I am trying to get one week earlier then current week of the year but my sql query is returning null. here is my query
select date_sub(yearweek('2014-01-01'),INTERVAL 1 week)
what is wrong with this query
If you want to get YEARWEEK of week prior to date, you can do this:
Note: YEARWEEK results in 6-digit number, first 4 digits are week year, trailing 2 digits are week number.
SELECT YEARWEEK('2014-01-01' - INTERVAL 1 WEEK)
If you need to get a date that is one week before a given date, then:
SELECT '2014-01-01' - INTERVAL 1 WEEK
Try this
select date_sub(date('2014-01-01'),INTERVAL 1 week)
Try this:-
DATE_SUB(date('2014-01-01'), INTERVAL 7 DAY)
or
SELECT '2014-01-01' - INTERVAL 1 WEEK
The problem is that DATE_SUB takes a date as the first arguement, but year week returns yyyyww i.e. not a date. So this:
SELECT yearweek('2014-01-01');
Returns 201352, this is then implicitly casted to a date, since it is not a date the result is null. You can replicate this by doing:
SELECT DATE(yearweek('2014-01-01'));
So if you subtract a week from NULL the result is also NULL.
The fix is to subtract the week first, then get the year week:
SELECT yearweek(date_sub('2014-01-01', INTERVAL 1 WEEK));
And the result is 201351.
Are you looking for week number??? If yes then plz try this if it will work for you
Select DatePart(Week, Date add(day,-7,convert(date time,'01-jan-2014')))
Pleas let me know if you are looking for something else.
SELECT * FROM [table] WHERE WEEKOFYEAR(date) = WEEKOFYEAR(NOW()) - 1;
Related
How to get last day of the month in MySQL by providing month and year as input.
Similar example, To get last day of the month by date as input
Eg: SELECT LAST_DAY(?) as lastDate
Format:
Input month and year: '07/2015'
Output Result: day /*Eg: 30,31*/
I have tried with date Format
SELECT LAST_DAY(DATE_FORMAT('09/2015','%m/%Y')) as lastDate but it din't work
You just need to convert your values to a date. Here is one way:
select last_day(date(concat_ws('-', year, month, 1)))
Get last day of current month
SELECT LAST_DAY('2020-06-18') AS 'Result';
Get last day of next month
SELECT LAST_DAY(CURDATE() + INTERVAL 1 MONTH);
Get Last day of previous month
SELECT LAST_DAY(CURDATE() - INTERVAL 1 MONTH);
set #year:=2010;
set #month:=10;
select last_day(concat(#year,'-',#month,'-01')) as last_day;
SELECT DATE_FORMAT(LAST_DAY('2015-9-1'),'%d')
just add an any number date (day) in the query, and then get last day only using date format %d
I'm trying to solve a task: I have a table containing information about ships' battles. Battle is made of name and date. The problem is to get the last friday of the month when the battle occurred.
WITH num(n) AS(
SELECT 0
UNION ALL
SELECT n+1 FROM num
WHERE n < 31),
dat AS (
SELECT DATEADD(dd, n, CAST(battles.date AS DATE)) AS day,
dateadd(dd, 0, cast(battles.date as date)) as fight,
name FROM num,battles)
SELECT name, fight, max(day) FROM dat WHERE DATENAME(dw, day) = 'friday'
I thought there must be a maximum of date or something, but my code is wrong.
The result should look like this:
Please, help!!
P.S. DATE_FORMAT is not available
Possible problem: as spencer7593 noticed - and as I should have done and didn't - your original query is not MySQL at all. If you're porting a query that's OK. Otherwise this answer will not be helpful, as it makes use of MySQL functions.
The day you want is number 4 (0 being Sunday in MySQL).
So you want the last day of the month if the last day of the month is a 4; if the day of the month is a 5 you want a date which is 1 day earlier; if the day of the month is a 3 you want a date which is 1 day later, but that's impossible (the month ends), so you really need a date six days earlier.
This means that if the daynumber difference is negative, you want it modulo seven.
You can then build this expression (#DATE is your date; I use a fake date for testing)
SET #DATE='2015-02-18';
DATE_SUB(LAST_DAY(#DATE), INTERVAL ((WEEKDAY(LAST_DAY(#DATE))+7-4))%7 DAY);
It takes the last day of the month (LASTDAY(#DATE)), then it computes its weekday, getting a number from 0 to 6. Adds seven to ensure positivity after subtracting; then subtract the desired daynumber, in this case 4 for Friday.
The result, modulo seven, is the difference (always positive) from the last day's daynumber to the wanted daynumber. Since DATE_SUB(date, 0) returns the argument date, we needn't use IF.
SET #DATE='1962-10-20';
SELECT DATE_SUB(LAST_DAY(#DATE), INTERVAL ((WEEKDAY(LAST_DAY(#DATE))+7-4))%7 DAY) AS friday;
+------------+
| friday |
+------------+
| 1962-10-26 |
+------------+
Your query then would become something like:
SELECT `name`, `date`,
DATE_SUB(LAST_DAY(`date`),
INTERVAL ((WEEKDAY(LAST_DAY(`date`))+7-4))%7 DAY) AS friday
FROM battles;
I store a date in my database as a string like this:
03/08/2013 --> 8th of march
I'm trying to select only the rows that are the same day as the current day:
SELECT * FROM wp_aerezona_booking WHERE DATE_SUB(CURDATE(),INTERVAL 1
DAY) <= STR_TO_DATE(date, '%m/%d/%Y')
The above is what I tried, but it is returning a lot of results and should only return 1.
This should work already:
SELECT * FROM wp_aerezona_booking
WHERE STR_TO_DATE('03/08/2013', '%m/%d/%Y') = CURDATE();
By using the DATE_SUB you are subtracting1 day from the current day. You're not looking at today but yesterday. Also the <= makes you look at yesterday and all days before that.
Then you don't want <=, but you want =. The former will get all results if date is less than or equal to yesterday's date. I'm not sure that you even want the DATE_SUB either.
If you want the same date as today's date then you have to use "=" operator with.
SELECT *
FROM wp_aerezona_booking
WHERE STR_TO_DATE(date, '%m/%d/%Y')= CURDATE()
For "2012-07-12", how can I get the start of the week, i.e., "2012-07-08", and start of the month, i.e., "2012-07-01"?
First day of the month:
SELECT DATE_FORMAT('2007-07-12', '%Y-%m-01');
output: 2007-07-01
First day of the week:
SELECT DATE_SUB('2007-07-12', INTERVAL DAYOFWEEK('2007-07-12')-1 DAY);
output: 2007-07-08
MySQL reference: http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_date-add
Same answer as Borophyll's, but I have changed the behavior of the first day of the month to return a date, not just a string which avoids date formatting/parsing mentioned in user151220's answer.
First day of the month:
SELECT DATE_SUB('2007-07-12', INTERVAL DAYOFMONTH('2007-07-12') - 1 DAY);
output: 2007-07-01
First day of the week:
SELECT DATE_SUB('2007-07-12', INTERVAL DAYOFWEEK('2007-07-12') - 1 DAY);
output: 2007-07-08
MySQL reference: http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_date-add
For those who need Monday as the first day of the week:
SELECT DATE_SUB('2007-07-12', INTERVAL WEEKDAY('2007-07-12') DAY);
output: 2007-07-09
This relies on the WEEKDAY function, which starts with Monday instead of DAYOFWEEK, which starts with Sunday.
The DATE_FORMAT reply from Borophyll is very good, but gives a string rather than a date. So can't be compared easily.
If you need to use this as a comparison to a date field, use str_to_date to reverse it back to date rather than string.
select x from y where date >= str_to_date( DATE_FORMAT(now()-interval 12 month,'Y-%m-01'), '%Y-%m-%d')
If you are (say) looking at 12 months sales figures, but you want to always start off from the 1st of a month.
This will work if you want to just code it and forget about it, it will use datetime now and always return MTD results-
where date_completed between date_sub(date(now()), INTERVAL dayofmonth(now()) -1 day) and now()
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!