Ranking with Echo on Php [duplicate] - php

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?

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);

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.

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!

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.

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;
}
?>

Simple and Easy Answer will be:
$Day = 3;
echo date("S", mktime(0, 0, 0, 0, $Day, 0));
//OUTPUT - rd

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';
}

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';
}

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';
}
?>

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.

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.

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;
}

Here is the correct solution
$numberFormatter = new NumberFormatter('en_US', NumberFormatter::ORDINAL);
echo $numberFormatter->format(11);

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';
}
?>

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');
}

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.

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

Related

Too many Ifs to check variable really contains a valid, 4 digit year

The following if statement works but it's ugly and I have a feeling the same could be achieved with a preg_match REGEX. How?
Basically I don't know how to limit the amount of digits to 4 in regex yet. And once the 4 numerical digit criteria is met, I would like to limit the return of true to between 2019 and 2025.
Sample regex:
/* digits only, no dots */
function is_digits($element) {
return !preg_match ("/[^0-9]/", $element);
}
Sample PHP
// SET $TARGET_YEAR TO current year in [yyyy] format.
// Is set, Is not empty, Is numeric, Is 4 digits long?
if(isset($_GET["year"]) && !empty($_GET["year"]) && is_numeric($_GET["year"]) && strlen($_GET["year"]) == 4) {
// WE HAVE A REAL YEAR.
// SET $TARGET_YEAR TO target year in [yyyy] format.
self::$TARGET_YEAR = $_GET["year"];
} else {
// WE DON'T HAVE A REAL YEAR.
// SET $TARGET_YEAR TO current year in [yyyy] format.
self::$TARGET_YEAR = date('Y');
}
Regular expressions are unnecessary for such a simple check. Your existing code could be reduced to this:
$year = (int)$_GET["year"] ?? 0;
if ($year >= 2019 && $year <= 2025) {
// do stuff
}
Just cast the string as an integer and then check the range. If the string is not a valid number the cast will turn it to 0. If the value isn't set, the null coalesce will also turn it to zero.
This should do the trick. When you put in curly braces number 4 it will allow only 4 numbers if you use 0,4 it will allow 0 to 4 digits.
/* digits only, no dots */
function is_digits($element) {
$element = preg_match ("/^[0-9]{4}$/", $element);
if($element > 2019 && $element < 2025) {
return true; //Or $element
} else {
return false; //Or something else you want
}
}
You can use a built-in validate filter:
$opts = ['options' => ['min_range' => 2019, 'max_range' => 2025]];
if($date = filter_input(INPUT_GET, 'year', FILTER_VALIDATE_INT, $opts)) {
//yes
}

Format number to N significant digits in PHP

I would like to format (round) float (double) numbers to lets say 2 significant digits for example like this:
1 => 1
11 => 11
111 => 110
119 => 120
0.11 => 0.11
0.00011 => 0.00011
0.000111 => 0.00011
So the arbitrary precision remains same
I expect there is some nice function for it already built in, but could not find any so far
I was pointed to How to round down to the nearest significant figure in php, which is close but doesn't work for N significant digits and I'm not sure what it does with 0.000XXX numbers
To get a number rounded to n significant figures you need to find the size of the number in powers of ten, and subtract that from n.
This works fine for simple rounding:
function sigFig($value, $digits)
{
if ($value == 0) {
$decimalPlaces = $digits - 1;
} elseif ($value < 0) {
$decimalPlaces = $digits - floor(log10($value * -1)) - 1;
} else {
$decimalPlaces = $digits - floor(log10($value)) - 1;
}
$answer = round($value, $decimalPlaces);
return $answer;
}
This will give the following:
0.0001234567 returns 0.0001235
123456.7 returns 123500
However a value such as 10 to four significant figures should strictly be represented as 10.00 to signify the precision to which the value is known.
If this is the desired output you can use the following:
function sigFig($value, $digits)
{
if ($value == 0) {
$decimalPlaces = $digits - 1;
} elseif ($value < 0) {
$decimalPlaces = $digits - floor(log10($value * -1)) - 1;
} else {
$decimalPlaces = $digits - floor(log10($value)) - 1;
}
$answer = ($decimalPlaces > 0) ?
number_format($value, $decimalPlaces) : round($value, $decimalPlaces);
return $answer;
}
Now 1 is displayed as 1.000
With little modification to possible duplicate, answer by Todd Chaffee:
public static function roundRate($rate, $digits)
{
$mod = pow(10, intval(round(log10($rate))));
$mod = $mod / pow(10, $digits);
$answer = ((int)($rate / $mod)) * $mod;
return $answer;
}
To make sigFig(0.9995, 3) output 1.00, use
if(floor(log10($value)) !== floor(log10(round($value, $decimalPlaces)))) {$decimalPlaces--;}
Said line of code should be placed before declaring $answer.
If input $value is negative, set a flag and remove the sign at the beginning of the function, like this:
if($value < 0){$flag = 1;}
$value = ltrim($value, "-");
Then right before returning $answer, detect if the flag is set and if so restore the negative sign, like this:
if(isset($flag)){$answer = "-".$answer;}
Finally, for result values with ambiguous number of significant digits (e.g., 1000, 12000,...), express the result in scientific notation to the desired number of significant digits using sprintf or printf.

Finding the difference between a number variable within a repeating range?

I'm trying to process a numeric variable as follows-
if number is between 1-6 I want to count the difference between it and 7,
if number is between 15-20 I want to count the difference between it and 21,
if number is between 29-34 I want to count the difference between it and 35
and so on....
In other words, grouping it in multiples of 14, i need to find the difference between the first 6 numbers and the 7th number in each group.
I hope that makes sense. i find it hard to explain in writing, my code isn't much better but i managed to get it to do what i want, but I know it isn't ideal so what I am looking for is a simpler solution to this.
$total = 'A NUMBER'; /* the input number (example: 16 )*/
$fill =($total / 14);
$fill = $fill - floor($fill);
if($fill > 0 && $fill < 0.5)
{
$fill = $total;
while ($fill >= 7)
{
$fill = ($fill - 7);
}
$fill = (7 - $fill);
} else {
$fill=0;
}
echo $fill; /* the output (example answer: 5) */
Try the following function:
function MyProcess( $in ) {
$quotient = (int) ($in / 7);
if( ($quotient % 2 === 0) and ($in % 7 !== 0) )
return 7 - ($in % 7);
return 0;
}

Generate an array in PHP of random number not close to the X previous element

I want to generate in PHP an array of random numbers, but each number should not be the same as any of the X (for example 2 ) numbers bofore it and not even close to any of them by a define range (for example 5).
So for example:
I need numbers between 1 and 100
i've set my "range" to 5
the first two generated number are 20 and 50.
the third number will be a random number between 1 and 100, excluding all the numbers between 15 and 25, and between 45 and 55.
I can't figure out a function to achieve it. Ideally I want to call something like this:
getRandomNumbers( $min, $max, $previous, $range);
where $previous is the number of previous elements to take in consideration when generating the next one and $range is the "proximity" to those number where I don't want the next number to be.
I hope I explained in a decent way my request. :) Please, add a comment if you have any question about it.
I just came up with this:
function getRandomNumbers($min, $max, $previous, $range) {
static $generated = array();
$chunk = array_slice($generated, -$previous);
// Added this infinite loop check to save you some headache.
if (((($max - $min + 1) / (($range * 2) + 1)) + 1) <= $previous) {
die("Values set have the potential of running into an infinite loop. Min: $min, Max: $max, Previous: $previous, Range: $range");
}
while(true) {
$number = rand($min, $max);
$found = true;
foreach ($chunk as $value) {
if (in_array($number, range($value-$range, $value+$range))) {
$found = false;
}
}
if ($found) {
$generated[] = $number;
return $number;
}
}
}
Test it using this:
for ($i = 1; $i < 25; $i++) {
echo getRandomNumbers(1, 100, 5, 5) . "<br />";
}
PHPFiddle Link: http://phpfiddle.org/main/code/51ke-4qzs
Edit: Added a check to prevent a possible infinite loop. For example: if you set the following values:
$min = 1;
$max = 100;
$previous = 5;
$range = 12;
echo getRandomNumbers($min, $max, $previous, $range);
Then let's say, in a really unfortunate situation it would generate 13, 38, 63 and 88. So the 5th number cannot be anything between 1 and 25, 26 and 50, 51 and 75, 76 and 100. So it would result in an infinite loop. I've updated the PHPFiddle link as well.
getRandomNumbers( $previous, $range ) {
//I'm assuming that previous will be an array of your previous X that you don't want to be close to
$num = getRandomNumber() //However you are doing this now
foreach( $previous as $key => $value ) {
if ( ( $value - $range ) > $num && ( $value + $range ) < $num ) {
return getRandomNumbers($previous, $range);
}
}
//You need to also replace a value in previous
return num;
}

PHP date(): minutes without leading zeros

I'd like to know if there is a formatting letter for PHP's date() that allows me to print minutes without leading zeros, or whether I have to manually test for and remove leading zeros?
Use:
$minutes = intval(date('i'));
For times with more information than just minutes:
ltrim() - Strip whitespace (or other characters) from the beginning of a string
ltrim(date('i:s'), 0);
returns:
8:24
According to the PHP Documentation, the date() function does not have a placeholder for minutes without leading zeros.
You could, however, get that information by simply multiplying the dates, with a leading zero, by 1, turning it into an integer.
$minutesWithoutZero = 1* date( 'i' );
I tried to find this for seconds as well, gave up the search and just casting the result as a int like this:
echo (int)date("s");
That will get rid of the leading zero's in a fast efficient way.
Doesn't look like it, but you could do something like...
echo date('g:') . ltrim(date('i'), '0');
Alternately, you could cast the second call to date() with (int).
This also works
$timestamp = time(); // Or Your timestamp. Skip passing $timestamp if you want current time
echo (int)date('i',$timestamp);
I use this format if I need a XXmXXs format:
//Trim leading 0's and the 'm' if no minutes
ltrim(ltrim(gmdate("i\ms\s", $seconds), '0'), 'm');
This will output the following:
12m34s
1m23s
12s
i just did this one line solution
$min = intval(date('i',strtotime($date)));
Using ltrim method may remove all the leading zeroes.For ex if '00' min.In this case this will remove all the zeroes and gives you empty result.
My solution:
function seconds2string($seconds) {
if ($seconds == 0) {
return '-';
}
if ($seconds < 60) {
return date('0:s', $seconds);
}
if ($seconds < 3600) {
return ltrim(date('i:s', $seconds), 0);
}
return date('G:i:s', $seconds);
}
This will output:
0 seconds: -
10 seconds: 0:10
90 seconds: 1:30
301 seconds: 5:01
1804 seconds: 30:04
3601 seconds: 1:00:01
Just use this:
(int) date('i');
Or in mySQL just multiply it by 1, like such:
select f1, ..., date_format( fldTime , '%i' ) * 1 as myTime, ..., ...
$current_date = Date("n-j-Y");
echo $current_date;
// Result m-d-yy
9-10-2012
A quickie from me. Tell me what you think:
<?php function _wo_leading_zero($n) {
if(!isset($n[1])) return $n;
if(strpos($n, '.') !== false) {
$np = explode('.', $n); $nd = '.';
}
if(strpos($n, ',') !== false) {
if(isset($np)) return false;
$np = explode(',', $n); $nd = ',';
}
if(isset($np) && count($np) > 2) return false;
$n = isset($np) ? $np[0] : $n;
$nn = ltrim($n, '0');
if($nn == '') $nn = '0';
return $nn.(isset($nd) ? $nd : '').(isset($np[1]) ? $np[1] : '');
}
echo '0 => '._wo_leading_zero('0').'<br/>'; // returns 0
echo '00 => '._wo_leading_zero('00').'<br/>'; // returns 0
echo '05 => '._wo_leading_zero('05').'<br/>'; // returns 5
echo '0009 => '._wo_leading_zero('0009').'<br/>'; //returns 9
echo '01 => '._wo_leading_zero('01').'<br/>'; //returns 1
echo '0000005567 => '._wo_leading_zero('0000005567').'<br/>'; //returns 5567
echo '000.5345453 => '._wo_leading_zero('000.5345453').'<br/>'; //returns 0.5345453
echo '000.5345453.2434 => '._wo_leading_zero('000.5345453.2434').'<br/>'; //returns false
echo '000.534,2434 => '._wo_leading_zero('000.534,2434').'<br/>'; //returns false
echo date('m').' => '._wo_leading_zero(date('m')).'<br/>';
echo date('s').' => '._wo_leading_zero(date('s')).'<br/>'; ?>
use PHP's absolute value function:
abs( '09' ); // result = 9
abs( date( 'i' ) ); // result = minutes without leading zero
My Suggestion is Read this beautiful documentation it have all details of php date functions
Link of Documentation
And as per your question you can use i - Minutes with leading zeros (00 to 59) Which return you minutes with leading zero(0).
And Also introducing [Intval()][2] function returns the integer value of a variable. You can not use the intval() function on an object

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