Question

I have a mySQL table to indicate weather the shop is open or closed, based on time frames:

 shift table
--------------
 shift_id
 day_of_week    (1-7 : Monday-Sunday)
 open_time      (time, default NULL)
 close_time     (time, default NULL)
 type           (1 || 2)

Explanation of field type:
1 indicates the first timeframe within the same day and 2 indicates that the timeframe is the second timeframe within the same day.
For example shop opens on 13:00 – 15:00 AND 18:00 – 01:00.

I am getting the current day using the method below:

$jd = cal_to_jd(CAL_GREGORIAN,date("m"),date("d"),date("Y"));
    $day = jddayofweek($jd, 0);

    switch($day){
        case 0:
            $curDay = 7;
        break;
        default:
            $curDay = $day;
        break;
    }
    return $curDay;

I am getting the current time using the method below:

$timeString = "%H:%i:%s";
$curTime = mdate($timeString, time());

Complex example follows:
Current day: Monday
Current time: 02:00.
Sunday timeframe: 15:00 – 18:00 and 21:00 – 02:30.
Monday timeframe: 08:30 – 15:30.

Clearly the shift is OPEN but based on Sunday s timeframe and not on Monday s which is the current day.

What is the best way to query my table (or alter my table if needed) or the 100% accurate method to define each time if the shift is open or closed, in a given time of a given day ?

Answer 1

This is what I was thinking:

CREATE TABLE Shifts (
  shift_id INT AUTO_INCREMENT PRIMARY KEY,
  start_day CHAR(9) NOT NULL,
  start_time TIME NOT NULL,
  end_day CHAR(9) NOT NULL,
  end_time TIME NOT NULL
) ENGINE = InnoDB;

Now that I think about it, I’m not really sure you even need end_day. However, using:

INSERT INTO Shifts (
  start_day,
  start_time,
  end_day,
  end_time
) VALUES
( LOWER('Sunday'), '15:00:00', LOWER('Sunday'), '18:30:00' ),
( LOWER('Sunday'), '21:00:00', LOWER('Monday'), '26:30:00' ),
( LOWER('Monday'), '12:00:00', LOWER('Monday'), '18:00:00' ),
( LOWER('Tuesday'), '10:00:00', LOWER('Tuesday'), '20:45:00' ),
( LOWER('Wednesday'), '16:00:00', LOWER('Wednesday'), '19:30:00' ),
( LOWER('Thursday'), '10:00:00', LOWER('Thursday'), '17:00:00' ),
( LOWER('Thursday'), '19:00:00', LOWER('Friday'), '25:30:00' ),
( LOWER('Friday'), '16:00:00', LOWER('Saturday'), '24:30:00' ),
( LOWER('Saturday'), '15:00:00', LOWER('Saturday'), '20:00:00' ),
( LOWER('Saturday'), '18:30:00', LOWER('Sunday'), '27:00:00' )

You could then use something like the following:

SELECT DAYNAME(NOW()), start_day
FROM Shifts
WHERE (start_day = LOWER(DAYNAME(NOW()))
       AND start_time < CURTIME()
       AND end_time > CURTIME())
   OR (start_day = LOWER(DAYNAME(DATE_SUB(NOW(), INTERVAL 1 DAY)))
       AND start_time < ADDTIME('24:00:00', CURTIME())
       AND end_time > ADDTIME('24:00:00', CURTIME()))

Which you can test out here:

http://sqlfiddle.com/#!2/41a26/34

And try it with different test values:

SELECT DAYNAME(NOW()), start_day
FROM Shifts
WHERE (start_day = LOWER('Friday')
       AND start_time < '01:12:35'
       AND end_time > '01:12:35')
   OR (start_day = LOWER('Thursday')
       AND start_time < ADDTIME('24:00:00', '01:12:35')
       AND end_time > ADDTIME('24:00:00', '01:12:35'))

http://sqlfiddle.com/#!2/41a26/33