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()
This question already has answers here:
Test if number is odd or even
(20 answers)
Closed 7 years ago.
How can I get if a number is even or odd or neither (have decimal, like 1.5) with PHP? I know that there are operators like *, /, but they did not work.
Here's a try (of course it did not) (work that's just to find if it's a even number):
function even($n) {
return (($n/2)*2 == $n);
}
echo even(1); // true (should be false)
echo even(2); // true
How about
function even($n) {
if (!is_int($n)) {return 'n';}
return !($n % 2);
}
even(1); // false;
even(2); // true;
even(1.5); // 'n'
The danger here is that 'n' will evaluate as false if used as a boolean. It might be better to return some specific constants instead of true or false. The OP didn't specify what the return values should be.
It is pretty simple. modulo (%) is the operator you want, it determines if there would be a remainder if x is divided by y... for example (3 % 2 = 1) and (4 % 2 = 0).
This has been asked before too - pretty common question - you really just need to see if your number, $n % 2 is equal to 0.
php test if number is odd or even
Check if given number is integer first. And bitwise & to check if it is even or odd. Here is an example...
if (is_int($n)) {
if ($n & 1) {
echo 'Odd!';
} else {
echo 'Even!';
}
} else {
echo "Not a Integer!";
}
Hope this is helpful.
Use the modulo operator (%) to determine whether the integer is divisible by 2. You also need abs() to handle negative numbers, and is_int() to handle the fact that the modulo operator doesn't correctly handle floating point numbers. An example implementation follows:
function is_even($num) {
return is_int($num) && abs($num % 2) == 0;
}
function is_odd($num) {
return is_int($num) && abs($num % 2) == 1;
}
// this last one seems self-explanatory, but if you want it, here it is
function is_neither_even_nor_odd($num) {
return !is_even($num) && !is_odd($num);
}
// Tests: The following should all output true:
var_dump(
is_even(0),
is_even(2),
is_even(-6),
is_even(51238238),
is_odd(1),
is_odd(-1),
is_odd(57),
is_neither_even_nor_odd(1.5),
is_neither_even_nor_odd(2.5),
is_neither_even_nor_odd(-0.5),
is_neither_even_nor_odd(0.00000001)
);
Here's a demo.
is_numeric returns true if the given variable is a number
is_int returns true if the given variable is an integer
The modulor operator % can be used to determine if an integer is even or odd:
$num % 2 == 0 // returns true if even, false if odd
I have a function that does this...
function coefficient_display($coeff){
if ($coeff == 1){
return '';
} elseif ($coeff == -1){
return '-';
} else {
return $coeff;
}
}
It seems that, if you feed it a "formatted" number, it fails. Why?
coefficient_display(1200) gives 1200.
coefficient_display('1200') gives 1200.
coefficient_display(1,200) fails.
Because no some "formatted" numbers here: foo(1,200) only 2 arguments 1 and 200.
Try coefficient_display("1,200") or coefficient_display("1.200")
Here is some explanation for what you are getting.
coefficient_display('1200')gives 1200.
PHP data-types are interchangeable so '1200'(string) != 1 or -1. thus executes the else part.
coefficient_display(1,200) fails.
This is not actually failing this returns the blank string. As the first parameter is 1 if ($coeff == 1) is true and it executes return ''; line. the second parameter 200 is ignored.
Hope this helps.
1,200 is not a number, try coefficient_display('1,200')
And change all '==' to '==='
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 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');