Display numbers with ordinal suffix in PHP

Question

I want to display numbers as follows

• 1 as 1st,
• 2 as 2nd,
• ...,
• 150 as 150th.

How should I find the correct ordinal suffix (st, nd, rd or th) for each number in my code?

Answer 1

from wikipedia:

$ends = array('th','st','nd','rd','th','th','th','th','th','th');
if (($number %100) >= 11 && ($number%100) <= 13)
   $abbreviation = $number. 'th';
else
   $abbreviation = $number. $ends[$number % 10];

Where $number is the number you want to write. Works with any natural number.

As a function:

function ordinal($number) {
    $ends = array('th','st','nd','rd','th','th','th','th','th','th');
    if ((($number % 100) >= 11) && (($number%100) <= 13))
        return $number. 'th';
    else
        return $number. $ends[$number % 10];
}
//Example Usage
echo ordinal(100);

Answer 2

PHP has built-in functionality for this. It even handles internationalization!

$locale = 'en_US';
$nf = new NumberFormatter($locale, NumberFormatter::ORDINAL);
echo $nf->format($number);

Note that this functionality is only available in PHP 5.3.0 and later.

Answer 3

This can be accomplished in a single line by leveraging similar functionality in PHP's built-in date/time functions. I humbly submit:

Solution:

function ordinalSuffix( $n )
{
  return date('S',mktime(1,1,1,1,( (($n>=10)+($n>=20)+($n==0))*10 + $n%10) ));
}

Detailed Explanation:

The built-in date() function has suffix logic for handling nth-day-of-the-month calculations. The suffix is returned when S is given in the format string:

date( 'S' , ? );

Since date() requires a timestamp (for ? above), we'll pass our integer $n as the day parameter to mktime() and use dummy values of 1 for the hour, minute, second, and month:

date( 'S' , mktime( 1 , 1 , 1 , 1 , $n ) );

This actually fails gracefully on values out of range for a day of the month (i.e. $n > 31) but we can add some simple inline logic to cap $n at 29:

date( 'S', mktime( 1, 1, 1, 1, ( (($n>=10)+($n>=20))*10 + $n%10) ));

The only positive value(May 2017) this fails on is $n == 0, but that's easily fixed by adding 10 in that special case:

date( 'S', mktime( 1, 1, 1, 1, ( (($n>=10)+($n>=20)+($n==0))*10 + $n%10) ));

Update, May 2017

As observed by @donatJ, the above fails above 100 (e.g. "111st"), since the >=20 checks are always returning true. To reset these every century, we add a filter to the comparison:

date( 'S', mktime( 1, 1, 1, 1, ( (($n>=10)+($n%100>=20)+($n==0))*10 + $n%10) ));

Just wrap it in a function for convenience and off you go!

Answer 4

Here is a one-liner:

$a = <yournumber>;
echo $a.substr(date('jS', mktime(0,0,0,1,($a%10==0?9:($a%100>20?$a%10:$a%100)),2000)),-2);

Probably the shortest solution. Can of course be wrapped by a function:

function ordinal($a) {
  // return English ordinal number
  return $a.substr(date('jS', mktime(0,0,0,1,($a%10==0?9:($a%100>20?$a%10:$a%100)),2000)),-2);
}

Regards, Paul

EDIT1: Correction of code for 11 through 13.

EDIT2: Correction of code for 111, 211, ...

EDIT3: Now it works correctly also for multiples of 10.

Answer 5

from http://www.phpro.org/examples/Ordinal-Suffix.html

<?php

/**
 *
 * @return number with ordinal suffix
 *
 * @param int $number
 *
 * @param int $ss Turn super script on/off
 *
 * @return string
 *
 */
function ordinalSuffix($number, $ss=0)
{

    /*** check for 11, 12, 13 ***/
    if ($number % 100 > 10 && $number %100 < 14)
    {
        $os = 'th';
    }
    /*** check if number is zero ***/
    elseif($number == 0)
    {
        $os = '';
    }
    else
    {
        /*** get the last digit ***/
        $last = substr($number, -1, 1);

        switch($last)
        {
            case "1":
            $os = 'st';
            break;

            case "2":
            $os = 'nd';
            break;

            case "3":
            $os = 'rd';
            break;

            default:
            $os = 'th';
        }
    }

    /*** add super script ***/
    $os = $ss==0 ? $os : '<sup>'.$os.'</sup>';

    /*** return ***/
    return $number.$os;
}
?> 

Answer 6

Simple and Easy Answer will be:

$Day = 3; 
echo date("S", mktime(0, 0, 0, 0, $Day, 0));

//OUTPUT - rd

Answer 7

I wrote this for PHP4. It's been working ok & it's pretty economical.

function getOrdinalSuffix($number) {
    $number = abs($number) % 100;
    $lastChar = substr($number, -1, 1);
    switch ($lastChar) {
        case '1' : return ($number == '11') ? 'th' : 'st';
        case '2' : return ($number == '12') ? 'th' : 'nd';
        case '3' : return ($number == '13') ? 'th' : 'rd'; 
    }
    return 'th';  
}

Answer 8

you just need to apply given function.

function addOrdinalNumberSuffix($num) {
  if (!in_array(($num % 100),array(11,12,13))){
    switch ($num % 10) {
      // Handle 1st, 2nd, 3rd
      case 1:  return $num.'st';
      case 2:  return $num.'nd';
      case 3:  return $num.'rd';
    }
  }
  return $num.'th';
}

Answer 9

Generically, you can use that and call echo get_placing_string(100);

<?php
function get_placing_string($placing){
    $i=intval($placing%10);
    $place=substr($placing,-2); //For 11,12,13 places

    if($i==1 && $place!='11'){
        return $placing.'st';
    }
    else if($i==2 && $place!='12'){
        return $placing.'nd';
    }

    else if($i==3 && $place!='13'){
        return $placing.'rd';
    }
    return $placing.'th';
}
?>

Answer 10

I made a function that does not rely on the PHP's date(); function as it's not necessary, but also made it as compact and as short as I think is currently possible.

The code: (121 bytes total)

function ordinal($i) { // PHP 5.2 and later
  return($i.(($j=abs($i)%100)>10&&$j<14?'th':(($j%=10)>0&&$j<4?['st', 'nd', 'rd'][$j-1]:'th')));
}

More compact code below.

It works as follows:

printf("The %s hour.\n",    ordinal(0));   // The 0th hour.
printf("The %s ossicle.\n", ordinal(1));   // The 1st ossicle.
printf("The %s cat.\n",     ordinal(12));  // The 12th cat.
printf("The %s item.\n",    ordinal(-23)); // The -23rd item.

Stuff to know about this function:

• It deals with negative integers the same as positive integers and keeps the sign.
• It returns 11th, 12th, 13th, 811th, 812th, 813th, etc. for the -teen numbers as expected.
• It does not check decimals, but will leave them in place (use floor($i), round($i), or ceil($i) at the beginning of the final return statement).
• You could also add format_number($i) at the beginning of the final return statement to get a comma-separated integer (if you're displaying thousands, millions, etc.).
• You could just remove the $i from the beginning of the return statement if you only want to return the ordinal suffix without what you input.

This function works commencing PHP 5.2 released November 2006 purely because of the short array syntax. If you have a version before this, then please upgrade because you're nearly a decade out of date! Failing that, just replace the in-line ['st', 'nd', 'rd'] with a temporary variable containing array('st', 'nd', 'rd');.

The same function (without returning the input), but an exploded view of my short function for better understanding:

function ordinal($i) {
  $j = abs($i); // make negatives into positives
  $j = $j%100; // modulo 100; deal only with ones and tens; 0 through 99

  if($j>10 && $j<14) // if $j is over 10, but below 14 (so we deal with 11 to 13)
    return('th'); // always return 'th' for 11th, 13th, 62912th, etc.

  $j = $j%10; // modulo 10; deal only with ones; 0 through 9

  if($j==1) // 1st, 21st, 31st, 971st
    return('st');

  if($j==2) // 2nd, 22nd, 32nd, 582nd
    return('nd'); // 

  if($j==3) // 3rd, 23rd, 33rd, 253rd
    return('rd');

  return('th'); // everything else will suffixed with 'th' including 0th
}

Code Update:

Here's a modified version that is 14 whole bytes shorter (107 bytes total):

function ordinal($i) {
  return $i.(($j=abs($i)%100)>10&&$j<14?'th':@['th','st','nd','rd'][$j%10]?:'th');
}

Or for as short as possible being 25 bytes shorter (96 bytes total):

function o($i){return $i.(($j=abs($i)%100)>10&&$j<14?'th':@['th','st','nd','rd'][$j%10]?:'th');}

With this last function, simply call o(121); and it'll do exactly the same as the other functions I listed.

Code Update #2:

Ben and I worked together and cut it down by 38 bytes (83 bytes total):

function o($i){return$i.@(($j=abs($i)%100)>10&&$j<14?th:[th,st,nd,rd][$j%10]?:th);}

We don't think it can possibly get any shorter than this! Willing to be proven wrong, however. :)

Hope you all enjoy.

Answer 11

An even shorter version for dates in the month (up to 31) instead of using mktime() and not requiring pecl intl:

function ordinal($n) {
    return (new DateTime('Jan '.$n))->format('jS');
}

or procedurally:

echo date_format(date_create('Jan '.$n), 'jS');

This works of course because the default month I picked (January) has 31 days.

Interestingly enough if you try it with February (or another month without 31 days), it restarts before the end:

...clip...
31st
1st
2nd
3rd

so you could count up to this month's days with the date specifier t in your loop: number of days in the month.

Answer 12

function ordinal($number){

    $last=substr($number,-1);
    if( $last>3 || $last==0 || ( $number >= 11 && $number <= 19 ) ){
      $ext='th';
    }else if( $last==3 ){
      $ext='rd';
    }else if( $last==2 ){
      $ext='nd';
    }else{
      $ext='st';
    }
    return $number.$ext;
  }

Answer 13

Here is the correct solution

$numberFormatter = new NumberFormatter('en_US', NumberFormatter::ORDINAL);

echo $numberFormatter->format(11);

Answer 14

Found an answer in PHP.net

<?php
function ordinal($num)
{
    // Special case "teenth"
    if ( ($num / 10) % 10 != 1 )
    {
        // Handle 1st, 2nd, 3rd
        switch( $num % 10 )
        {
            case 1: return $num . 'st';
            case 2: return $num . 'nd';
            case 3: return $num . 'rd';  
        }
    }
    // Everything else is "nth"
    return $num . 'th';
}
?>

Answer 15

Here's another very short version using the date functions. It works for any number (not constrained by days of the month) and takes into account that *11th *12th *13th does not follow the *1st *2nd *3rd format.

function getOrdinal($n)
{
    return $n . date_format(date_create('Jan ' . ($n % 100 < 20 ? $n % 20 : $n % 10)), 'S');
}

Answer 16

I realise this is an ancient post, however it's probably worth adding that as of PHP 8, the match control structure allows a more concise:

$ordinal = match(in_array(abs($position)%100, [11, 12, 13]) ? 0 : abs($position)%10) { 1 => 'st', 2 => 'nd', 3 => 'rd', default => 'th' };

If you need to append that to the position to get a place:

$place = $position.match(in_array(abs($position)%100, [11, 12, 13]) ? 0 : abs($position)%10) { 1 => 'st', 2 => 'nd', 3 => 'rd', default => 'th' };

You can obviously simplify that out if you know your position will always be positive or less than 11 or whatever.

Answer 17

I fond this small snippet

<?php

  function addOrdinalNumberSuffix($num) {
    if (!in_array(($num % 100),array(11,12,13))){
      switch ($num % 10) {
        // Handle 1st, 2nd, 3rd
        case 1:  return $num.'st';
        case 2:  return $num.'nd';
        case 3:  return $num.'rd';
      }
    }
    return $num.'th';
  }

?>

HERE