I need to check in PHP if user entered a decimal number (US way, with decimal point: X.XXX)
Any reliable way to do this?
You can get most of what you want from is_float, but if you really need to know whether it has a decimal in it, your function above isn't terribly far (albeit the wrong language):
function is_decimal( $val )
{
return is_numeric( $val ) && floor( $val ) != $val;
}
if you want "10.00" to return true check Night Owl's answer
If you want to know if the decimals has a value you can use this answer.
Works with all kind of types (int, float, string)
if(fmod($val, 1) !== 0.00){
// your code if its decimals has a value
} else {
// your code if the decimals are .00, or is an integer
}
Examples:
(fmod(1.00, 1) !== 0.00) // returns false
(fmod(2, 1) !== 0.00) // returns false
(fmod(3.01, 1) !== 0.00) // returns true
(fmod(4.33333, 1) !== 0.00) // returns true
(fmod(5.00000, 1) !== 0.00) // returns false
(fmod('6.50', 1) !== 0.00) // returns true
Explanation:
fmod returns the floating point remainder (modulo) of the division of the arguments, (hence the (!== 0.00))
Modulus operator - why not use the modulus operator? E.g. ($val % 1 != 0)
From the PHP docs:
Operands of modulus are converted to integers (by stripping the decimal part) before processing.
Which will effectively destroys the op purpose, in other languages like javascript you can use the modulus operator
If all you need to know is whether a decimal point exists in a variable then this will get the job done...
function containsDecimal( $value ) {
if ( strpos( $value, "." ) !== false ) {
return true;
}
return false;
}
This isn't a very elegant solution but it works with strings and floats.
Make sure to use !== and not != in the strpos test or you will get incorrect results.
another way to solve this: preg_match('/^\d+\.\d+$/',$number); :)
The function you posted is just not PHP.
Have a look at is_float [docs].
Edit: I missed the "user entered value" part. In this case you can actually use a regular expression:
^\d+\.\d+$
I was passed a string, and wanted to know if it was a decimal or not. I ended up with this:
function isDecimal($value)
{
return ((float) $value !== floor($value));
}
I ran a bunch of test including decimals and non-decimals on both sides of zero, and it seemed to work.
is_numeric returns true for decimals and integers. So if your user lazily enters 1 instead of 1.00 it will still return true:
echo is_numeric(1); // true
echo is_numeric(1.00); // true
You may wish to convert the integer to a decimal with PHP, or let your database do it for you.
This is a more tolerate way to handle this with user input. This regex will match both "100" or "100.1" but doesn't allow for negative numbers.
/^(\d+)(\.\d+)?$/
// if numeric
if (is_numeric($field)) {
$whole = floor($field);
$fraction = $field - $whole;
// if decimal
if ($fraction > 0)
// do sth
else
// if integer
// do sth
}
else
// if non-numeric
// do sth
i use this:
function is_decimal ($price){
$value= trim($price); // trim space keys
$value= is_numeric($value); // validate numeric and numeric string, e.g., 12.00, 1e00, 123; but not -123
$value= preg_match('/^\d$/', $value); // only allow any digit e.g., 0,1,2,3,4,5,6,7,8,9. This will eliminate the numeric string, e.g., 1e00
$value= round($value, 2); // to a specified number of decimal places.e.g., 1.12345=> 1.12
return $value;
}
$lat = '-25.3654';
if(preg_match('/./',$lat)) {
echo "\nYes its a decimal value\n";
}
else{
echo 'No its not a decimal value';
}
A total cludge.. but hey it works !
$numpart = explode(".", $sumnum);
if ((exists($numpart[1]) && ($numpart[1] > 0 )){
// it's a decimal that is greater than zero
} else {
// its not a decimal, or the decimal is zero
}
the easy way to find either posted value is integer and float so this will help you
$postedValue = $this->input->post('value');
if(is_numeric( $postedValue ) && floor( $postedValue ))
{
echo 'success';
}
else
{
echo 'unsuccess';
}
if you give 10 or 10.5 or 10.0 the result will be success if you define any character or specail character without dot it will give unsuccess
How about (int)$value != $value?
If true it's decimal, if false it's not.
I can't comment, but I have this interesting behaviour.
(tested on v. 7.3.19 on a website for php testing online)
If you multiply 50 by 1.1 fmod gives different results than expected.
If you do by 1.2 or 1.3 it's fine, if you do another number (like 60 or 40) is also fine.
$price = 50;
$price = $price * 1.1;
if(strpos($price,".") !== false){
echo "decimal";
}else{
echo "not a decimal";
}
echo '<br />';
if(fmod($price, 1) !== 0.00){
//echo fmod($price, 1);
echo "decimal";
} else {
echo "not a decimal";
}//end if
Simplest solution is
if(is_float(2.3)){
echo 'true';
}
If you are working with form validation. Then in this case form send string.
I used following code to check either form input is a decimal number or not.
I hope this will work for you too.
function is_decimal($input = '') {
$alphabets = str_split($input);
$find = array('0','1','2','3','4','5','6','7','8','9','.'); // Please note: All intiger numbers are decimal. If you want to check numbers without point "." then you can remove '.' from array.
foreach ($alphabets as $key => $alphabet) {
if (!in_array($alphabet, $find)) {
return false;
}
}
// Check if user has enter "." point more then once.
if (substr_count($input, ".") > 1) {
return false;
}
return true;
}
function is_decimal_value( $a ) {
$d=0; $i=0;
$b= str_split(trim($a.""));
foreach ( $b as $c ) {
if ( $i==0 && strpos($c,"-") ) continue;
$i++;
if ( is_numeric($c) ) continue;
if ( stripos($c,".") === 0 ) {
$d++;
if ( $d > 1 ) return FALSE;
else continue;
} else
return FALSE;
}
return TRUE;
}
Known Issues with the above function:
1) Does not support "scientific notation" (1.23E-123), fiscal (leading $ or other) or "Trailing f" (C++ style floats) or "trailing currency" (USD, GBP etc)
2) False positive on string filenames that match a decimal: Please note that for example "10.0" as a filename cannot be distinguished from the decimal, so if you are attempting to detect a type from a string alone, and a filename matches a decimal name and has no path included, it will be impossible to discern.
Maybe try looking into this as well
!is_int()
Related
I need to check in PHP if user entered a decimal number (US way, with decimal point: X.XXX)
Any reliable way to do this?
You can get most of what you want from is_float, but if you really need to know whether it has a decimal in it, your function above isn't terribly far (albeit the wrong language):
function is_decimal( $val )
{
return is_numeric( $val ) && floor( $val ) != $val;
}
if you want "10.00" to return true check Night Owl's answer
If you want to know if the decimals has a value you can use this answer.
Works with all kind of types (int, float, string)
if(fmod($val, 1) !== 0.00){
// your code if its decimals has a value
} else {
// your code if the decimals are .00, or is an integer
}
Examples:
(fmod(1.00, 1) !== 0.00) // returns false
(fmod(2, 1) !== 0.00) // returns false
(fmod(3.01, 1) !== 0.00) // returns true
(fmod(4.33333, 1) !== 0.00) // returns true
(fmod(5.00000, 1) !== 0.00) // returns false
(fmod('6.50', 1) !== 0.00) // returns true
Explanation:
fmod returns the floating point remainder (modulo) of the division of the arguments, (hence the (!== 0.00))
Modulus operator - why not use the modulus operator? E.g. ($val % 1 != 0)
From the PHP docs:
Operands of modulus are converted to integers (by stripping the decimal part) before processing.
Which will effectively destroys the op purpose, in other languages like javascript you can use the modulus operator
If all you need to know is whether a decimal point exists in a variable then this will get the job done...
function containsDecimal( $value ) {
if ( strpos( $value, "." ) !== false ) {
return true;
}
return false;
}
This isn't a very elegant solution but it works with strings and floats.
Make sure to use !== and not != in the strpos test or you will get incorrect results.
another way to solve this: preg_match('/^\d+\.\d+$/',$number); :)
The function you posted is just not PHP.
Have a look at is_float [docs].
Edit: I missed the "user entered value" part. In this case you can actually use a regular expression:
^\d+\.\d+$
I was passed a string, and wanted to know if it was a decimal or not. I ended up with this:
function isDecimal($value)
{
return ((float) $value !== floor($value));
}
I ran a bunch of test including decimals and non-decimals on both sides of zero, and it seemed to work.
is_numeric returns true for decimals and integers. So if your user lazily enters 1 instead of 1.00 it will still return true:
echo is_numeric(1); // true
echo is_numeric(1.00); // true
You may wish to convert the integer to a decimal with PHP, or let your database do it for you.
This is a more tolerate way to handle this with user input. This regex will match both "100" or "100.1" but doesn't allow for negative numbers.
/^(\d+)(\.\d+)?$/
// if numeric
if (is_numeric($field)) {
$whole = floor($field);
$fraction = $field - $whole;
// if decimal
if ($fraction > 0)
// do sth
else
// if integer
// do sth
}
else
// if non-numeric
// do sth
i use this:
function is_decimal ($price){
$value= trim($price); // trim space keys
$value= is_numeric($value); // validate numeric and numeric string, e.g., 12.00, 1e00, 123; but not -123
$value= preg_match('/^\d$/', $value); // only allow any digit e.g., 0,1,2,3,4,5,6,7,8,9. This will eliminate the numeric string, e.g., 1e00
$value= round($value, 2); // to a specified number of decimal places.e.g., 1.12345=> 1.12
return $value;
}
$lat = '-25.3654';
if(preg_match('/./',$lat)) {
echo "\nYes its a decimal value\n";
}
else{
echo 'No its not a decimal value';
}
A total cludge.. but hey it works !
$numpart = explode(".", $sumnum);
if ((exists($numpart[1]) && ($numpart[1] > 0 )){
// it's a decimal that is greater than zero
} else {
// its not a decimal, or the decimal is zero
}
the easy way to find either posted value is integer and float so this will help you
$postedValue = $this->input->post('value');
if(is_numeric( $postedValue ) && floor( $postedValue ))
{
echo 'success';
}
else
{
echo 'unsuccess';
}
if you give 10 or 10.5 or 10.0 the result will be success if you define any character or specail character without dot it will give unsuccess
How about (int)$value != $value?
If true it's decimal, if false it's not.
I can't comment, but I have this interesting behaviour.
(tested on v. 7.3.19 on a website for php testing online)
If you multiply 50 by 1.1 fmod gives different results than expected.
If you do by 1.2 or 1.3 it's fine, if you do another number (like 60 or 40) is also fine.
$price = 50;
$price = $price * 1.1;
if(strpos($price,".") !== false){
echo "decimal";
}else{
echo "not a decimal";
}
echo '<br />';
if(fmod($price, 1) !== 0.00){
//echo fmod($price, 1);
echo "decimal";
} else {
echo "not a decimal";
}//end if
Simplest solution is
if(is_float(2.3)){
echo 'true';
}
If you are working with form validation. Then in this case form send string.
I used following code to check either form input is a decimal number or not.
I hope this will work for you too.
function is_decimal($input = '') {
$alphabets = str_split($input);
$find = array('0','1','2','3','4','5','6','7','8','9','.'); // Please note: All intiger numbers are decimal. If you want to check numbers without point "." then you can remove '.' from array.
foreach ($alphabets as $key => $alphabet) {
if (!in_array($alphabet, $find)) {
return false;
}
}
// Check if user has enter "." point more then once.
if (substr_count($input, ".") > 1) {
return false;
}
return true;
}
function is_decimal_value( $a ) {
$d=0; $i=0;
$b= str_split(trim($a.""));
foreach ( $b as $c ) {
if ( $i==0 && strpos($c,"-") ) continue;
$i++;
if ( is_numeric($c) ) continue;
if ( stripos($c,".") === 0 ) {
$d++;
if ( $d > 1 ) return FALSE;
else continue;
} else
return FALSE;
}
return TRUE;
}
Known Issues with the above function:
1) Does not support "scientific notation" (1.23E-123), fiscal (leading $ or other) or "Trailing f" (C++ style floats) or "trailing currency" (USD, GBP etc)
2) False positive on string filenames that match a decimal: Please note that for example "10.0" as a filename cannot be distinguished from the decimal, so if you are attempting to detect a type from a string alone, and a filename matches a decimal name and has no path included, it will be impossible to discern.
Maybe try looking into this as well
!is_int()
I want to check if number is numeric. ctype_digit() is a good function but I want 2.0 to return true and 2.1 to return false.
var_dump(ctype_digit(1)); // true
var_dump(ctype_digit(1.0)); // false - I want true
var_dump(ctype_digit(1.1)); // false
If you want to check whether a variable represents an integer, while allowing 0 decimals, you can use:
floor($number) == $number
Examples:
floor(1) == 1 // true
floor(1.0) == 1.0 // true
floor(1.1) == 1.1 // false
Note that instead of floor($number) (round down), you could use ceil($number) (round up), or round($number) (round to the nearest integer).
Not as easy as one would think, here is my way of doing it.
$numbers = array(1.0, '2', 2.1, '2.1');
foreach($numbers as $number){
$numberArray = explode('.', $number);
if($numberArray[1] == 0){
echo $number . '<br>';
}
}
I want to get the length of integer values for validation in PHP.
Example:
Mobile numbers should be only 10 integer values. It should not be more than 10 or less than 10 and also it should not be included of alphabetic characters.
How can I validate this?
$num_length = strlen((string)$num);
if($num_length == 10) {
// Pass
} else {
// Fail
}
if (preg_match('/^\d{10}$/', $string)) {
// pass
} else {
// fail
}
This will work for almost all cases (except zero) and easily coded in other languages:
$length = ceil(log10(abs($number) + 1));
In my opinion, the best way is:
$length = ceil(log10($number))
A decimal logarithm rounded up is equal to length of a number.
If you are using a web form, make sure you limit the text input to only hold 10 characters as well to add some accessibility (users don't want to input it wrong, submit, get a dialog about their mistake, fix it, submit again, etc.)
Use intval function in loop,
See this example
<?php
$value = 16432;
$length=0;
while($value!=0) {
$value = intval($value/10);
$length++
}
echo "Length of Integer:- ".$length;
?>
$input = "03432 123-456"; // A mobile number (this would fail)
$number = preg_replace("/^\d/", "", $number);
$length = strlen((string) $number);
if ($number == $input && $length == 10) {
// Pass
} else {
// Fail
}
If you are evaluating mobile numbers (phone numbers) then I would recommend not using an int as your chosen data type. Use a string instead because I cannot forsee how or why you would want to do math with these numbers. As a best practice, use int, floats, etc, when you want/need to do math. Use strings when you don't.
From your question, "You want to get the lenght of an integer, the input will not accept alpha numeric data and the lenght of the integer cannot exceed 10. If this is what you mean; In my own opinion, this is the best way to achieve that:"
<?php
$int = 1234567890; //The integer variable
//Check if the variable $int is an integer:
if (!filter_var($int, FILTER_VALIDATE_INT)) {
echo "Only integer values are required!";
exit();
} else {
// Convert the integer to array
$int_array = array_map('intval', str_split($int));
//get the lenght of the array
$int_lenght = count($int_array);
}
//Check to make sure the lenght of the int does not exceed or less than10
if ($int_lenght != 10) {
echo "Only 10 digit numbers are allow!";
exit();
} else {
echo $int. " is an integer and its lenght is exactly " . $int_lenght;
//Then proceed with your code
}
//This will result to: 1234556789 is an integer and its lenght is exactly 10
?>
By using the assertion library of Webmozart Assert we can use their build-in methods to validate the input.
Use integerish() to validate that a value casts to an integer
Use length() to validate that a string has a certain number of characters
Example
Assert::integerish($input);
Assert::length((string) $input, 10); // expects string, so we type cast to string
As all assertions in the Assert class throw an Webmozart\Assert\InvalidArgumentException if they fail, we can catch it and communicate a clear message to the user.
Example
try {
Assert::integerish($input);
Assert::length((string) $input, 10);
} catch (InvalidArgumentException) {
throw new Exception('Please enter a valid phone number');
}
As an extra, it's even possible to check if the value is not a non-negative integer.
Example
try {
Assert::natural($input);
} catch (InvalidArgumentException) {
throw new Exception('Please enter a valid phone number');
}
I hope it helps 🙂
A bit optimazed answer in 2 or 3 steps depends if we allow negative value
if(is_int($number)
&& strlen((string)$number) == 10)
{
// 1 000 000 000 Executions take from 00:00:00.153200 to 00:00:00.173900
//Code
}
Note that will allow negative up to 9 numbers like -999999999
So if we need skip negatives we need 3rd comparision
if(is_int($number)
&& $number >= 0
&& strlen((string)$number) == 10)
{
// 1 000 000 000 Executions take from 00:00:00.153200
// to 00:00:00.173900 over 20 tests
}
Last case when we want from -1 000 000 000 to 1 000 000 000
if(is_int($number)
&& $number >= 0
&& strlen(str_replace('-', '', (string)$number)) == 10)
{
// 1 000 000 000 Executions take from 00:00:00.153200
// to 00:00:00.173900 over 20 tests
}
For compare
First naswer with regex
if (preg_match('/^\d{10}$/', $number)) {
// Fastest test with 00:00:00.246200
}
** Tested at PHP 8.0.12
** XAMPP 3.3.0
** Ryzen 7 2700
** MSI Radeon RX 5700 8G
Tested like
$function = function($number)
{
if(is_int($number)
&& $number >= 0
&& strlen((string)$number) == 10)
{
return true;
}
}
$number = 1000000000;
$startTime = DateTime::createFromFormat('U.u', microtime(true);
for($i = 0; $i < 1000000000; $i++)
{
call_user_func_array($function, $args);
}
$endTime = DateTime::createFromFormat('U.u', microtime(true);
echo $endTime->diff($startTime)->format('%H:%I:%S.%F');
I'm using this function to check if binary is correct, I know it looks sloppy.. I'm not sure how to write the function that well.. but it doesn't seem to work!
If binary = 10001000 it says malformed, even though it's not.. what is wrong in my function?..
function checkbinary($bin) {
$binary = $bin;
if(!strlen($binary) % 8 == 0){
return 1;
}
if (strlen($binary) > 100) {
return 1;
}
if (!preg_match('#^[01]+$#', $binary)){ //Tried without !
return 1;
}
if (!is_numeric($binary)) {
return 1;
}
}
if (checkbinary("10001000") != 1) {
echo "Correct";
} else {
echo "Binary incorrect";
}
Why does this function always say 10001000 is incorrect?
if(!strlen($binary) % 8 == 0){ should be
if( strlen($binary) % 8 !== 0 ){
edit and btw: Since you're already using preg_match() you can simplify/shorten the function to
function checkbinary($binary) {
return 1===preg_match('#^(?:[01]{8}){0,12}$#', $binary);
}
This allows 0 - 12 groups of 8 0/1 characters which includes all the tests you currently have in your function:
strlen()%8 is covered by {8} in the inner group
strlen() > 100 is covered by {0,12} since any string longer than 8*12=96 characters would trigger either the first if or the >100 test
0/1 test is obvious
is_numeric is kinda superfluous
edit2: The name checkbinary might not be a perfect choice for the function. I wouldn't necessarily expect it to check for 8bit/byte alignment and strlen()<=100.
function checkbinary($bin) {
return preg_match('#^[01]+$#', $bin);
}
Some tests: http://www.ideone.com/3D9SQetX and http://www.ideone.com/1HCCtxVV.
I'm not entirely sure what you're attempting to achieve, but you might be able to get there via the following...
if($binaryString == base_convert($binaryString, 2, 2))...
In essence, this is comparing the results of a conversion from binary to binary - hence if the resultant output is identical, the input must be a valid binary string.
I have this PHP code:
$entityElementCount = (-($highScore-$totalKeywordCount))/0.29;
What i want to know is, how to check whether $entityElementCount is a whole number (2, 6, ...) or partial (2.33, 6.2, ...).
Thank you!
if (floor($number) == $number)
I know this is old, but I thought I'd share something I just found:
Use fmod and check for 0
$entityElementCount = (-($highScore-$totalKeywordCount))/0.29;
if (fmod($entityElementCount,1) !== 0.0) {
echo 'Not a whole number!';
} else {
echo 'A whole number!';
}
fmod is different from % because if you have a fraction, % doesn't seem to work for me (it returns 0...for example, echo 9.4 % 1; will output 0). With fmod, you'll get the fraction portion. For example:
echo fmod(9.4, 1);
Will output 0.4
$entityElementCount = (-($highScore-$totalKeywordCount))/0.29;
if (ctype_digit($entityElementCount) ){
// (ctype_digit((string)$entityElementCount)) // as advised.
print "whole number\n";
}else{
print "not whole number\n";
}
I would use intval function like this:
if($number === intval($number)) {
}
Tests:
var_dump(10 === intval(10)); // prints "bool(true)"
var_dump("10" === intval("10")); // prints "bool(false)"
var_dump(10.5 === intval(10.5)); // prints "bool(false)"
var_dump("0x539" === intval("0x539")); // prints "bool(false)"
Other solutions
1)
if(floor($number) == $number) { // Currently most upvoted solution:
Tests:
$number = true;
var_dump(floor($number) == $number); // prints "bool(true)" which is incorrect.
2)
if (is_numeric($number) && floor($number) == $number) {
Corner case:
$number = "0x539";
var_dump(is_numeric($number) && floor($number) == $number); // prints "bool(true)" which depend on context may or may not be what you want
3)
if (ctype_digit($number)) {
Tests:
var_dump(ctype_digit("0x539")); // prints "bool(false)"
var_dump(ctype_digit(10)); // prints "bool(false)"
var_dump(ctype_digit(0x53)); // prints "bool(false)"
The basic way, as Chacha said is
if (floor($number) == $number)
However, floating point types cannot accurately store numbers, which means that 1 might be stored as 0.999999997. This will of course mean the above check will fail, because it will be rounded down to 0, even though for your purposes it is close enough to 1 to be considered a whole number. Therefore try something like this:
if (abs($number - round($number)) < 0.0001)
If you know that it will be numeric (meaning it won't ever be a an integer cast as a string, like "ten" or "100", you can just use is_int():
$entityElementCount = (-($highScore-$totalKeywordCount))/0.29;
$entityWholeNumber = is_int($entityElementCount);
echo ($entityWholeNumber) ? "Whole Number!" : "Not a whole number!";
I tested all the proposed solutions with many problematic values mentioned, they all fail for at least one of the test cases. Start checking if $value is a number using is_numeric($value) reduces the number of failures for many solutions, but does not turn any solution into an ultimate one:
$test_cases = array(0.29, 2, 6, 2.33, 6.2, '10.00', 1.4, 10, "10", 10.5, "0x539", true,
false, 0x53, 9.4, "ten", "100", 1, 0.999999997, 0, 0.0001, 1.0, 0.9999999,
(-(4.42-5))/0.29);
function is_whole_number($value) {
// Doing this prevents failing for values like true or "ten"
if (!is_numeric($value)) {
return false;
}
// #ghostdog74's solution fails for "10.00"
// return (ctype_digit((string) $value));
// Both #Maurice's solutions fails for "10.00"
// return ((string) $value === (string) (int) $value);
// return is_int($value);
// #j.hull's solution always returns true for numeric values
// return (abs($value) % 1 == 0 ? true : false);
// # MartyIX's solution fails for "10.00"
// return ($value === intval($value));
// This one fails for (-(4.42-5))/0.29
// return (floor($value) == $value);
// This one fails for 2
// return ctype_digit($value);
// I didn't understand Josh Crozier's answer
// #joseph4tw's solution fails for (-(4.42-5))/0.29
// return !(fmod($value, 1) != 0);
// If you are unsure about the double negation, doing this way produces the same
// results:
// return (fmod($value, 1) == 0);
// Doing this way, it always returns false
// return (fmod($value, 1) === 0);
// #Anthony's solution fails for "10.00"
// return (is_numeric($value) && is_int($value));
// #Aistina's solution fails for 0.999999997
// return (abs($value - round($value)) < 0.0001);
// #Notinlist's solution fails for 0.999999997
// return (round($value, 3) == round($value));
}
foreach ($test_cases as $test_case) {
var_dump($test_case);
echo ' is a whole number? ';
echo is_whole_number($test_case) ? 'yes' : 'no';
echo "\n";
}
I think that solutions like the ones proposed by #Aistina and #Notinlist are the best ones, because they use an error threshold to decide whether a value is a whole number. It is important to note that they worked as expected for the expression (-(4.42-5))/0.29, while all the others failed in that test case.
I decided to use #Notinlist's solution because of its readability:
function is_whole_number($value) {
return (is_numeric($value) && (round($value, 3) == round($value)));
}
I need to test if values are whole numbers, currency or percentage, I think 2 digits of precision is enough, so #Notinlist's solution fits my needs.
Running this test:
$test_cases = array(0.29, 2, 6, 2.33, 6.2, '10.00', 1.4, 10, "10", 10.5, "0x539", true,
false, 0x53, 9.4, "ten", "100", 1, 0.999999997, 0, 0.0001, 1.0, 0.9999999,
(-(4.42-5))/0.29);
function is_whole_number($value) {
return (is_numeric($value) && (round($value, 3) == round($value)));
}
foreach ($test_cases as $test_case) {
var_dump($test_case);
echo ' is a whole number? ';
echo is_whole_number($test_case) ? 'yes' : 'no';
echo "\n";
}
Produces the following output:
float(0.29)
is a whole number? no
int(2)
is a whole number? yes
int(6)
is a whole number? yes
float(2.33)
is a whole number? no
float(6.2)
is a whole number? no
string(5) "10.00"
is a whole number? yes
float(1.4)
is a whole number? no
int(10)
is a whole number? yes
string(2) "10"
is a whole number? yes
float(10.5)
is a whole number? no
string(5) "0x539"
is a whole number? yes
bool(true)
is a whole number? no
bool(false)
is a whole number? no
int(83)
is a whole number? yes
float(9.4)
is a whole number? no
string(3) "ten"
is a whole number? no
string(3) "100"
is a whole number? yes
int(1)
is a whole number? yes
float(0.999999997)
is a whole number? yes
int(0)
is a whole number? yes
float(0.0001)
is a whole number? yes
float(1)
is a whole number? yes
float(0.9999999)
is a whole number? yes
float(2)
is a whole number? yes
if(floor($number) == $number)
Is not a stable algorithm. When a value is matematically 1.0 the numerical value can be 0.9999999. If you apply floor() on it it will be 0 which is not equals to 0.9999999.
You have to guess a precision radius for example 3 digits
if(round($number,3) == round($number))
$num = 2.0000000000001;
if( $num == floor( $num ) ){
echo('whole');
}else{
echo('fraction');
}
EX:
2.0000000000001 | fraction
2.1 | fraction
2.00 | whole
2 | whole
Another hacky way I came up with is ceil($value) === floor($value). If a number is a whole number, this should always be true, even if comparing 10 with 10.000 and will even work with numbers cast in string, for example ceil("10.0") === floor(10).
(string)floor($pecahformat[3])!=(string)$pecahformat[3]
floor($entityElementCount) == $entityElementCount
This will be true if this is a whole number
This is not an attempt to answer this question so much. Their are plenty of answer already. If you are doing statistics as the question implies I suspect #antonio-vinicius-menezes-medei answer will suite you best. However I needed this answer for input validation. I found this check more reliable for validating an input string is a whole number:
is_numeric($number) && preg_match('/^[0-9]+$/', $number)
The 'is_numeric' simply corrects for "true" converting to "1" in preg_match.
So playing off of #antonio-vinicius-menezes-medei answer. I wrote a script to test this below. Note the ini_set('precision', 20). preg_match will convert the argument to a string. If your precicion is set below the length of the float values they will simply round at the given precision. Similar to #antonio-vinicius-menezes-medei answer this precision setting will force a similar estimation length.
ini_set('precision', 20);
$test_cases = array(0.29, 2, 6, 2.33, 6.2, '10.00', 1.4, 10, "10", 10.5, "0x539", true,
false, 0x53, 9.4, "ten", "100", 1, 0.999999997, 0, 0.0001, 1.0, 0.9999999,
(-(4.42-5))/0.29);
foreach ($test_cases as $number)
{
echo '<strong>';
var_dump($number);
echo '</strong>';
echo boolFormater(is_numeric($number) && preg_match('/^[0-9]+$/', $number));
echo '<br>';
}
function boolFormater($value)
{
if ($value)
{
return 'Yes';
}
return 'No';
}
Which produces this output:
float(0.28999999999999998002) No
int(2) Yes
int(6) Yes
float(2.3300000000000000711) No
float(6.2000000000000001776) No
string(5) "10.00" No
float(1.3999999999999999112) No
int(10) Yes
string(2) "10" Yes
float(10.5) No
string(5) "0x539" No
bool(true) No
bool(false) No
int(83) Yes
float(9.4000000000000003553) No
string(3) "ten" No
string(3) "100" Yes
int(1) Yes
float(0.99999999699999997382) No
int(0) Yes
float(0.00010000000000000000479) No
float(1) Yes
float(0.99999990000000005264) No
float(2.0000000000000004441) No
improved version of #Tyler Carter's solution, which handles edge cases better than the original:
function is_whole_number($number){
return (is_float(($f=filter_var($number,FILTER_VALIDATE_FLOAT))) && floor($f)===$f);
}
(Tyler's code fail to recognize that the string "123foobar" is not a whole number. this improved version won't make that mistake. credits to #Shafizadeh in the comments for discovering the bug. also this is php7 strict_types=1-compatible)
What seems a simple approach would be to use modulus (%) to determine if a value is whole or not.
x = y % 1
if y is anything other then a whole number the result is not a zero (0). A test then would be:
if (y % 1 == 0) {
// this is a whole number
} else {
// this is not a whole number
}
var isWhole = (y % 1 == 0? true: false); // to get a boolean return.
Granted this will view a negative number as a whole number, then then just wrap ABS() around y to always test on the positive.
I always use typecasting to check if variables contain a whole number, handy when you don't know the origin or type of the value.
if ((string) $var === (string) (int) $var) {
echo 'whole number';
} else {
echo 'whatever it is, it\'s something else';
}
In your particular case, I would use is_int()
if (is_int($var) {
echo 'integer';
}
A simple solution for positive whole numbers only. This may not work for everything.
$string = '0x539';
$ceil = ceil($string);
if($ceil < 1){
$ceil = FALSE; // or whatever you want i.e 0 or 1
}
echo $ceil; // 1337
You can use floor() instead of ceil() if so desired.
function isInteger($value)
{
// '1' + 0 == int, '1.2' + 0 == float, '1e2' == float
return is_numeric($value) && is_int($value + 0);
}
function isWholeNumber($value)
{
return is_numeric($value)
&& (is_int($value + 0)
|| (intval($value + 0) === intval(ceil($value + 0))));
}
If you want to check for both whole and decimal numbers, you can do the following:
if (isInteger($foo))
{
// integer as int or string
}
if (isWholeNumber($foo))
{
// integer as int or string, or float/double with zero decimal part
}
else if (is_numeric($foo))
{
// decimal number - still numeric, but not int
}
This will correctly check your number without rounding it, casting it to int (which in the case of a decimal number will lose the decimal part), or doing any math. If, however, you want to treat 1.00 as a whole number, then that's a whole another story.
I know this is a super old post but this is a simple function that will return a valid whole number and cast it to an int. Returns false if it fails.
function isWholeNumber($v)
{
if ($v !='' && is_numeric($v) && strpos($v, '.') === false) {
return (int)$v;
}
return false;
}
Usage :
$a = 43;
$b = 4.3;
$c = 'four_three';
isWholeNumber($a) // 43
isWholeNumber($b) // false
isWholeNumber($c) // false
Just to share my solution with localized string/number, this combo worked like a charm for me.
public static function isWholeNumber ($input, $decimalDelimiter = ',')
{
if (is_string($input)){
$input = str_replace($decimalDelimiter, '.', $input);
$input = floatval($input);
}
if (fmod($input,1) !== 0.0) {
return false;
}
return true;
}
$entityElementCount = (-($highScore-$totalKeywordCount))/0.29;
Method 1-
By using ctype_digit() function.
if ( ctype_digit($entityElementCount )) {
echo "Whole Number\n";
} else {
echo "Not a whole Number\n";
}
Method 2-
By using is_float() function.
if (is_float($entityElementCount )) {
echo "Not a Whole Number\n";
} else {
echo "Whole Number\n";
}
Method 3-
By using is_int() function.
if (is_int($entityElementCount )) {
echo "Whole Number\n";
} else {
echo "Not a whole Number\n";
}
Method 5-
By using fmod() function.
It needs 2 parameters one dividend and other is divisor
Here $dividend=$entityElementCount and divisor=1
if (fmod($dividend,$divisor) !== 0.0) {
echo 'Not a whole number!';
} else {
echo 'A whole number!';
}
there are some more function like intval(), floor(),... can be used to check it`enter code here`