Question

I need to generate an html code of a list of 10 open days following the current day, by open days I mean business days (m,t,w,t and f), I’m using the following function to translate the date to french:

function f_date() {
    $temps = time();
    $jours = array('Dimanche', 'Lundi', 'Mardi', 'Mercredi', 'Jeudi', 'Vendredi', 'Samedi');
    $jours_numero = date('w', $temps);
    $jours_complet = $jours[$jours_numero];
    $NumeroDuJour = date('d', $temps);
    $mois = array(' ', 'Janvier', 'Février', 'Mars', 'Avril', 'Mai', 'Juin', 'Juillet', 'Août', 'Septembre', 'Octobre', 'Novembre', 'Décembre');
    $mois_numero = date("n", $temps);
    $mois_complet = $mois[$mois_numero];
    $an = date('Y', $temps);
    $fr_temps = "$jours_complet, $NumeroDuJour $mois_complet $an";
    return "$fr_temps";
}
echo "<br/>".f_date();

And i want to generate the following result:

<select name="ladate">
    <option selected="selected" value="Mardi, 29 mai 2012">29 mai 2012</option>
    <option value="Mercredi, 30 mai 2012">30 mai 2012</option></select>
    ....
    <option value="Vendredi, 15 juin 2012">15 juin 2012</option></select>
</select>

Please tell me if you need further information.

Thank you.

Answer 1

Just create a loop, increase the days up to ten and ignore all non open days (Sat, Sun). date('N') is your friend to detect the weekday of a given date.

<?php
$i = $openDay = 0;
while($openDay < 10) {
  $i++;
  $time = strtotime('+'.$i.' days');
  $day = date('N', $time);

  if ($day == 6 or $day == 7) { // ignore Saturdays and Sundays
    continue;
  }

  echo f_date($time).'<br>';
  $openDay++;
}

You also have to modify your date_f() function to use $temps as parameter.

<?php
function f_date($temps = null) {
  // $temps = time();
  // ...
}