PHP - Loop while 'false', stop when it is 'true' - php

Question

I want to get tomorrow date. But if tomorrow is Saturday or Sunday. It will get date in Monday. It will skip Saturday and Sunday. Also if tomorrow is holiday, it will skip the date and get the next date. I have list of holiday date in my database.
I've trying this code. Suppose today is Friday, May 24.
$today = "2013-05-24";
$tommorow = date('Y-m-d', strtotime($today . ' + 1 day'));
$valid = check_valid($tommorow);
while (!$valid){
$tommorow = date('Y-m-d', strtotime($today . ' + 1 day'));
$valid = check_valid($tommorow);
if($valid){
break;
}
}
function check_valid($date){
return true;
$timestamp = strtotime($date);
$day = date('D', $timestamp);
if ($day == "Sat" || $day == "Sun"){
return false;
}
$mysql= mysql_query("SELECT * FROM holiday_data WHERE date = '$date'");
if (mysql_num_rows($mysql) >= 1){
return false;
}
}
I'm using while clause because i assume that there is holiday date that appears in two or three days in a row. For example, holiday is on May 27 and 28, then tomorrow date should be in Wednesday.
Any ideas?
Could you help me please?
If you have another approach to achieve this, I also want to know it.
Thank you,

Answer 1

Here is the code, doing, what you need:
It uses DateTime class and do-while loop:
<?php
function not_valid($date){
$weekday = $date->format('l');
if ($weekday == "Saturday" || $weekday == "Sunday"){
return true;
}
$str_date = $date->format('Y-m-d');
$mysql= mysql_query("SELECT * FROM holiday_data WHERE date = '$str_date'");
if (mysql_num_rows($mysql) >= 1){
return true;
}
return false;
}
$today = new DateTime("2013-05-24");
do{
//add 1 day
$today->add(new DateInterval('P1D'));
}while(not_valid($today));
print $today->format('Y-m-d');
?>
At the beginning it is 24.05.2013 - Friday. Then it adds 1 to it. 25 and 26 are Saturday and Sunday, so it skips them and prints 2013-05-27.

Answer 2

As Syjin have already said, your check_valid() will always return true because you've got an return true; at the beginning of your function.
Move it to the end of your function to return true if the function hasn't returned false before.

Answer 3

This function checks (with your logic) if a timestamp is valid and if not searches for the next valid timestamp. it will return a timestamp of the first valid timestamp possible.
function getNextValidDay($timestamp) {
$day = date('D', $timestamp);
if ($day == "Sat" || $day == "Sun"){
$timestamp += 60*60*24;
return getNextValidDay($timestamp);
}
$mysql= mysql_query("SELECT * FROM holiday_data WHERE date = '$timestamp'");
if (mysql_num_rows($mysql) >= 1){
$timestamp += 60*60*24;
return getNextValidDay($timestamp);
}
return $timestamp;
}

Answer 4

The problem is in the function check_valid(). You have returned true in the very beginning of the function so it will not allow to process remaining code so it should be something like this,
function check_valid($date){
$timestamp = strtotime($date);
$day = date('D', $timestamp);
if ($day == "Sat" || $day == "Sun"){
return false;
}
$mysql= mysql_query("SELECT * FROM holiday_data WHERE date = '$date'");
if (mysql_num_rows($mysql) >= 1){
return false;
}
return true;
}

Answer 5

you should update/increment '+ 1 day' in each loop. change your code to be something like this
$today = "2013-05-24";
$ii_day = 1;
$tommorow = date('Y-m-d', strtotime($today . ' + '.($ii_day).' day'));
$valid = check_valid($tommorow);
while (!$valid){
$ii_day++;
$tommorow = date('Y-m-d', strtotime($today . ' + '.($ii_day).' day'));
$valid = check_valid($tommorow);
if($valid){
break;
}
}
and then on your check_valid function, change your code a bit.
function check_valid($date){
//return true;
$timestamp = strtotime($date);
$day = date('D', $timestamp);
if ($day == "Sat" || $day == "Sun"){
return false;
}
else {
$mysql= mysql_query("SELECT * FROM holiday_data WHERE date = '$date'");
if (mysql_num_rows($mysql) >= 1){
return false;
}
else
return true;
}
}
hopefully it works..:)

Answer 6

You can try this solution, it's based on my working days code that I use in a couple of my projects. It uses Swedish holidays but you can adjust these to the holidays that are observed in your country. Remember that many holidays depend on the current year.
function next_valid_day() {
$year = date('Y');
$easter_date = easter_date($year);
$holidays = array(
mktime(0, 0, 0, 1, 1, $year), // Nyårsdagen
mktime(0, 0, 0, 1, 6, $year), // Trettondedag jul
//strtotime('-3 days', $easter_date), // Skärtorsdagen
strtotime('-2 days', $easter_date), // Långfredagen
strtotime('-1 day', $easter_date), // Påskafton
$easter_date, // Påskdagen
strtotime('+1 day', $easter_date), // Annandag påsk
mktime(0, 0, 0, 5, 1, $year), // 1:a maj
strtotime('+39 days', $easter_date), // Kristi himmelsfärdsdag
mktime(0, 0, 0, 6, 6, $year), // Sveriges nationaldag
strtotime('-1 day', strtotime('next Saturday', mktime(0, 0, 0, 6, 19, $year))), // Midsommarafton
mktime(0, 0, 0, 12, 24, $year), // Julafton
mktime(0, 0, 0, 12, 25, $year), // Juldagen
mktime(0, 0, 0, 12, 26, $year), // Annandag jul
mktime(0, 0, 0, 12, 31, $year) // Nyårsafton
);
$valid_date = false;
$d = 0;
while ($valid_date === false) {
$date = strtotime('+' . ++$d . ' days');
if (!in_array($date, $holidays) && !(date('N', $date) > 5)) {
$valid_date = true;
}
}
return date('Y-m-d', $date);
}
var_dump(next_valid_day());