Hello,
I researched a lot and tried many times but I was not successful.
Example
Today is the day of the week 4
Today today date is 2024-04-18
$weekdays = array(4); // If there is today of the week in the series, give today's date.
// Output: 2024-04-18
$weekdays = array(1,3,6); // If today's week does not exist in the series, give the date of the day in the series after today.
// Output: 2024-04-20
Thank you from now
If there is not a later day in the array, do you want to go to the following week? For example, if $weekdays = array(1,3); and today is 4, do you want to use next monday, the 22nd, since it corresponds to the 1 in the array, as the result?
Thank you for the answer
Yes, In such a case, the first number in the array should give the date of the 1st day of the week?
That is, array(2,3). In this case, it should give the dates of 2 days of the next week.
I ran such a code, but I don’t know how accurate or professional the coding is.
It looks like amateur coding
$haftanin_gunu = array(2,3);
$haftaningunu=date('N');
// If the foreach loop does not give any results, return the first value in the array.
$sonraki_gun = $haftanin_gunu[0];
foreach($haftanin_gunu as $number) {
if($number > $haftaningunu) {
$sonraki_gun = $number;
break;
}
}
if($sonraki_gun=='1'){
$h_tarihi=date('d.m.Y', strtotime('noon monday')); // Pazartesi
}elseif($sonraki_gun=='2'){
$h_tarihi=date('d.m.Y', strtotime('noon tuesday')); // Salı
}elseif($sonraki_gun=='3'){
$h_tarihi=date('d.m.Y', strtotime('noon wednesday')); // Çarşamba
}elseif($sonraki_gun=='4'){
$h_tarihi=date('d.m.Y', strtotime('noon thursday')); // Perşembe
}elseif($sonraki_gun=='5'){
$h_tarihi=date('d.m.Y', strtotime('noon friday')); // Cuma
}elseif($sonraki_gun=='6'){
$h_tarihi=date('d.m.Y', strtotime('noon saturday')); // Cumartesi
}elseif($sonraki_gun=='7'){
$h_tarihi=date('d.m.Y', strtotime('noon sunday')); // Pazar
}
echo $h_tarihi;
Almost the same -
// define week day number to day name, for the strtotime('next day_name') expression
$days[1] = 'mon';
$days[2] = 'tue';
$days[3] = 'wed';
$days[4] = 'thu';
$days[5] = 'fri';
$days[6] = 'sat';
$days[7] = 'sun';
$weekdays = array(4); // If there is today of the week in the series, give today's date.
// Output: 2024-04-18
$weekdays = array(1,3,6); // If today's week does not exist in the series, give the date of the day in the series after today.
// Output: 2024-04-20
$weekdays = array(2,3);
// 2024-04-23
// get today's week day number (1-7)
$wd = intval(date('N'));
// if today's week day number is in the array, use today's date
if(in_array($wd,$weekdays))
{
$date = date('Y-m-d');
}
else
{
// find the next week day number and use it's date
$found = false;
foreach($weekdays as $day)
{
// if a day > today is found
if($day > $wd)
{
$found = true;
// exit the loop, $day is the found entry
break;
}
}
if(!$found)
{
// a day > today not found, use the first entry
$day = $weekdays[0];
}
echo "Found: $day<br>";
$date = date('Y-m-d',strtotime("next ".$days[$day]));
}
echo $date;Thank you very much,
I think your coding is more professional than my coding. 