PHP "is_numeric" accepts "E" as a number

Question

I noticed PHP is_numeric() accepts "E" as a number. I have a string: "88205052E00" and I want the result to be: NOT numeric.

Here is the code which I tested.

<?php

$notnumber = '88205052E00';

if(is_numeric($notnumber)) {
    echo $notnumber . ' is a number';
} else {
    echo $notnumber . ' is NOT a number';
}

?>

The Code above gives result:

88205052E00 is a number

How can I get the result to be: 88205052E00 is NOT a number?

`E` is used in scientific notation, so yes, it is considered a valid number. — trincot, Dec 19 '18 at 19:01
You might want to look into a regex that checks for non `0-9` characters. And handles decimals `.` if you need them. — Tim Lewis, Dec 19 '18 at 19:02
It is a number as long as PHP is concerned ([demo](https://3v4l.org/Jk9Lk)). — Álvaro González, Dec 19 '18 at 19:04
Your request is impossible because `88205052E00` **is** a number. It is `88205052*10^0` which is exactly `88205052`. — axiac, Dec 19 '18 at 20:01
@nerdlyist No, I want only to check for digits, negative or decimals are not expected. — Cyborg, Dec 20 '18 at 07:43

nerdlyist · Accepted Answer · 2018-12-19T20:00:29.920

8

I will keep the answer incase it helps but as pointed out there are shortcomings with ctype_digit in that it does not like - or ..

More likely then you want to use ctype_digit which checks if all of the characters in the provided string, text, are numerical.

Where as is_numeric — Finds whether a variable is a number or a numeric string

<?php

$s = "88205052E00";

if(ctype_digit($s)){
    echo "Yes";
} else {
    echo "No";
}

returns no.

edited Dec 19 '18 at 20:00

answered Dec 19 '18 at 19:18

nerdlyist

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It is good if you want `3948394873281738173172493274871218201832081239183` to return true and false for `-1`. – Salman A Dec 19 '18 at 19:23
I knew there was a reason I did not use this before. Working on a tweak – nerdlyist Dec 19 '18 at 19:31
1

Since I need to check for digits only (I dont need ` - ` or ` . `), this answer solves my problem. Thank you for helping. – Cyborg Jan 07 '19 at 04:13

miken32 · Answer 2 · 2018-12-19T19:57:49.037

2

Just use a regular expression:

<?php
if (preg_match("/^\-?[0-9]*\.?[0-9]+\z/", $notnumber)) {
    echo "$notnumber is numeric\n";
} else {
    echo "$notnumber is not numeric\n";
}

Results:

   1234 is numeric
1234E56 is not numeric
  -1234 is numeric
  .1234 is numeric
 -.1234 is numeric
 -12.34 is numeric

edited Dec 19 '18 at 19:57

answered Dec 19 '18 at 19:07

miken32

42,008
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Using `trim()` means that `\t\t12345\n\t\t\n` (several whitespace characters on either side) is numeric. – salathe Dec 19 '18 at 19:54
Yes, I'd assume the use of trim on the original variable. Probably best not to assume though. – miken32 Dec 19 '18 at 19:55

Jordan Soltman · Answer 3 · 2018-12-19T19:06:23.983

1

E is valid because of floating point numbers (http://php.net/manual/en/language.types.float.php).

If you don't want to allow E for whatever reason, you could check for it independently:

if (strpos(strtolower($notnumber), 'e') === false && is_numeric($notnumber))

This makes sure that there isn't an E and that it is also numeric.

edited Dec 19 '18 at 19:06

answered Dec 19 '18 at 19:03

Jordan Soltman

3,795
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2

Should be `===`, not `!==` – trincot Dec 19 '18 at 19:05

score 1 · Answer 4 · answered Dec 19 '18 at 19:09

1

Assuming you want to ensure that the string is a valid integer you could use filter_var:

$tests = ['1', '1.1', '1e0', '0x1'];
foreach($tests as $str) {
    $int = filter_var($str, FILTER_VALIDATE_INT);
    if ($int === false) {
        echo $str . ' is not an integer' . PHP_EOL;
    } else {
        echo $str . ' is an integer' . PHP_EOL;
    }
}

Result:

1 is an integer
1.1 is not an integer
1e0 is not an integer
0x1 is not an integer

answered Dec 19 '18 at 19:09

Salman A

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If you just want to check if the variable is an integer, couldn't you just use [`is_int`](http://php.net/manual/en/function.is-int.php)? – Cave Johnson Dec 19 '18 at 19:14
No. `is_int('1234')` returns false. – Salman A Dec 19 '18 at 19:15
ah, my mistake. I could have sworn there was a built-in for this. – Cave Johnson Dec 19 '18 at 19:16
1

I guess that's what filter_var is for. – Cave Johnson Dec 19 '18 at 19:16
@KodosJohnson There is [`ctype_digit()`](https://secure.php.net/manual/fr/function.ctype-digit.php), though (and the very interesting first comment on the documentation page). – Blackhole Dec 19 '18 at 19:17
1

All functions (including the above) have gotchas. `ctype_digit('239284923942734924729749327')` is true but obviously it is too big for integer. – Salman A Dec 19 '18 at 19:18

score 0 · Answer 5 · answered Dec 20 '18 at 08:47

Let's check the definition:

Finds whether the given variable is numeric. Numeric strings consist of optional sign, any number of digits, optional decimal part and optional exponential part. Thus +0123.45e6 is a valid numeric value. Hexadecimal (e.g. 0xf4c3b00c) and binary (e.g. 0b10100111001) notation is not allowed.

The relationship with floating point literals is clear but, how does it all relate to integer literals?

Octal notation is considered numeric.
Hexadecimal notation stopped being considered in PHP/7.0.
Binary notation has never been.
Integer overflow is not a direct concern but will lead to precision loss.

The manual doesn't state explicitly what the use cases are but in general it's more a helper tool (to ensure that you can feed stringified data to functions that expect numeric values) than a proper form validator. If input is collected in an environment where 88205052E00 is not expected then it might be a good idea to expect (and generate) localised data and implement a locale-aware solution.

PHP "is_numeric" accepts "E" as a number

5 Answers5