My problem is, I have a form named 'weekly-off setting' in which I am selecting a one or two days as a weekly off.If I have weekly off on saturday then I want to find out the first saturdays date which is coming in the first week of the selected month.So can anyone tell me the mysql query for this problem.
Thanks in advance.
It's a little unclear from your statement, but you're trying to find the next Saturday?
select date_add(now(), interval 7-dayofweek(now()) day);
Which unfortunately will return today if you're on a Saturday, so the sequence becomes:
SET #OFFSET = 7-dayofweek(now());
SET #OFFSET = IF(#OFFSET = 0, 7, #OFFSET);
select date_add(now(), interval #OFFSET day);
which can be combined into one:
select date_add(now(), interval IF(7-dayofweek(now()) = 0, 7, 7-dayofweek(now())) day) as next;
Related
I want to know is it possible in mysql query.. when I say give me date when it is 9am.. the return answer is depends upon current time when it is 8am it give me today's date. when it is 10pm it gives me tomorrow date. how it is possible in mysql query.
You can use SUBSTRING_INDEX(CURTIME(), ':', 1) to get the hours of current time.
As I understood you want to get tomorrow date, if it is 10pm or later
Example given:
SELECT
CASE
WHEN SUBSTRING_INDEX(CURTIME(), ':', 1) >= 22
THEN DATE_ADD(CURDATE(), INTERVAL 1 DAY)
ELSE CURDATE()
END
Source: http://www-db.deis.unibo.it/courses/TW/DOCS/w3schools/sql/sql_dates.asp.html
You can get the hour value from a given datetime expression, using HOUR function. CURDATE() function is used to return the current date. You can add/subtract 'integers' to it get the date corresponding to current date +/- 'integer days' . Assuming that the time >= 10 pm returns next day:
SELECT IF(HOUR(`datetime_field`) > 22, CURDATE(), CURDATE() + 1);
You could just add 2 hours
SELECT DATE(DATE_ADD(NOW(), INTERVAL 2 HOUR));
This will then return tomorrow’s date for anytime after 10pm.
How do i extract all rows greater then 7 days of a start date?, I'm trying to use this query in MySQL. Below is my statement.
SELECT * from v_polygons a
INNER JOIN tblProjectData z
on z.Project_ID = a.Project_ID
WHERE DATE_ADD(z.FlyDate, INTERVAL 7 DAY) > NOW() + INTERVAL rge DAY
I have a start date z.FlyDate, So i give it +7 days, then i check to see if that date is greater then NOW()
is this correct or have i messed it up?
You can just do:
WHERE DATE_ADD(DATE(z.FlyDate), INTERVAL 7 DAY) < DATE(NOW());
This will ignore the time part. You can remove DATE function call if you want to consider the time as well.
Here is my problem, I want to fetch next 30 days records from the table. I have a field in my table. For ex: In my table I have resource_date, In this column I have many records from 2013-02-05 to 2015-10-10. Say, If I logged into the website today(Today's Date is- 16/01/2015, It should fetch record for next 15 days and so on). How to do this? Thanks in advance
One way to do it
SELECT *
FROM table1
WHERE resource_date >= CURDATE() + INTERVAL 1 DAY -- skip today
AND resource_date < CURDATE() + INTERVAL 17 DAY -- 15 days starting tomorrow
Here is a SQLFiddle demo
In MySQL, you can use the NOW() function to get the current DATETIME, and the INTERVAL keyword to get intervals of time.
So, to get the records where resource_date is within the next 30 days, you would use:
SELECT *
FROM `my_table_name`
WHERE `resource_date` >= NOW()
AND `resource_date` < NOW() + INTERVAL 1 MONTH
;
In practice, you should rarely use SELECT *, and you should consider adding a LIMIT to this query to prevent your application from returning a result set that is "too large".
http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html
...
WHERE
'resource_date'> NOW() AND
'resource_date'< DATE_ADD(NOW(), INTERVAL 31 DAY);
Careful I think now() does minutes and hours so you miss a portion of a day.
WHERE resource_date >= CURDATE() AND resource_date <= DATE_ADD(CURDATE(), interval 15 DAY)
Suppose I have 2011-01-03 and I want to get the first of the week, which is sunday, which is 2011-01-02, how do I go about doing that?
The reason is I have this query:
select
YEAR(date_entered) as year,
date(date_entered) as week, <-------This is what I want to change to select the first day of the week.
SUM(1) as total_ncrs,
SUM(case when orgin = picked_up_at then 1 else 0 end) as ncrs_caught_at_station
from sugarcrm2.ncr_ncr
where
sugarcrm2.ncr_ncr.date_entered > date('2011-01-01')
and orgin in(
'Silkscreen',
'Brake',
'Assembly',
'Welding',
'Machining',
'2000W Laser',
'Paint Booth 1',
'Paint Prep',
'Packaging',
'PEM',
'Deburr',
'Laser ',
'Paint Booth 2',
'Toolpath'
)
and date_entered is not null
and orgin is not null
AND(grading = 'Minor' or grading = 'Major')
and week(date_entered) > week(current_timestamp) -20
group by year, week(date_entered)
order by year asc, week asc
And yes, I realize that origin is spelled wrong but it was here before I was so I can't correct it as too many internal apps reference it.
So, I am grouping by weeks but I want this to populate my chart, so I can't have all the beginning of weeks looking like different dates. How do I fix this?
If the week starts on Sunday do this:
DATE_ADD(mydate, INTERVAL(1-DAYOFWEEK(mydate)) DAY)
If the week starts on Monday do this:
DATE_ADD(mydate, INTERVAL(-WEEKDAY(mydate)) DAY);
more info
If you need to handle weeks which start on Mondays, you could also do it that way. First define a custom FIRST_DAY_OF_WEEK function:
DELIMITER ;;
CREATE FUNCTION FIRST_DAY_OF_WEEK(day DATE)
RETURNS DATE DETERMINISTIC
BEGIN
RETURN SUBDATE(day, WEEKDAY(day));
END;;
DELIMITER ;
And then you could do:
SELECT FIRST_DAY_OF_WEEK('2011-01-03');
For your information, MySQL provides two different functions to retrieve the first day of a week. There is DAYOFWEEK:
Returns the weekday index for date (1 = Sunday, 2 = Monday, …, 7 = Saturday). These index values correspond to the ODBC standard.
And WEEKDAY:
Returns the weekday index for date (0 = Monday, 1 = Tuesday, … 6 = Sunday).
If week starts on Monday
SELECT SUBDATE(mydate, weekday(mydate));
If week starts on Sunday
SELECT SUBDATE(mydate, dayofweek(mydate) - 1);
Example:
SELECT SUBDATE('2018-04-11', weekday('2018-04-11'));
2018-04-09
SELECT SUBDATE('2018-04-11', dayofweek('2018-04-11') - 1);
2018-04-08
Week starts day from sunday then get First date of the Week and Last date of week
SELECT
DATE("2019-03-31" + INTERVAL (1 - DAYOFWEEK("2019-03-31")) DAY) as start_date,
DATE("2019-03-31" + INTERVAL (7 - DAYOFWEEK("2019-03-31")) DAY) as end_date
Week starts day from Monday then get First date of the Week and Last date of week
SELECT
DATE("2019-03-31" + INTERVAL ( - WEEKDAY("2019-03-31")) DAY) as start_date,
DATE("2019-03-31" + INTERVAL (6 - WEEKDAY("2019-03-31")) DAY) as end_date
select '2011-01-03' - INTERVAL (WEEKDAY('2011-01-03')+1) DAY;
returns the date of the first day of week. You may look into it.
This is a much simpler approach than writing a function to determine the first day of a week.
Some variants would be such as
SELECT DATE_ADD((SELECT CURDATE() - INTERVAL (WEEKDAY(CURDATE())+1)DAY),INTERVAL 7 DAY) (for the ending date of a query, such as between "beginning date" and "ending date").
SELECT CURDATE() - INTERVAL (WEEKDAY(CURDATE())+1) DAY (for the beginning date of a query).
This will return all values for the current week. An example query would be as follows:
SELECT b.foo FROM bar b
WHERE b.datefield BETWEEN
(SELECT CURDATE() - INTERVAL (WEEKDAY(CURDATE())+1) DAY)
AND (SELECT DATE_ADD((SELECT CURDATE() - INTERVAL (WEEKDAY(CURDATE())+1)DAY),INTERVAL 7 DAY))
This works form me
Just make sure both dates in the below query are the same...
SELECT ('2017-10-07' - INTERVAL WEEKDAY('2017-10-07') Day) As `mondaythisweek`
This query returns: 2017-10-02 which is a monday,
But if your first day is sunday, then just subtract a day from the result of this and wallah!
If the week starts on Monday do this:
DATE_SUB(mydate, INTERVAL WEEKDAY(mydate) DAY)
SELECT MIN(DATE*given_date*) FROM *table_name*
This will return when the week started at for any given date.
Keep the good work going!
I want to get first day of every corresponding month of current year. For example, if user selects '2010-06-15', query demands to run from '2010-06-01' instead of '2010-06-15'.
Please help me how to calculate first day from selected date. Currently, I am trying to get desirable using following mysql select query:
Select
DAYOFMONTH(hrm_attendanceregister.Date) >=
DAYOFMONTH(
DATE_SUB('2010-07-17', INTERVAL - DAYOFMONTH('2010-07-17') + 1 DAY
)
FROM
hrm_attendanceregister;
Thanks
Is this what you are looking for:
select CAST(DATE_FORMAT(NOW() ,'%Y-%m-01') as DATE);
You can use the LAST_DAY function provided by MySQL to retrieve the last day of any month, that's easy:
SELECT LAST_DAY('2010-06-15');
Will return:
2010-06-30
Unfortunately, MySQL does not provide any FIRST_DAY function to retrieve the first day of a month (not sure why). But given the last day, you can add a day and subtract a month to get the first day. Thus you can define a custom function:
DELIMITER ;;
CREATE FUNCTION FIRST_DAY(day DATE)
RETURNS DATE DETERMINISTIC
BEGIN
RETURN ADDDATE(LAST_DAY(SUBDATE(day, INTERVAL 1 MONTH)), 1);
END;;
DELIMITER ;
That way:
SELECT FIRST_DAY('2010-06-15');
Will return:
2010-06-01
There is actually a straightforward solution since the first day of the month is simply today - (day_of_month_in_today - 1):
select now() - interval (day(now())-1) day
Contrast that with the other methods which are extremely roundabout and indirect.
Also, since we are not interested in the time component, curdate() is a better (and faster) function than now(). We can also take advantage of subdate()'s 2-arity overload since that is more performant than using interval. So a better solution is:
select subdate(curdate(), (day(curdate())-1))
This is old but this might be helpful for new human web crawlers XD
For the first day of the current month you can use:
SELECT LAST_DAY(NOW() - INTERVAL 1 MONTH) + INTERVAL 1 DAY;
You can use EXTRACT to get the date parts you want:
EXTRACT( YEAR_MONTH FROM DATE('2011-09-28') )
-- 201109
This works well for grouping.
You can use DATE_FORMAT() function in order to get the first day of any date field.
SELECT DATE_FORMAT(CURDATE(),'%Y-%m-01') as FIRST_DAY_CURRENT_MONTH
FROM dual;
Change Curdate() with any other Date field like:
SELECT DATE_FORMAT(purchase_date,'%Y-%m-01') AS FIRST_DAY_SALES_MONTH
FROM Company.Sales;
Then, using your own question:
SELECT *
FROM
hrm_attendanceregister
WHERE
hrm_attendanceregister.Date) >=
DATE_FORMAT(CURDATE(),'%Y-%m-01')
You can change CURDATE() with any other given date.
There are many ways to calculate the first day of a month, and the following are the performance in my computer (you may try this on your own computer)
And the winner is LAST_DAY(#D - interval 1 month) + interval 1 day
set #D=curdate();
select BENCHMARK(100000000, subdate(#D, (day(#D)-1))); -- 33 seconds
SELECT BENCHMARK(100000000, #D - INTERVAL (day(#D) - 1) DAY); -- 33 seconds
SELECT BENCHMARK(100000000, cast(DATE_FORMAT(#D, '%Y-%m-01') as date)); -- 29 seconds
SELECT BENCHMARK(100000000, LAST_DAY(#D - interval 1 month) + interval 1 day); -- 26 seconds
I'm surprised no one has proposed something akin to this (I do not know how performant it is):
CONCAT_WS('-', YEAR(CURDATE()), MONTH(CURDATE()), '1')
Additional date operations could be performed to remove formatting, if necessary
use date_format method and check just month & year
select * from table_name where date_format(date_column, "%Y-%m")="2010-06"
SELECT LAST_DAY(date) as last_date, DATE_FORMAT(date,'%Y-%m-01') AS fisrt_date FROM table_name
date=your column name
The solutions that use last_day() and then add/subtract a month and a day are not interchangeable.
Example:
date_sub(date_add(last_day(curdate()), interval 1 day), interval 3 month)
always works for any supplied number of months you want to go back
date_add(date_sub(last_day(now()), interval 3 month), interval 1 day)
will fail in some cases, for instance if your current month has 30 days and the month you're subtracting back to (and then adding a day) has 31.
date_add(subdate(curdate(), interval day(?) day), interval 1 day)
change the ? for the corresponding date
This works fine for me.
date(SUBDATE("Added Time", INTERVAL (day("Added Time") -1) day))
** replace "Added Time" with column name
Use Cases:
If you want to reset all date fields except Month and Year.
If you want to retain the column format as "date". (not as "text" or "number")
Slow (17s):
SELECT BENCHMARK(100000000, current_date - INTERVAL (day(current_date) - 1) DAY);
SELECT BENCHMARK(100000000, cast(DATE_FORMAT(current_date, '%Y-%m-01') as date));
If you don't need a date type this is faster:
Fast (6s):
SELECT BENCHMARK(100000000, DATE_FORMAT(CURDATE(), '%Y-%m-01'));
SELECT BENCHMARK(100000000, DATE_FORMAT(current_date, '%Y-%m-01'));
select big.* from
(select #date := '2010-06-15')var
straight_join
(select * from your_table where date_column >= concat(year(#date),'-',month(#date),'-01'))big;
This will not create a full table scan.