显示基于多个日期的内容

Question

I am making a website for a store that wants to show the next open sunday dynamically based on date. I managed to make a script that does so but only tests one date. I want to test 10 dates without putting them in a database. This is my script.

<?

    $mydate = '2012-12-23';
    $curdate = date('Y-m-d');

    if($curdate == $mydate)
    {
        echo 'Vandaag';
    }
    elseif($curdate > $mydate)
    {
        echo '26 december';
    }
    elseif($curdate < $mydate)
    {
        echo '23 december';
    }
    else
    {
        echo 'Koopzondag niet ingeladen';
    }

?>

Answer 1

Maybe this is what you are looking for:

<?php

$mydate = '2012-12-23';
echo date('d F', strtotime(getNextSunday($mydate)));

function getNextSunday($date) {
    $timestamp = strtotime($date);
    if (date('w', $timestamp) == 0) {
         $sunday = date('Y-m-d', $timestamp); // today
    }
    else {
        $sunday = date('Y-m-d', strtotime('next Sunday', $timestamp)); // next sunday
    }
    return $sunday;
}