for business purposes, if the day is a Saturday or a Sunday increment
-- until we reach Monday
WHILE ( DAYNAME(tempdate) = 'Saturday' OR DAYNAME(tempdate) = 'Sunday' ) DO
BEGIN
SET tempdate = DATE_ADD(DATE(tempdate), INTERVAL '1 9' DAY_HOUR);
END;
END WHILE;
END;
END IF;
END;
END WHILE;
-- return all the inserted date and times
SELECT * FROM calendarhours ORDER BY caldaytime;
END
This procedure will then loop through the two dates, starting from 9 am each day and finishing at 5pm each day (17:00). When the time reaches 18:00, the procedure increments to the next day and starts again at 9 am.
If you are doing a standard business week timesheet, then if the day is equal to Saturday or Sunday, it will increment until it reaches Monday.
To test this I used the following statements:
CALL `timesheetdays`(NOW(), DATE_ADD(DATE(NOW()), INTERVAL '5 0' DAY_HOUR));
SELECT * FROM `calendarhours`;
This tests the procedure from today to 5 days from today and shows the hours as required. The first statement adds the records to the table and then returns the records, the second statement returns the records from the table.
you can use a temporary table in a stored procedure.
DELIMITER ;;
DROP PROCEDURE IF EXISTS ListHours ;;
CREATE PROCEDURE ListHours()
BEGIN
DECLARE curDT DATETIME;
DECLARE today DATETIME ;
DECLARE nextSaturday DATETIME;
DECLARE nextSunday DATETIME;
DECLARE iterDate DATETIME;
DECLARE iterDateTime DATETIME;
DECLARE iterBound DATETIME;
DECLARE resDate DATETIME;
DECLARE resTime DATETIME;
DECLARE delta INT;
DROP TABLE IF EXISTS tempNextWeek;
CREATE TEMPORARY TABLE IF NOT EXISTS tempNextWeek
(
client_date VARCHAR(20),
client_time VARCHAR(20)
);
DELETE FROM tempNextWeek;
SET curDT = NOW();
SET today = ADDTIME(SUBTIME(curDT , TIME(curDT)) , '9:0:0');
SET delta = 8 - DAYOFWEEK(today);
SET nextSunday = ADDDATE(today , INTERVAL delta DAY);
SET nextSaturday = ADDTIME(nextSunday , '6 0:0:0');
-- select today , delta , nextSaturday , nextSunday ;
SET iterDate = nextSunday;
WHILE iterDate <= nextSaturday DO
SET iterDateTime = iterDate;
SET iterBound = ADDTIME(iterDateTime, '8:0:0');
WHILE iterDateTIme <= iterBound DO
INSERT tempNextWeek (client_date, client_time) VALUE ( DATE_FORMAT(iterDateTime, '%Y-%m-%d'), DATE_FORMAT(iterDateTime, '%H:%i:%s') );
SET iterDateTime = ADDTIME(iterDateTime , '1:0:0');
END WHILE;
SET iterDate = ADDTIME(iterDate , '1 0:0:0');
END WHILE ;
SELECT * FROM tempNextWeek;
-- drop table if exists tempNextWeek;
END;;
DELIMITER ;
CALL ListHours();