在 PHP 中显示带有序数后缀的数字
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Display numbers with ordinal suffix in PHP
提问by ArK
I want to display numbers as follows
我想显示数字如下
- 1 as 1st,
- 2 as 2nd,
- ...,
- 150 as 150th.
- 1 为 1，
- 2 作为 2，
- ...,
- 150 为 150。
How should I find the correct ordinal suffix (st, nd, rd or th) for each number in my code?
我应该如何为代码中的每个数字找到正确的序数后缀（st、nd、rd 或 th）？
回答by Iacopo
$ends = array('th','st','nd','rd','th','th','th','th','th','th');
if (($number %100) >= 11 && ($number%100) <= 13)
$abbreviation = $number. 'th';
else
$abbreviation = $number. $ends[$number % 10];
Where $numberis the number you want to write. Works with any natural number.
$number你想写的数字在哪里。适用于任何自然数。
As a function:
作为一个函数：
function ordinal($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);
回答by Jeremy Kauffman
PHP has built-in functionality for this. It even handles internationalization!
PHP 具有为此的内置功能。它甚至可以处理国际化！
$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.
请注意，此功能仅在 PHP 5.3.0 及更高版本中可用。
回答by Andrew Kozak
This can be accomplished in a single line by leveraging similar functionality in PHP's built-in date/time functions. I humbly submit:
这可以通过利用 PHP 内置日期/时间函数中的类似功能在一行中完成。我谦虚地提出：
Solution:
解决方案：
function ordinalSuffix( $n )
{
return date('S',mktime(1,1,1,1,( (($n>=10)+($n>=20)+($n==0))*10 + $n%10) ));
}
Detailed Explanation:
详细说明：
The built-in date()function has suffix logic for handling nth-day-of-the-month calculations. The suffix is returned when Sis given in the format string:
内置date()函数具有用于处理每月第 n 天计算的后缀逻辑。后缀S在格式字符串中给出时返回：
date( 'S' , ? );
Since date()requires a timestamp (for ?above), we'll pass our integer $nas the dayparameter to mktime()and use dummy values of 1for the hour, minute, second, and month:
由于date()需要的时间戳（对于?以上），我们将通过我们的整数$n作为day参数mktime()和使用假人值1的hour，minute，second，和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 $nat 29:
这实际上在一个月中的某一天（即$n > 31）超出范围的值时正常失败，但我们可以添加一些简单的内联逻辑以将上限设为$n29：
date( 'S', mktime( 1, 1, 1, 1, ( (($n>=10)+($n>=20))*10 + $n%10) ));
The only positive value(May 2017) this fails on is $n == 0, but that's easily fixed by adding 10 in that special case:
唯一的正值（2017 年 5 月）这在 is 上失败$n == 0，但在这种特殊情况下通过添加 10 很容易解决：
date( 'S', mktime( 1, 1, 1, 1, ( (($n>=10)+($n>=20)+($n==0))*10 + $n%10) ));
Update, May 2017
更新，2017 年 5 月
As observed by @donatJ, the above fails above 100 (e.g. "111st"), since the >=20checks are always returning true. To reset these every century, we add a filter to the comparison:
正如@donatJ 所观察到的，上面的失败超过 100（例如“111st”），因为>=20检查总是返回 true。为了每个世纪重置这些，我们在比较中添加了一个过滤器：
date( 'S', mktime( 1, 1, 1, 1, ( (($n>=10)+($n%100>=20)+($n==0))*10 + $n%10) ));
Just wrap it in a function for convenience and off you go!
为方便起见，只需将其包装在一个函数中即可！
回答by Paul
Here is a one-liner:
这是一个单行：
$a = <yournumber>;
echo $a.substr(date('jS', mktime(0,0,0,1,($a%10==0?9:($a%100>20?$a%10:$a%100)),2000)),-2);
Probably the shortest solution. Can of course be wrapped by a function:
可能是最短的解决方案。当然可以用函数包装：
function ordinal($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
问候， 保罗
EDIT1: Correction of code for 11 through 13.
EDIT1：更正 11 到 13 的代码。
EDIT2: Correction of code for 111, 211, ...
EDIT2：更正 111、211、...的代码
EDIT3: Now it works correctly also for multiples of 10.
EDIT3：现在它也适用于 10 的倍数。
回答by John Boker
from http://www.phpro.org/examples/Ordinal-Suffix.html
来自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;
}
?>
回答by WhiteHorse
Simple and Easy Answer will be:
简单易行的答案将是：
$Day = 3;
echo date("S", mktime(0, 0, 0, 0, $Day, 0));
//OUTPUT - rd
回答by iletras
I wrote this for PHP4. It's been working ok & it's pretty economical.
我为 PHP4 写了这个。它一直工作正常，而且非常经济。
function getOrdinalSuffix($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';
}
回答by Chintan Thummar
you just need to apply given function.
你只需要应用给定的功能。
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 $num.'th';
}
回答by Uzair Bin Nisar
Generically, you can use that and call echo get_placing_string(100);
通常，您可以使用它并调用echo get_placing_string(100);
<?php
function get_placing_string($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 $placing.'th';
}
?>
回答by nxasdf
I made a function that does not rely on the PHP's date();function as it's not necessary, but also made it as compact and as short as I think is currently possible.
我创建了一个不依赖 PHPdate();函数的函数，因为它不是必需的，但也使它尽可能紧凑和简短。
The code: (121 bytes total)
代码：（共 121 字节）
function ordinal($i) { // PHP 5.2 and later
return($i.(($j=abs($i)%100)>10&&$j<14?'th':(($j%=10)>0&&$j<4?['st', 'nd', 'rd'][$j-1]:'th')));
}
More compact code below.
下面更紧凑的代码。
It works as follows:
它的工作原理如下：
printf("The %s hour.\n", ordinal(0)); // The 0th hour.
printf("The %s ossicle.\n", ordinal(1)); // The 1st ossicle.
printf("The %s cat.\n", ordinal(12)); // The 12th cat.
printf("The %s item.\n", ordinal(-23)); // The -23rd item.
Stuff to know about this function:
有关此功能的信息：
- It deals with negative integers the same as positive integers and keeps the sign.
- It returns 11th, 12th, 13th, 811th, 812th, 813th, etc. for the -teennumbers as expected.
- It does not check decimals, but will leave them in place (use
floor($i),round($i), orceil($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
$ifrom the beginning of the return statement if you only want to return the ordinal suffix without what you input.
- 它与正整数一样处理负整数并保持符号。
- 它按预期为-teen数字返回第 11、第 12、第 13、第 811、第 812、第 813 等。
- 它不检查小数，但会将它们留在原处（使用
floor($i),round($i), 或ceil($i)在最终 return 语句的开头）。 - 您还可以
format_number($i)在最终 return 语句的开头添加以获取逗号分隔的整数（如果您要显示千、百万等）。 $i如果您只想在没有输入内容的情况下返回序数后缀，则可以从 return 语句的开头删除。
This function works commencing PHP 5.2 released November 2006 purely because of the short array syntax. If you have a version before this, then please upgrade because you're nearly a decade out of date! Failing that, just replace the in-line ['st', 'nd', 'rd']with a temporary variable containing array('st', 'nd', 'rd');.
这个函数从 2006 年 11 月发布的 PHP 5.2 开始工作，纯粹是因为短数组语法。如果您有此之前的版本，那么请升级，因为您已经过时了近十年！如果失败，只需将内联替换为['st', 'nd', 'rd']包含array('st', 'nd', 'rd');.
The same function (without returning the input), but an exploded view of my short function for better understanding:
相同的函数（不返回输入），但我的简短函数的分解图以便更好地理解：
function ordinal($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:
代码更新：
Here's a modified version that is 14 whole bytes shorter (107 bytes total):
这是一个修改后的版本，缩短了 14 个字节（总共 107 个字节）：
function ordinal($i) {
return $i.(($j=abs($i)%100)>10&&$j<14?'th':@['th','st','nd','rd'][$j%10]?:'th');
}
Or for as short as possible being 25 bytes shorter (96 bytes total):
或者尽可能短 25 个字节（总共 96 个字节）：
function o($i){return $i.(($j=abs($i)%100)>10&&$j<14?'th':@['th','st','nd','rd'][$j%10]?:'th');}
With this last function, simply call o(121);and it'll do exactly the same as the other functions I listed.
使用最后一个函数，只需调用o(121);它，它将与我列出的其他函数完全相同。
Code Update #2:
代码更新#2：
Benand I worked together and cut it down by 38 bytes (83 bytes total):
Ben和我一起工作并将其减少了 38 个字节（总共 83 个字节）：
function o($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. :)
我们认为它不可能比这更短！然而，愿意被证明是错误的。:)
Hope you all enjoy.
希望大家喜欢。
