How to get all fridays from current month

Question

I have a problem,

$fridays = array();
$fridays[0] = date('Y-m-d', strtotime('first friday of this month'));
$fridays[1] = date('Y-m-d', strtotime('second friday of this month'));
$fridays[2] = date('Y-m-d', strtotime('third friday of this month'));
$fridays[3] = date('Y-m-d', strtotime('fourth friday of this month'));
$fridays[4] = date('Y-m-d', strtotime('fifth friday of this month'));

but there is no fifth friday. Some months have fifth fridays. How to check and not set the last item array?

Answer 1

$fifth = strtotime('fifth friday of this month');

if (date('m') === date('m', $fifth)) {
  $fridays[4] = date('Y-m-d', $fifth);
}

Answer 2

You can do this using the PHP date function. Get the month you want in $timestamp and then do something like this:

<?php
function fridays_get($month, $stop_if_today = true) {

$timestamp_now = time();

for($a = 1; $a < 32; $a++) {

    $day = strlen($a) == 1 ? "0".$a : $a;
    $timestamp = strtotime($month . "-$day");
    $day_code = date("w", $timestamp);
    if($timestamp > $timestamp_now)
        break;
    if($day_code == 5)
        @$fridays++;

}

return $fridays;
}

echo fridays_get('2011-02');

You can find a similar post about this: In PHP, how to know how many mondays have passed in this month uptil today?

Answer 3

I have a function to count fridays in a month to my very own application.... it could be helpful for someone.

 function countFridays($month,$year){ $ts=strtotime('first friday of '.$year.'-'.$month.'-01'); $ls=strtotime('last day of '.$year.'-'.$month.'-01'); $fridays=array(date('Ym-d', $ts)); while(($ts=strtotime('+1 week', $ts))<=$ls){ $fridays[]=date('Ym-d', $ts); }return $fridays; }

Answer 4

I'm not on a machine that I could test this but what about something along the lines of....

if(date('Y-m-d', strtotime('fifth friday of this month')) > ""){
$fridays[4] = date('Y-m-d', strtotime('fifth friday of this month'));
}

The link below does exactly the same thing and will cover exactly what you are wanting todo without clunky if statements like above..

VERY Similar reading: read more...

Answer 5

只有当月有 30 天且第一个星期五是该月的第一天或第二天，或者如果有 31 天且第一个星期五是该月的第一天、第二天或第三天时，该月才会有 5 个星期五，因此您可以进行条件语句根据第 5 个星期五的计算

Answer 6

for($i=0;$i<=5;$i++)
{
       echo date("d/m/y", strtotime('+'.$i.' week friday september 2012'));
}

Answer 7

I used this as the basis for my own solution:

$fridays = array();
$fridays[0] = date('d',strtotime('first fri of this month'));
$fridays[1] = $fridays[0] + 7;
$fridays[2] =  $fridays[0] + 14;
$fridays[3] =  $fridays[0] + 21;
$fridays['last'] = date('d',strtotime('last fri of this month'));

if($fridays[3] == $fridays['last']){
  unset($fridays['last']);
}
else {
  $fridays[4] = $fridays['last'];
  unset($fridays['last']);
}

print_r($fridays);

I needed to get an array of every Friday in a month, even if there were 5 and this seemed to do the trick using the original question as my basis.

Answer 8

<?php //php 7.0.8

$offDays = array();

$date = date('2020-02');
$day= 'Friday';
$offDays[0] = date('d',strtotime("first {$day} of ".$date));
$offDays[1] = $offDays[0] + 7;
$offDays[2] =  $offDays[0] + 14;
$offDays[3] =  $offDays[0] + 21;
$offDays['last'] = date('d',strtotime("last {$day} of ".$date));

if($offDays[3] == $offDays['last']){
  unset($offDays['last']);
}
else {
  $offDays[4] = $offDays['last'];
  unset($offDays['last']);
}

foreach($offDays as $off){
    echo date('Y-m-d-D',strtotime(date($date."-".$off)));
    echo "\n";
}

?>