PHP 'NumberFormatter' 'SPELLOUT' is not providing required format in english - php

Question

I Am using PHP NumberFormatter in my code to convert price value in words
for example, if it is rupees 125 then it should be 'One hundred and twenty five' instead of 'one hundred twenty five'.
I have tried the other examples like checking each digit unit value and replace the words
$numberFormatterClass = new \NumberFormatter("en", \NumberFormatter::SPELLOUT);
echo str_replace('-', ' ', $numberFormatterClass->format($number));
expecting for 125 = "One hundred and twenty five"

Answer 1

When the number is above 99 you can generate the spellout for the last two digits only. You then know where to insert the "and". In code:
$number = 125;
$numberFormatter = new \NumberFormatter('en', \NumberFormatter::SPELLOUT);
$fullSpellout = str_replace('-', ' ', $numberFormatter->format($number));
if ($number > 100) {
$lastTwoSpellout = str_replace('-', ' ', $numberFormatter->format(substr($number, -2)));
$hunderdsLength = strlen($fullSpellout) - strlen($lastTwoSpellout);
$fullSpellout = substr($fullSpellout, 0, $hunderdsLength) . 'and ' . $lastTwoSpellout;
}
echo $fullSpellout;
This outputs:
one hundred and twenty five
This is certainly not the only possible solution. There are many ways to insert the "and", and if the last two digits always generate two words you could also use that to detect where to insert the "and".
Here's a version based on words and using an array to insert the 'and':
$number = 125;
$numberFormatter = new \NumberFormatter('en', \NumberFormatter::SPELLOUT);
$spellout = str_replace('-', ' ', $numberFormatter->format($number));
if ($number > 100) {
$words = explode(' ', $spellout);
array_splice($words, -2, 0, ['and']);
$spellout = implode(' ', $words);
}
echo $spellout;