PHP Generate day range

Solution:

You should pass correct timestamp as the second parameter of date function and usej (without leading zero) day format:

list($m, $y) = explode('-', date('m-Y'));
for($d = 1; $d <= 28; $d++)
{
    $arrayRange[] = date('jS', mktime(0, 0, 0, $m, $d, $y));
}
echo "<pre>";
print_r($arrayRange);
echo "</pre>";
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Solution:

Try this, it uses the contextual date functionality of '+1 day' etc to register your integer as a day :)

To answer your second question - as in what you're doing wrong - you're passing an integer to a function that expects a string.

<?php
for($i = 0; $i <= 27; $i++)
{
//The February is there to keep '1st' a '1st' even on days when it's not 
//the '31st'
$arrayRange[] = date('dS', strtotime("1st february +".$i.' day'));
}

echo "<pre>";
print_r($arrayRange);
echo "</pre>";
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Output:

Array
(
    [0] => 01st
    [1] => 02nd
    ...
    [28] => 28th
)ВЁ
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Edit:
To remove the 0s, you can use ltrim() like this:

$arrayRange[$i] = ltrim(date('dS', strtotime("1st february +".$i.' day')), "0");
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Which will give you an output like this

    [0] => 1st
    [1] => 2nd
    ...
    [28] => 28th
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Edit 2: Fixed it. Props to MLF for noticing the mistake.

Solution:

You can use something like this:

$arrayRange[] = (new DateTime('Aug '.$i))->format('jS');
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