How to find current day no. in month in php

date, php

Solution

Credit for the Math part goes to Jon (above)

In combination with your code, full solution can be implemented as follows

$t=date('d-m-Y');
$dayName = strtolower(date("D",strtotime($t)));
$dayNum = strtolower(date("d",strtotime($t)));
echo floor(($dayNum - 1) / 7) + 1

or else as a function with optional date

PHP Fiddle here

This just return the number you are requesting.

function dayNumber($date=''){
    if($date==''){
        $t=date('d-m-Y');
    } else {
        $t=date('d-m-Y',strtotime($date));
    }

    $dayName = strtolower(date("D",strtotime($t)));
    $dayNum = strtolower(date("d",strtotime($t)));
    $return = floor(($dayNum - 1) / 7) + 1;
    return $return;
}


echo dayNumber('2014-01-27');

Problem

Today's date is 27-01-2014 so I got day name using following function: ``` $t=date('d-m-Y'); $day = strtolower(date("D",strtotime($t))); ``` So now the day name is `mon`. How to find that this Monday is the forth Monday of current month? In other words, I am trying to find the 1st, 2nd, 3rd, 4th of a particular day (eg. Monday) of a month?

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