I want to extract week from datetime, the output I want is 'YY/week', where week is the week of the year (eg '201724' is the 24th week in 2017).
The term "week of the year" is too ambiguous.
The week may start from Sunday, Monday or another weekday
The weeks enumeration in the year may start from 0 or 1
The weeks enumeration in the year may start from the week which includes January, 1 (and hence may be partial) or from first complete week of the year
The last week of the year, if it is partial, may be counted or not
Each DBMS has its own functions (sometimes original, always with original names) that can return the number of the week in the year on a given date. But they can not always take into account the above features.
Important addition provided by jarlh:
ISO 8601 (#4.3.4):
The first calendar week of a year is the one that includes the first Thursday of that year.
The last calendar week of a calendar year is the week immediately preceding the first calendar week of the next calendar year.
Week 1 is the first week of a year.
A calendar week starts on a Monday.
ISO 9075 doesn't even mention weeks.
SELECT TO_CHAR(TO_DATE('19-FEB-22') , 'IW') from DUAL;
To get the corresponding four-digit year, use
SELECT TO_CHAR(TO_DATE('19-FEB-22'), 'IYYY') FROM DUAL;
TO_CHAR() having so many options like this read more in Oracle manual or extract portation of date Extract Portion of Date Time Value
OutPut
You can use the following Mysql type query to extract.
SELECT DATE_FORMAT(BirthDate, " %u %Y") FROM Employees;
where the BirthDate date column in the database and the Employees is the table name.
This will result
49 1968
08 1952
35 1963
Week and the year.
in postgresql:
SELECT to_char('2016-12-31 13:30:15'::timestamp without time zone, 'yy/ww') ;
result:
16/53
Related
I am having problems with date formats, was trying many formatting solutions but non of them was working.
I do have table with dates and I am summing repeating dates:
SELECT
tt.time,
DATE_FORMAT(tt.time, '%x-%v') AS time_label,
SUM(value) AS value
FROM
time_table tt
GROUP BY DATE_FORMAT(tt.time, '%x-%v')
ORDER BY time ASC
As we can see it's formatting end year date as a new year date. w3school %x is saying
Year for the week where Monday is the first day of the week. Used with %V
and for %v
Week where Monday is the first day of the week (01 to 53). Used with %X
The first week of the year 2020 started on Dec 30, 2019. See Week Numbers for 2020.
Therefore 2019-12-30 12:42:53 is being formatted correctly as 2020-01.
just wondering if anyone can shed some light on why why the following query (Note: today's date is 1 May 2018)
SELECT WEEKOFYEAR(NOW()); gives a result of 18.
BUT the query:
SELECT WEEK('2018-05-1') (using all the different modes 0 -7) gives a result of 17?
Shouldn't the week of year for NOW() technically be the same as the week of the year for: 2018-05-01 since today is the 5th of May 2018?
Very curious as to why its not giving the same result.
Many thanks
In weekoftheyear() function, Week number will start from the 1st day of the year.
In Week() function, Week number will start from 1st Monday of the given year.
I want to calculate start date and end date for the particular given week in the month , (assuming my week starts on monday and ends on sunday).
So, What I want is if I select any week number lets say 1,2 ,3 or 4 I should be able to calculate the start date for that particular week and end date .
In addition to the week number for that particular month , I will also give an input of year and month.
selection of year month and week number will be from HTML page through a drop down select option.
And I am looking for some efficient way of doing this .
Use Moment.js
Get the value of week number
var weeknumber = moment("07-27-2017", "MM-DD-YYYY").week(); console.log(weeknumber);
I have the following table 'Schedule' in my database:
**Schedule**
PK Weekday (ie., November 11, 2012)
PK Punch-In (ie., 1:00 PM)
PK Punch-Out (ie., 9:00 PM)
I am tasked with totaling the number of hours worked per week.
For my example Sunday is Day 1 of the week and Saturday is Day 7.
My question is how do I determine which days are in a "week." Is there an SQL command for this? In other words, how do I determine if November 10 is the same week as November 11.
You can use the WEEK() function, which returns week number so WHERE WEEK(date1) = WEEK(date2)
http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_week
http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_week
Week function should return week number, if this is what you want?
We have customers that currently have defined weeks starting either on Sat, Sun or Monday. Came across these DATE_FORMAT options which nicely handle the week starting on Sunday and Monday but can't find a way to do same for week starting on Saturday. Any suggestions?
%U Week (00..53), where Sunday is the first day of the week
%u Week (00..53), where Monday is the first day of the week
I had a similar issue: I needed to calculate week numbers based on the following rules:
Week starts on Friday
The remainder days of an year (all the days after the last Friday of the year that do not complete a week) should be counted in the
first week of the next year.
For example:
27/12/2012 (Thursday) should be Week 52 of 2012
28/12/2012 (Friday) should be Week 1 of 2013
Week 1 2013 goes from 28/12/2012 to 3/1/2013
I made this statement that calculates both the YEAR and WEEKNUMBER based on these rules that you can easily adapt to your circunstance:
SELECT IF(ceil(( dayofyear(current_date) + dayofweek(date_format(current_date, '%Y-01-01'))+1 )/7) > 52, YEAR(current_date)+1, YEAR(current_date)),
IF(ceil(( dayofyear(current_date) + dayofweek(date_format(current_date, '%Y-01-01'))+1 )/7) > 52, 1, ceil(( dayofyear(current_date) + dayofweek(date_format(current_date, '%Y-01-01'))+1 )/7));
The tricky part is just this expression:
ceil(( dayofyear(current_date) + dayofweek(date_format(current_date, '%Y-01-01'))+1 )/7)
The rest (If clauses) are just for adapting the result of the expression to make year+1 and week = 1 on week 53.
I'll try to explain the expression as best as I can. The following expression gives you the week number pure simple (the day of the year divided by 7 days in a week rounded up):
ceil(( dayofyear(current_date))/7)
But now you want to make it start on Friday (or any other day). To do this you need to add to the current day, the days of the first week that were part of the previous year (it's like your current actually started a few days before, because your first week contains days from the previous year).
This expression calculates that offset based on the weekday on Jan/1:
dayofweek(date_format(current_date, '%Y-01-01'))+OFFSET
The offset is the difference between 7 and the weekdaynumber you want the week to start:
0 for Saturday
1 for Friday
2 for Thursday
3 for Wednesday
...
So now you just have to add it to the previous one resulting in the above mentioned expression that calculates the week numbers starting on any weekday and assuming week 1 to start on the previous year:
ceil(( dayofyear(current_date) + dayofweek(date_format(current_date, '%Y-01-01'))+OFFSET )/7)
Then I just added an IF that turns week 53 into week 1 and another to add 1 to the year if it's week 53.
It took me a while to think on this question.
ISO standard defines the first week to start on Monday and contain 4th day of the year.
MySQL's functions provide much more choices.
date_format() flags %U and %u are using notation where first week is the one where Sunday or Monday is first met. As this is not according to the ISO, I will provide both variants.
If you want to count week numbers starting from Saturday and first year's week is the one containing Saturday, you can use one of the following expressions:
SELECT sign(dayofweek(current_date) - 7) + ceil(dayofyear(current_date)/7);
SELECT ceil((dayofyear(current_date)+
(dayofweek(date_format(current_date, '%Y-01-01'))%7-7))/7);
If first year's week is the one where 4th day of the year falls into, use:
SELECT ceil((dayofyear(current_date)+
(dayofweek(date_format(current_date, '%Y-01-04'))%7-4+1))/7);
The very first expression is quite straightforward.
I will elaborate on the 2nd and 3rd ones. I calculate week number by taking current day of the year, dividing by 7 and ceiling up, quite simple. Week number needs to be adjusted based on the situation at the beginning of the year though.
for the first case (first week starts with the first Saturday), I take day-of-week for the Jan/1 of the year in subject, make Saturday as the day 0 and then adjust day-of-year by the difference. This this makes all days before first saturday yielding negative adjustment number and it ceils up to zero;
for the second case (first week is the one where 4 day of the year falls in), I take day-of-week for the Jan/4 of the year in subject, make Saturday as the day 0. The -4+1 formula gives adjustment to the first Saturday before Jan/4, +1 is used as days of the year starts from 1, not from 0. Negative adjustment means 1st day of the year is not in the first week of the year.
Here're some test dates on the SQL Fiddle.
If you want to count weeks from any other day, you just have to change the formula, making that day being 0 in the sequence. Say, to count weeks starting from Wednesday, use:
SELECT ceil((dayofyear(current_date)+
((dayofweek(date_format(current_date, '%Y-01-04'))+3)%7-4+1))/7);
+3 is used as it complements dayofweek() value for Wednesday to the 7.
Make an adjustment based on DAYOFWEEK().