Hi,

I was wanting to write some code to work out whether I needed st, nd rd or
th in my dates.
While the code below works, I am certain that there is a better way....

if (($daterep == "1") || ($daterep == "21") || ($daterep == "31")){
$letters = "st";
} elseif (($daterep == "2") || ($daterep == "22")){
$letters = "nd";
} elseif (($daterep == "3") || ($daterep == "23")){
$letters = "rd";
} else {
$letters = "th";
}

I tried:

if ($daterep == "1" || "21" || "31") { ... etc

but it did not work... I don't understand why though?
I cast $daterep into a string, but now I find that I could have left it
alone....
Any suggestions as to how to get this code working a bit smoother?

Robert

*** Robert escribió/wrote (Sat, 18 Sep 2004 13:03:06 GMT):[color=blue]
> I was wanting to write some code to work out whether I needed st, nd rd or
> th in my dates.
> While the code below works, I am certain that there is a better way....[/color]

A switch may suit:

switch($daterep){
case '1':
case '21':
case '31':
$letters='st';
break;
case '2':
case '22':
$letters='nd';
break;
case '3':
case '23':
$letters='rd';
break;
default:
$letters='th';
}

[color=blue]
> if ($daterep == "1" || "21" || "31") { ... etc
>
> but it did not work... I don't understand why though?[/color]

"21" is not a valid expression. It should be:

if( ($daterep == "1") || ($daterep == "21") || ($daterep == "31") ){



In any case, check the manual page for date(), you have format characters
that make what you want:

S
English ordinal suffix for the day of the month, 2 characters
st, nd, rd or th.

--
-+ Álvaro G. Vicario - Burgos, Spain
+- http://www.demogracia.com (la web de humor barnizada para la intemperie)
++ Las dudas informáticas recibidas por correo irán directas a la papelera
-+ I'm not a free help desk, please don't e-mail me your questions
--

Thanks Alvaro,

The case method works well.
Also I didn't know about the S. It is definatley easy to use!

My text is missing it! I have found it on the php site now.

Robert



"Alvaro G. Vicario" <kAlvaroNOSPAMTHANKS@terra.es> wrote in message
news:6id4dd3pppcy$.1uhj805imblxe.dlg@40tude.net...[color=blue]
> *** Robert escribió/wrote (Sat, 18 Sep 2004 13:03:06 GMT):[color=green]
>> I was wanting to write some code to work out whether I needed st, nd rd
>> or
>> th in my dates.
>> While the code below works, I am certain that there is a better way....[/color]
>
> A switch may suit:
>
> switch($daterep){
> case '1':
> case '21':
> case '31':
> $letters='st';
> break;
> case '2':
> case '22':
> $letters='nd';
> break;
> case '3':
> case '23':
> $letters='rd';
> break;
> default:
> $letters='th';
> }
>
>[color=green]
>> if ($daterep == "1" || "21" || "31") { ... etc
>>
>> but it did not work... I don't understand why though?[/color]
>
> "21" is not a valid expression. It should be:
>
> if( ($daterep == "1") || ($daterep == "21") || ($daterep == "31") ){
>
>
>
> In any case, check the manual page for date(), you have format characters
> that make what you want:
>
> S
> English ordinal suffix for the day of the month, 2 characters
> st, nd, rd or th.
>
> --
> -+ Álvaro G. Vicario - Burgos, Spain
> +- http://www.demogracia.com (la web de humor barnizada para la
> intemperie)
> ++ Las dudas informáticas recibidas por correo irán directas a la papelera
> -+ I'm not a free help desk, please don't e-mail me your questions
> --[/color]

Robert wrote:[color=blue]
> if (($daterep == "1") || ($daterep == "21") || ($daterep == "31")){
> $letters = "st";
> } elseif (($daterep == "2") || ($daterep == "22")){
> $letters = "nd";
> } elseif (($daterep == "3") || ($daterep == "23")){
> $letters = "rd";
> } else {
> $letters = "th";
> }[/color]

Perhaps a lookup table:
$map[1] = 'st';
$map[21] = 'st';
$map[31] = 'st';
$map[2] = 'nd';
....

--
| Just another PHP saint |
Email: rrjanbiah-at-Y!com

Good Idea,

But then I'd have to put in an item for every day wouldn't I?

Say I wanted the leetters for the 15th.....

Robert

"R. Rajesh Jeba Anbiah" <ng4rrjanbiah@rediffmail.com> wrote in message
news:1095515362.325880.301070@k26g2000oda.googlegr oups.com...[color=blue]
> Robert wrote:[color=green]
>> if (($daterep == "1") || ($daterep == "21") || ($daterep == "31")){
>> $letters = "st";
>> } elseif (($daterep == "2") || ($daterep == "22")){
>> $letters = "nd";
>> } elseif (($daterep == "3") || ($daterep == "23")){
>> $letters = "rd";
>> } else {
>> $letters = "th";
>> }[/color]
>
> Perhaps a lookup table:
> $map[1] = 'st';
> $map[21] = 'st';
> $map[31] = 'st';
> $map[2] = 'nd';
> ...
>
> --
> | Just another PHP saint |
> Email: rrjanbiah-at-Y!com
>[/color]

On Sat, 18 Sep 2004 15:19:36 +0200, "Alvaro G. Vicario"
<kAlvaroNOSPAMTHANKS@terra.es> wrote:
[color=blue][color=green]
>> if ($daterep == "1" || "21" || "31") { ... etc
>>
>> but it did not work... I don't understand why though?[/color]
>
>"21" is not a valid expression. It should be:[/color]

"21" is certainly a valid expression, it's a non-empty, non-"0" string, which
in a Boolean context becomes bool(true).

if ($daterep == "1" || "21" || "31")
->
if ($daterep == true)
->
if ($daterep)

So if $daterep is non-zero and non-empty, the condition is always true.
[color=blue]
>if( ($daterep == "1") || ($daterep == "21") || ($daterep == "31") ){[/color]

Clearly this is what the OP meant to do, but "21" is still an expression.

--
Andy Hassall / <andy@andyh.co.uk> / <http://www.andyh.co.uk>
<http://www.andyhsoftware.co.uk/space> Space: disk usage analysis tool

On Sat, 18 Sep 2004 13:03:06 GMT, "Robert" <lonerngraus@yahoo.com.au> wrote:
[color=blue]
>I was wanting to write some code to work out whether I needed st, nd rd or
>th in my dates.
>While the code below works, I am certain that there is a better way....
>
>if (($daterep == "1") || ($daterep == "21") || ($daterep == "31")){
> $letters = "st";
>} elseif (($daterep == "2") || ($daterep == "22")){
> $letters = "nd";
>} elseif (($daterep == "3") || ($daterep == "23")){
> $letters = "rd";
>} else {
> $letters = "th";
>}
>
>I tried:
>
>if ($daterep == "1" || "21" || "31") { ... etc
>
>but it did not work... I don't understand why though?
>I cast $daterep into a string, but now I find that I could have left it
>alone....
>Any suggestions as to how to get this code working a bit smoother?[/color]

<?php
function ordinal($n)
{
$ordinalSuffix = array('th', 'st', 'nd', 'rd', 'th',
'th', 'th', 'th', 'th', 'th');
if ($n >= 10 && $n <= 19)
return $n . 'th';
else
return $n . $ordinalSuffix[$n % 10];
}

for ($i = 1; $i <= 31; $i++)
{
print ordinal($i) . '<br>';
}
?>

--
Andy Hassall / <andy@andyh.co.uk> / <http://www.andyh.co.uk>
<http://www.andyhsoftware.co.uk/space> Space: disk usage analysis tool

*** Andy Hassall escribió/wrote (Sat, 18 Sep 2004 15:20:28 +0100):[color=blue]
> "21" is certainly a valid expression[/color]

Of course, you are right.

--
-+ Álvaro G. Vicario - Burgos, Spain
+- http://www.demogracia.com (la web de humor barnizada para la intemperie)
++ Las dudas informáticas recibidas por correo irán directas a la papelera
-+ I'm not a free help desk, please don't e-mail me your questions
--

.oO(Robert)
[color=blue]
>Good Idea,
>
>But then I'd have to put in an item for every day wouldn't I?
>
>Say I wanted the leetters for the 15th.....[/color]

With the modulo-operator you can get the last digit (mod 10):

$last = $value % 10;

For $value = 15 this would be 5. A lookup-table could look like this:

$map[0] = 'th';
$map[1] = 'st';
$map[2] = 'nd';
$map[3] = 'rd';

You would simply have to check if there's a key in that table for the
last digit, if not use the element with the key 0 ('th').

Only little problem are numbers ending on 11, 12 or 13, they need a
special treatment.

Another simple algorithm is described in

http://en.wikipedia.org/wiki/How_to_...rdinal_numbers

A possible quick 'n dirty implementation, using both algorithms:

$lookup = array('th', 'st', 'nd', 'rd');
$letters = ($value / 10 % 10) != 1
? isset($lookup[$last = $value % 10]) ? $lookup[$last] : $lookup[0]
: $lookup[0];

This should work for all numbers, but if you only have to take care of
dates the date() function should be fine.

Micha

Robert wrote:[color=blue]
> Hi,
>
> I was wanting to write some code to work out whether I needed st, nd rd or
> th in my dates.
> While the code below works, I am certain that there is a better way....
>
> if (($daterep == "1") || ($daterep == "21") || ($daterep == "31")){
> $letters = "st";
> } elseif (($daterep == "2") || ($daterep == "22")){
> $letters = "nd";
> } elseif (($daterep == "3") || ($daterep == "23")){
> $letters = "rd";
> } else {
> $letters = "th";
> }
>
> I tried:
>
> if ($daterep == "1" || "21" || "31") { ... etc
>
> but it did not work... I don't understand why though?
> I cast $daterep into a string, but now I find that I could have left it
> alone....
> Any suggestions as to how to get this code working a bit smoother?
>
> Robert
>
>[/color]
A little math is in order:
$daterepl = $daterep % 10;
switch ($daterepl) {
case 1:
$letters="st";
break;
case 2:
$letters="nd";
break;
case 3:
$letters="rd";
break;
default:
$letters="th";
}
A nice side effect of this is that it will work for any (positive) integer.

On Sat, 18 Sep 2004 13:02:44 -0400, Robert Stearns <rstearns1241@charter.net>
wrote:
[color=blue]
>A little math is in order:
>$daterepl = $daterep % 10;
>switch ($daterepl) {
> case 1:
> $letters="st";
> break;
> case 2:
> $letters="nd";
> break;
> case 3:
> $letters="rd";
> break;
> default:
> $letters="th";
>}
>A nice side effect of this is that it will work for any (positive) integer.[/color]

11st, 12nd, 13rd...

--
Andy Hassall / <andy@andyh.co.uk> / <http://www.andyh.co.uk>
<http://www.andyhsoftware.co.uk/space> Space: disk usage analysis tool

"Robert" <lonerngraus@yahoo.com.au> wrote in message
news:<emW2d.35062$D7.7240@news-server.bigpond.net.au>...[color=blue]
>
> I was wanting to write some code to work out whether I needed st,
> nd rd or th in my dates.[/color]

$date = strtotime('2004-09-22');
echo date('F jS', $date); // prints "September 22nd"

Cheers,
NC

Robert wrote:[color=blue]
> Hi,
>
> I was wanting to write some code to work out whether I needed st, nd rd or
> th in my dates.
> While the code below works, I am certain that there is a better way....
>
> if (($daterep == "1") || ($daterep == "21") || ($daterep == "31")){
> $letters = "st";
> } elseif (($daterep == "2") || ($daterep == "22")){
> $letters = "nd";
> } elseif (($daterep == "3") || ($daterep == "23")){
> $letters = "rd";
> } else {
> $letters = "th";
> }
>
> I tried:
>
> if ($daterep == "1" || "21" || "31") { ... etc
>
> but it did not work... I don't understand why though?
> I cast $daterep into a string, but now I find that I could have left it
> alone....
> Any suggestions as to how to get this code working a bit smoother?
>
> Robert
>
>[/color]

here's my 2cents worth:

function
numToOrdinal($num)
{
$n = $num % 100;
$suff = array("th", "st", "nd", "rd", "th"); // suff for suffix
$ord = $n<21?($n<4 ? $suff[$n]:$suff[0]): ($n%10>4 ? $suff[0] : $suff[$n%10]);
return $num . $ord;
}


for($i = 0; $i < 200; $i++)
{
echo numToOrdinal($i) . "<br>";
}


handles 0th and all positive integers.

[error checking should be implemented]

Fox
***************