Adding Ordinal Contractions to $i

Question

Possible Duplicate:
php display number with ordinal suffix

I'm attempting to add ordinal contractions i.e.(st/nd/rd/th) to an increment.

Somehow I need to get the last digit of $i to test it against my if statements...

Here is my code so far:

    $i = 1;
    while($i < 101 ){

    if($i == 1){$o_c = "st";}else{$o_c = "th";}
    if($i == 2){$o_c = "nd";}
    if($i == 3){$o_c = "rd";}

    echo $i.$o_c."<br/>";
    $i++;

    }

score 0 · Answer 1 · answered Feb 24 '11 at 01:35

0

What about using the modulus operator: $i % 10?

answered Feb 24 '11 at 01:35

D.Shawley

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score 0 · Answer 2 · edited May 23 '17 at 12:02

0

Display numbers with ordinal suffix in PHP

(that thread has other solutions. I liked that one)

edited May 23 '17 at 12:02

Community

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1

answered Feb 24 '11 at 01:35

AlfaTeK

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score 0 · Accepted Answer · answered Feb 24 '11 at 01:36

0

You can use the modulus (%) operator to get the remainder when dividing by 10.

$i = 1;
while($i < 101 ){

$remainder = $i % 10;
if($remainder == 1){$o_c = "st";}else{$o_c = "th";}
if($remainder == 2){$o_c = "nd";}
if($remainder == 3){$o_c = "rd";}

echo $i.$o_c."<br/>";
$i++;

}

answered Feb 24 '11 at 01:36

Alex Deem

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That answer does not work unless 11st is correct (should be 11th). See the duplicate comment to your question. – Jacob Feb 24 '11 at 02:12
Good point. You'd need to special case when i%100 is between 11 & 13 – Alex Deem Feb 24 '11 at 02:31
1

@AlexDeem. 111st isn't correct either. You always need to check the last two digits. – TRiG May 18 '18 at 15:12

Adding Ordinal Contractions to $i

3 Answers3