How to get "start date" and "end date" of a week from MYSQL when week number is available?

Question

The week() function of MySql gives number of week from 0-53.

I have a weeknumber and I want to have the "start date" and "end date" of the week represented by weeknumber. How can I do that?

If there is no way of getting "start date" and "end date" of a week directly, how can I the "start week day" and "end week day" from the week number? I tried to get the first day (Monday) of the week by using the following query:-

$currentWeek = 7;
for($w = 0; $w <= $currentWeek; $w++)
{
    $actualWeek = intval($w + 1);
    if($w < 10)
        $queryWeek = '0'.$actualWeek;
    else
        $queryWeek = $actualWeek;
    $thisYearWeek = date('Y').$queryWeek;
    $weekMondayQuery = $this->CustomerPayment->query("SELECT STR_TO_DATE('$thisYearWeek', '%X%V %W')");
}

The Jan 1, 2018 was Monday. For the First week, ie. when $thisYearWeek = '201801', I am getting Monday date as 2018-01-08 instead of 2018-01-01.

Try this: stackoverflow.com/questions/16590500/…

Ahmad Rezk
– Ahmad Rezk

2018-02-10 19:38:06 +00:00
Commented Feb 10, 2018 at 19:38 — Ahmad Rezk
– Ahmad Rezk, Commented Feb 10, 2018 at 19:38

akarilimano · Accepted Answer · 2018-02-10 20:50:21Z

How about adding days to the first Monday in a year?

Assuming 2018-01-01 is a Monday in a first week and the number of week you need is $num_week:

SELECT '2018-01-01' + INTERVAL ($num_week-1)*7 DAY as start, 
       '2018-01-01' + INTERVAL $num_week*7-1 DAY as end;

Which for the 7th week gives:

>>> SELECT '2018-01-01' + INTERVAL 42 DAY as start, 
           '2018-01-01' + INTERVAL 48 DAY as end;

start       end
2018-02-12  2018-02-18

EDIT: Added correction if the first day of the year is not a Monday. But your implementation will depend on how you count weeks with a week() function:

>>> select '2016-01-01'+interval ($num_week-week('2016-01-01', 1))*7 - weekday('2016-01-01') day as start, 
           '2016-01-01'+interval ($num_week-week('2016-01-01', 1)+1)*7-1 - weekday('2016-01-01') day as end;

start       end
2016-02-15  2016-02-21

>>> select '2017-01-01'+interval ($num_week-week('2017-01-01', 1))*7 - weekday('2017-01-01') day as start, 
           '2017-01-01'+interval ($num_week-week('2017-01-01', 1)+1)*7-1 - weekday('2017-01-01') day as end;

start       end
2017-02-13  2017-02-19

sorak · Accepted Answer · 2018-02-10 22:27:15Z

Starting from the week number, we can do the following (most variables should be intuitive, but sow is Start Of Week):

SET @weeknum=42;
SET @week=DATE_ADD('2018-01-01', INTERVAL @weeknum WEEK);
SET @dow=DAYOFWEEK(@week)-1;
SET @sow=DATE_SUB(@week, INTERVAL @dow DAY);
SELECT @week, @dow, @sow;
SELECT @sow AS startDay, DATE_ADD(@sow, INTERVAL +6 DAY) AS endDay;

This tells us that week 42 for 2018 starts on 2018-10-21 and ends on 2018-10-27 ... remember that weeks of the year are zero-indexed using this technique. Days of the week as returned by MySQL are 1-indexed, hence the subtraction in @dow assignment.

The PHP equivalent of this is somewhat easier:

date('Y-m-d', strtotime('sunday last week '.date('Y-m-d', strtotime('2018-01-01 +42 weeks'))))

which returns: 2018-10-21 ... using "last week" because it considers monday the first day of the week.

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