A side note, finding out if an integer is even or odd, given the integer = $i ...
if (($i % 2) == 0) // This will test TRUE if the integer is even.
{
xxx insert code here xxx
}
is_numeric
(PHP 4, PHP 5)
is_numeric — 変数が数字または数値文字列であるかを調べる
説明
bool is_numeric ( mixed $var )指定した変数が数値であるかどうかを調べます。数値文字列は以下の要素から なります。（オプションの）符号、任意の数の数字、（オプションの）小数部、 そして（オプションの）指数部。つまり、+0123.45e6 は数値として有効な値です。16 進表記（0xFF）も 認められますが、この場合は符号や小数部、指数部を含めることはできません。
パラメータ
- var
評価する変数。
返り値
var が数値または数値文字列である場合に TRUE、それ以外の場合に FALSE を返します。
参考
| ctype_digit() |
| is_bool() |
| is_null() |
| is_float() |
| is_int() |
| is_string() |
| is_object() |
| is_array() |
is_numeric
Kenn
12-May-2007 03:45
12-May-2007 03:45
Q1712 at online dot ms
24-Apr-2007 05:33
24-Apr-2007 05:33
a regex doing the same as is_numeric() is:
"^((-?[0-9]*\.[0-9]+|[[:space:]]*[+-]?[0-9]+(\.[0-9]*)?)
([eE][+-]?[0-9]+)?|0x[0-9a-fA-F]+)$" (took me some testing to figure this out)
regex: "^(decimal|hex)$"
decimal: "(base1|base2)(exponent)?"
base1: "-?[0-9]*\.[0-9]+"
if there are no prepanded spaces and no "+" the
base may be any of "0", ".0", "0.0".
base2: "[[:space:]]*[+-]?[0-9]+(\.[0-9]*)?"
if there are empty spaces or a "+" prepanded the
base may only be "0" or "0.0" (no idea why)
exponent: "[eE][+-]?[0-9]+"
the exponent is opional
hex: "0x[0-9a-fA-F]+"
usage: preg_match("/regex/D", $str); or ereg("regex", $str);
Just to know what is_numeric() is doing. Of cause the regex should be much slower and is_numeric() also returns false if the number would be to big to fit into a double.
Elwin van Huissteden
27-Mar-2007 09:05
27-Mar-2007 09:05
To James holwell:
Maybe your function was more strickt, but profides FALSE to any numeric string that wasnt written in the English/American notition. To enable a person to use the both the English/American and the rest of the world's way:
<?php
function my_is_numeric($value)
{
return (preg_match ("/\A(-){0, 1}([0-9]+)((,|.)[0-9]{3, 3})*((,|.)[0-9]){0, 1}([0-9]*)\z/" ,$value) == 1);
}
?>
Other than that, i'd recommend using yours, if it works (i havent tested either yours or mine)
By using mine, there might be a slight chance to not being able to do calculations with the numeric string if it's not the English/American way.
(*Note:
-the E/A way of writing 1 million (with decimal for 1/50): 1,000,000.02
-the global way of writing 1 million (with decimal for 1/50): 1.000.000,02
james.holwell( at )exeter.ox.ac.uk
21-Mar-2007 07:07
21-Mar-2007 07:07
In reply to www.kigoobe.com, a more strict expression is
<?php
function my_is_numeric($value)
{
return (preg_match ("/^(-){0,1}([0-9]+)(,[0-9][0-9][0-9])*([.][0-9]){0,1}([0-9]*)$/", $value) == 1);
}
?>
This will not match strings like -6,77.8,8 which are matched by the below expression, and instead requires a single decimal point, with at least one character following, and only permits comma-separation when the right hand side is a triplet.
www.kigoobe.com
20-Mar-2007 10:55
20-Mar-2007 10:55
If you are looking for a way to check if the user input is a number, either positive or negative, or a decimal value (but not with an e, as in is_numeric), the following can be used -
<?php
mynumber = "3" // 3 is a number
$mynumber = "3.3" // 3.3 is a number
$mynumber = "-3.3" // -3.3 is a number
$mynumber = "-0,3" // -0,3 is a number
$mynumber = "-0,3x" // -0,3x is not a number
if (preg_match ("/^([0-9.,-]+)$/", $mynumber)) {
print $mynumber." is a number";
} else {
print $mynumber." is not a number";
}
?>
I have added comma (,) as in many european number system, decimal is expressed with a comma, and not with a dot. Like, $100 is expressed as $100,00 instead of $100.00
if you don't need that, remove the (,) and, you can add any additional value that you need to add after the series 0-9.,-
Stephen
26-Jan-2007 02:31
26-Jan-2007 02:31
The documentation is not completely precise here. is_numeric will also return true if the number begins with a decimal point and/or a space, provided a number follows (rather than a letter or punctuation). So, it doesn't necessarily have to start with a digit.
tenhamedo.net
20-Jan-2007 08:25
20-Jan-2007 08:25
For mariusads::at::helpedia.com
I believe this is mutch easyer:
function positive_number($nr) {
if(ereg("^[0-9]+$", $nr) && $nr > 0){
return true;
} else {
return false;
}
}
mariusads::at::helpedia.com
20-Jan-2007 06:56
20-Jan-2007 06:56
Test if a number is positive and contains only 0-9:
function is_number($number)
{
$text = (string)$number;
$textlen = strlen($text);
if ($textlen==0) return 0;
for ($i=0;$i < $textlen;$i++)
{ $ch = ord($text{$i});
if (($ch<48) || ($ch>57)) return 0;
}
return 1;
}
returns
0 : number contain character outside 0-9
1 : valid number.
meagar at gmail dot com
14-Jan-2007 12:52
14-Jan-2007 12:52
Miero: Your function doesn't match some special cases: '+1', '-0', '+0', all of which are valid integers. The easiest and most reliable way to get a definite integer match is with a regular expression:
function is_intval($value) {
return 1 === preg_match('/^[+-]?[0-9]+$/', $value);
}
This has two "problems" based on your input: it matches both '00' and '999999999999999999999999999999999' as valid integers.
I'm not sure why you wouldn't want to match "00". Regardless of whether somebody entered it in a form by accident or on purpose, it /is/ a valid integer, and in most instances you should accept it.
The second value, "999..." is also a valid integer, even if PHPs internal int type isn't precise enough to represent it.
alexander dot j dot summers at imperial dot ac dot uk
24-Oct-2006 01:05
24-Oct-2006 01:05
Note that the even simpler functions for checking if a variable contains an odd or even number below don't produce good results if you apply them to arguments which aren't numeric; I guess that was the idea of the originals.
e.g. using the functions defined in two posts below..
IS_ODD(null) returns false
IS_EVEN(null) returns true
is_odd(null) returns false
is_even(null) returns false
ja at krystof dot org
02-Sep-2006 10:10
02-Sep-2006 10:10
Here is a simple function to recognize whether the value is a natural number: (Zero is often exclude from the natural numbers, that's why there's the second parameter.)
<?php
function is_natural($val, $acceptzero = false) {
$return = ((string)$val === (string)(int)$val);
if ($acceptzero)
$base = 0;
else
$base = 1;
if ($return && intval($val) < $base)
$return = false;
return $return;
}
?>
moskalyuk at gmail dot com
22-Aug-2006 02:18
22-Aug-2006 02:18
is_numeric fails on the hex values greater than LONG_MAX, so having a large hex value parsed through is_numeric would result in FALSE being returned even though the value is a valid hex number
rontarrant at NO_SPAMsympatico dot ca
24-Jul-2006 11:15
24-Jul-2006 11:15
Here's an even simpler pair of functions for finding out if a number is odd or even:
function IS_ODD($number) { return($number & 1); }
function IS_EVEN($number) { return(!($number & 1)); }
Test:
$myNumber = 151;
if(IS_ODD($myNumber))
echo("number is odd\n");
else
echo("number is NOT odd\n");
if(IS_even($myNumber))
echo("number is even\n");
else
echo("number is NOT even\n");
Results:
number is odd
number is NOT even
Miero
09-Jul-2006 01:31
09-Jul-2006 01:31
function is_intval($a) {
return ((string)$a === (string)(int)$a);
}
true for ("123", "0", "-1", 0, 11, 9011, 00, 0x12, true)
false for (" ", "", 1.1, "123.1", "00", "0x123", "123a", "ada", "--1", "999999999999999999999999999999999", false, null, '1 ')
andrea dot vacondio at gmail dot com
14-Dec-2005 12:52
14-Dec-2005 12:52
Two simple functions using is_numeric:
<?php
function is_odd($num){
return (is_numeric($num)&($num&1));
}
function is_even($num){
return (is_numeric($num)&(!($num&1)));
}
//examples
echo "1: odd? ".(is_odd(1)? "TRUE": "FALSE")."<br />";
//is_numeric(0) returns true
echo "0: odd? ".(is_odd(0)? "TRUE": "FALSE")."<br />";
echo "6: odd? ".(is_odd(6)? "TRUE": "FALSE")."<br />";
echo "\"italy\": odd? ".(is_odd("italy")? "TRUE": "FALSE")."<br />";
echo "null: odd? ".(is_odd(null)? "TRUE": "FALSE")."<br /><br />";
echo "1: even? ".(is_even(1)? "TRUE": "FALSE")."<br />";
echo "0: even? ".(is_even(0)? "TRUE": "FALSE")."<br />";
echo "6: even? ".(is_even(6)? "TRUE": "FALSE")."<br />";
echo "\"italy\": even? ".(is_even("italy")? "TRUE": "FALSE")."<br />";
echo "null: even? ".(is_even(null)? "TRUE": "FALSE")."<br />";
?>
And here is the result:
1: odd? TRUE
0: odd? FALSE
6: odd? FALSE
"italy": odd? FALSE
null: odd? FALSE
1: even? FALSE
0: even? TRUE
6: even? TRUE
"italy": even? FALSE
null: even? FALSE
jamespam at hotmail dot com
09-Aug-2005 06:31
09-Aug-2005 06:31
Here's a function to determine if a variable represents a whole number:
function is_whole_number($var){
return (is_numeric($var)&&(intval($var)==floatval($var)));
}
just simple stuff...
is_whole_number(2.00000000001); will return false
is_whole_number(2.00000000000); will return true
namik at hub-cafe dot net
30-Jul-2005 10:11
30-Jul-2005 10:11
I needed a number_suffix function that takes numbers with thousand seperators (using number_format() function). Note that this doesn't properly handle decimals.
Example:
<?= number_suffix('1,021') ?> returns: 1,021st
Also, increasing the range above the condition statements increases efficiency. That's almost 20% of the numbers between 0 and 100 that get to end early.
<?
function number_suffix($number)
{
// Validate and translate our input
if ( is_numeric($number) )
{
// Get the last two digits (only once)
$n = $number % 100;
} else {
// If the last two characters are numbers
if ( preg_match( '/[0-9]?[0-9]$/', $number, $matches ) )
{
// Return the last one or two digits
$n = array_pop($matches);
} else {
// Return the string, we can add a suffix to it
return $number;
}
}
// Skip the switch for as many numbers as possible.
if ( $n > 3 && $n < 21 )
return $number . 'th';
// Determine the suffix for numbers ending in 1, 2 or 3, otherwise add a 'th'
switch ( $n % 10 )
{
case '1': return $number . 'st';
case '2': return $number . 'nd';
case '3': return $number . 'rd';
default: return $number . 'th';
}
}
?>
maninblack00 at mail dot ru
30-Jul-2005 08:23
30-Jul-2005 08:23
blazatek at wp dot pl wrote a function to check POST inputs for ASCII-keys or smth like that. there was an error while filtering $varvalue (it was only the value of the last filter 'addslashes')
here's the corrected function:
<?
function is_num ($num) {
// .....
}
function test_post($tab) {
$post = array();
$post = $tab;
foreach ($post as $varname => $varvalue) {
echo $tab[$varname];
if (empty($varvalue)) {
$post[$varname] = null;
}
elseif (is_num($varvalue)) {
$varvalue=trim($varvalue);
$varvalue=strip_tags($varvalue);
$varvalue=intval($varvalue);
$post[$varname]=addslashes($varvalue);
}
else {
$post[$varname]=NULL;
}
}//forech
return $post;
}
?>
hope that correction is correct ;)
sebu
28-Jul-2005 08:37
28-Jul-2005 08:37
Referring to previous post "Be aware if you use is_numeric() or is_float() after using set_locale(LC_ALL,'lang') or set_locale(LC_NUMERIC,'lang')":
This is totally wrong!
This was the example code:
-----
set_locale(LC_NUMERIC,'fr');
is_numeric(12.25); // Return False
is_numeric(12,25); // Return True
is_float(12.25); //Return False
is_float(12,25); //Return True
-----
This is nonsense!
- set_locale() does not exist, you must use setlocale() instead
- you have to enclose 12,25 with quotes; otherwise PHP will think that
the function gets _two_ arguments: 12 and 25 (depending on PHP version and setup you may additionally get a PHP warning)
- if you don't enclose 12,25 with quotes the first argument will be the inspected value (12), the second value (25) is discarded. And is_numeric(12) and is_float(12) is always TRUE
Corrected Example:
----
setlocale(LC_NUMERIC,'fr');
is_numeric(12.25); // Return True
is_numeric("12,25"); // Return False
is_float(12.25); //Return True
is_float("12,25"); //Return False
----
Remarks:
- is_float(12.25) is _always_ TRUE, 12.25 is a PHP language construct (a "value") and the way PHP interpretes files is definitely _not_ affected by the locale
- is_float("12,25") is _always_ FALSE, since is_float (other than is_numeric): if the argument is a string then is_float() always returns FALSE since it does a strict check for floats
And the corrected example shows: you get the _same_ results for every possible locale, is_numeric() does not depend on the locale.
ishmaelmakitla at yahoo dot com
20-May-2005 09:05
20-May-2005 09:05
//hello mates I just wanted to bring this forth so that you may see what I have been to
I tried to validate a certain field (name ) on my project web site design so that the only field acceptable is alphabetic the problem was that it has to include a space as in "Makitla M.I"
now ctype_alpha returned false as this includes a space
the same with
<?
if(!ctype_alpha($name)||!ctype_space($name))
?>
now this is what I did
<?
$name=explode(" ",$name);//to get rid of space charecter
$name=explode(".",$name)//to get rid of the dod
//finally I tested if the resulting $name is purely alphabetic
if(!ctype_alpha($name))
{
echo"the name entered contained some unacceptable charecters...please reenter";
exit;
}
//this solved my problem...hope this helps some of you!!
?>
Gregory Boshoff
15-May-2005 07:01
15-May-2005 07:01
As mentioned above use the ctype character type functions to determine a strings type as ctype is faster. ctype_digit has far better benchmarks than is_numeric.
// Example:
$num = '888';
if(ctype_digit($num) === TRUE):
echo 'The string variable $num contains the decimal value '.$num;
endif;
Titouthat
28-Apr-2005 04:47
28-Apr-2005 04:47
This function converts an input string into bool, int or float depending on its content.
<?php
function convert_type( $var )
{
if( is_numeric( $var ) )
{
if( (float)$var != (int)$var )
{
return (float)$var;
}
else
{
return (int)$var;
}
}
if( $var == "true" ) return true;
if( $var == "false" ) return false;
return $var;
}
?>
'90' return an int
'90.9' return a float
'true' return a bool
'90.0' return a int
lukesneeringer at gmail dot com
19-Mar-2005 01:32
19-Mar-2005 01:32
Regarding renimar at yahoo's function to yield ordinal numbers, the function lacks one thing. It accounts for numbers in the teens only if the number is below 100. If you used this function and gave 212 as the input, it would give 212nd, and not 212th. (Also, checking for numbers between 11 and 13 is sufficient, since 14-19 yield th either way.)
Therefore,
<?php if ($num >= 11 and $num <= 19) ?>
should be changed to...
<?php if ($num % 100>= 11 and $num % 100 <= 13) ?>
It will then work perfectly all the time.
Here's the entire function with the one line changed:
<?php
function ordinalize($num) {
if (!is_numeric($num))
return $num;
if ($num % 100 >= 11 and $num % 100 <= 13)
return $num."th";
elseif ( $num % 10 == 1 )
return $num."st";
elseif ( $num % 10 == 2 )
return $num."nd";
elseif ( $num % 10 == 3 )
return $num."rd";
else // $num % 10 == 0, 4-9
return $num."th";
}
?>
codeslinger at compsalot dot com
18-Feb-2005 02:00
18-Feb-2005 02:00
in version 4.3.10 I find the following
".73" TRUE
"0.73" TRUE
"+0.73" TRUE
"+.73" FALSE
I would not call it a bug, just something to be aware of.
-----
Also be aware that if you give php a huge number and then you convert it to a string you get
"INF"
if you pass that to mySQL etc. you could have a problem...
kiss dot pal at expert-net dot hu
06-Jan-2005 05:13
06-Jan-2005 05:13
function is_float_ex($pNum) {
$num_chars=("0123456789.,+-");
if (strlen(trim($pNum))==0) // empty $pNum -> null
return FALSE;
else {
$i=0;
$f=1; // modify
$v=strlen($num_chars)-$f;
while (($i<strlen($pNum)) && ($v>=0)) {
$v=strlen($num_chars)-$f;
while (($v>=0) && ($num_chars[$v]<>$pNum[$i]))
$v--;
if ($f==1) // Only first item + vagy -
$f=3;
if (($pNum[$i]=='.') ||
($pNum[$i]==','))
$f=5;
$i++;
}
if ($v<0)
return FALSE;
else
return TRUE;
}
}
drew at zitnay dot com
04-Jan-2005 11:07
04-Jan-2005 11:07
What php at thefriedmans dot net said below about .5 returning false and .0.5 returning true isn't true, at least not for me on PHP 5.0.0. The program:
<?php
echo (is_numeric('5') ? "true" : "false") . "\n";
echo (is_numeric('5.5') ? "true" : "false") . "\n";
echo (is_numeric('.5') ? "true" : "false") . "\n";
echo (is_numeric('.0.5') ? "true" : "false") . "\n";
?>
yields:
true
true
true
false
Just as I would expect.
Drew
mdallaire at virtuelcom dot com
18-Nov-2004 09:23
18-Nov-2004 09:23
Sometimes, we need to have no letters in the number and is_numeric does not quit the job.
You can try it this ways to make sure of the number format:
function new_is_unsigned_float($val) {
$val=str_replace(" ","",trim($val));
return eregi("^([0-9])+([\.|,]([0-9])*)?$",$val);
}
function new_is_unsigned_integer($val) {
$val=str_replace(" ","",trim($val));
return eregi("^([0-9])+$",$val);
}
function new_is_signed_float($val) {
$val=str_replace(" ","",trim($val));
return eregi("^-?([0-9])+([\.|,]([0-9])*)?$",$val);
}
function new_is_signed_integer($val) {
$val=str_replace(" ","",trim($val));
return eregi("^-?([0-9])+$",$val);
}
It returns 1 if okay and returns nothing "" if it's bad number formating.
php at thefriedmans dot net
14-Oct-2004 02:19
14-Oct-2004 02:19
is_numeric() in php5 returns false for strings with a leading decimal point:
<?php
is_numeric('5'); // true
is_numeric('5.5'); // true
is_numeric('.5'); // false
// but...
is_numeric('.0.5'); // true
?>
In certain situations, it may be useful to prepend a '0' to the string you're verifying with is_numeric():
<?php
if (is_numeric('0' . $user_input)) // ...
?>
renimar no spam at nospam yahoo dot com
03-May-2004 04:54
03-May-2004 04:54
A little function to ordinalize numbers using is_numeric() and accounting for the numbers in the teens.
<?php
function ordinalize($num) {
if (!is_numeric($num))
return $num;
if ($num >= 11 and $num <= 19)
return $num."th";
elseif ( $num % 10 == 1 )
return $num."st";
elseif ( $num % 10 == 2 )
return $num."nd";
elseif ( $num % 10 == 3 )
return $num."rd";
else
return $num."th";
}
// Demo
for ($i=1; $i<=25; $i++) {
print ordinalize($i) . " ";
}
// The loop returns:
// 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th 11th 12th
// 13th 14th 15th 16th 17th 18th 19th 20th 21st 22nd
// 23rd 24th 25th
?>
joe at kewlmail dot net
16-Jan-2004 08:12
16-Jan-2004 08:12
Here is a simple function that I found usefull for filtering user input into numbers. Basically, it attempts to fix fat fingering. For example:
$userString = "$654.4r5";
function numpass_filter($userString){
$money = explode(".", $userString);
//now $money[0] = "$645" and $money[1] = "4r5"
//next remove all characters save 0 though 9
//in both elements of the array
$dollars = eregi_replace("[^0-9]", null, $money[0]);
$cents = eregi_replace("[^0-9]", null, $money[1]);
//if there was a decimal in the original string, put it back
if((string)$cents!=null){
$cents = "." . $cents;
}
$result = $dollars . $cents;
return($result);
}
The output in this case would be '654.45'.
Please note that this function will work properly unless the user fat fingers an extra decimal in the wrong place.
kouber at saparev dot com
24-Nov-2003 08:05
24-Nov-2003 08:05
Note that this function is not appropriate to check if "is_numeric" for very long strings. In fact, everything passed to this function is converted to long and then to a double. Anything greater than approximately 1.8e308 is too large for a double, so it becomes infinity, i.e. FALSE. What that means is that, for each string with more than 308 characters, is_numeric() will return FALSE, even if all chars are digits.
However, this behaviour is platform-specific.
http://www.php.net/manual/en/language.types.float.php
In such a case, it is suitable to use regular expressions:
function is_numeric_big($s=0) {
return preg_match('/^-?\d+$/', $s);
}
blazatek at wp dot pl
21-Oct-2003 12:05
21-Oct-2003 12:05
the best way to check whether
variable is numeric is to check its ascii code :)
function is_num($var)
{
for ($i=0;$i<strlen($var);$i++)
{
$ascii_code=ord($var[$i]);
if ($ascii_code >=49 && $asci_code <=57)
continue;
else
return false;
}
return true;
}
this function can be usefull if you wont to chec eg $_POST
function test_post($tab)
{
$post = array();
$post=$tab;
echo $tab["user_name"];
foreach ($post as $varname => $varvalue)
{
if (empty($varvalue))
{
$post[$varname] = null;
}elseif (is_num($varvalue))
{
$post[$varname]=trim($varvalue);
$post[$varname]=strip_tags($varvalue);
$post[$varname]=intval($varvalue);
$post[$varname]=addslashes($varvalue);
}else
{
$post[$varname]=NULL;
}
}
}//forech
return $post;
}
stew_wilde at hotmail dot com
07-Aug-2003 05:25
07-Aug-2003 05:25
When using the exec() function in php to execute anther php script, any command line arguments passed the script will lose their type association, regardless of whether they are numeric or not, the same seems to hold true for strings as well.
ie : two scripts test.php:
<?php
$val = trim($argv[1]);
echo is_string($val);
?>
and testwrapper.php:
<?php
$tmp = 5;
exec("php ./test.php ".$tmp);
?>
Executing testwrapper.php on the command line will echo nothing (ie false), and false will be returned regardless of any escaping of parameters or other such attempts to overcome this. The solution then is to explicitly cast $val in test.php to be an int and then is_numeric will work. But as stated the same test was performed using a string for $val and the is_string() function and the same thing occurs. Not the end of the world, but something to be aware of :)
andi_nrw at hotmail dot com
13-Feb-2003 01:25
13-Feb-2003 01:25
<?php
/* This function is not useful if you want
to check that someone has filled in only
numbers into a form because for example
4e4 and 444 are both "numeric".
I used a regular expression for this problem
and it works pretty good. Maybe it is a good
idea to write a function and then to use it.
$input_number = "444"; // Answer 1
$input_number = "44 "; // Answer 2
$input_number = "4 4"; // Answer 2
$input_number = "4e4"; // Answer 2
$input_number = "e44"; // Answer 2
$input_number = "e4e"; // Answer 2
$input_number = "abc"; // Answer 2
*/
$input_number = "444";
if (preg_match ("/^([0-9]+)$/", $input_number)) {
print "Answer 1";
} else {
print "Answer 2";
}
?>
php dot net at davidmcarthur dot com
17-Jan-2003 05:02
17-Jan-2003 05:02
This is a little more explicit and won't break when the value can't be legally cast as an int
function is_intValued($var)
{
// Determines if a variable's value is an integer regardless of type
// meant to be an analogy to PHP's is_numeric()
if (is_int($var)) return TRUE;
if (is_string($var) and $var === (string)(int) $var) return TRUE;
if (is_float($var) and $var === (float)(int) $var) return TRUE;
else return FALSE;
}
redy dot r at madagascan dot com
11-Dec-2002 09:54
11-Dec-2002 09:54
Be aware if you use is_numeric() or is_float() after using set_locale(LC_ALL,'lang') or set_locale(LC_NUMERIC,'lang')
Example :
If at the beginning of your script, you declare :
<?php
set_locale(LC_NUMERIC,'fr');
?>
and after, you will get this :
<?php
is_numeric(12.25); // Return False
is_numeric(12,25); // Return True
is_float(12.25); //Return False
is_float(12,25); //Return True
?>
Because, for french language the decimal separator is ',' (Comma) instead of '.' (Dot).
martijnt at dataserve dot nl
02-Dec-2002 10:21
02-Dec-2002 10:21
If you want to make sure that a variable contains only one or more numbers as in the range 0-9, you could use this:
eregi("[^0-9]",$var)
This checks your variable for anything that is NOT in the range 0-9.
If you use is_numeric() while checking an IP address, you should not be surprised if the following is accepted as a valid IP: 1e13.3.4.12e3 -> is_numeric() considers 1e13 and 12e3 as valid numeric values (which is correct).
Have fun.
Martijn Tigchelaar,
DataServe.
eagleflyer2 at lycos dot com
07-Sep-2002 05:02
07-Sep-2002 05:02
Hi !
Many of you may have experienced that the 'is_numeric' function seems to fail always when form entries are checked against their variable type. So the function seems to return 'false' even if the form entry was aparently a number or numeric string.
The solution is pretty simple and no subroutines or fancy operations are necessary to make the 'is_numeric' function usable for form entry checks:
Simply strip off all (invisible) characters that may be sent along with the value when submitting a form entry.
Just use the 'trim' function before 'is_numeric'.
Example:
$variable = trim($variable);
if (is_numeric($variable)
{...#do something#...}
else
{...#do something else#...}
mjbraca at NOSPAM dot hotmail dot com
25-Oct-2001 02:36
25-Oct-2001 02:36
Some examples. Note that leading white space is OK, but not trailing white space, and there can't be white space between the "-" and the number.
is_numeric("1,000") = F
is_numeric("1e2") = T
is_numeric("-1e-2") = T
is_numeric("1e2.3") = F
is_numeric("1.") = T
is_numeric("1.2") = T
is_numeric("1.2.3") = F
is_numeric("-1") = T
is_numeric("- 1") = F
is_numeric("--1") = F
is_numeric("1-") = F
is_numeric("1A") = F
is_numeric(" 1") = T
is_numeric("1 ") = F
tvali at email dot ee
11-Oct-2001 04:53
11-Oct-2001 04:53
I changed function sent by
"ealma@hotmail.com 18-May-2001 12:02"
a little bit (it should be faster now).
Here is some code that will detect large values for numeric and for PHP3.
function is_num($s) {
for ($i=0; $i<strlen($s); $i++) {
if (($s[$i]<'0') or ($s[$i]>'9')) {return false;}
}
return true;
}
You can also check only the first char in string (($s[0])<'9')and($s[0]>'0')), since php converts string to numeric only from beginning to the last digit.
aulbach at unter dot franken dot de
18-Sep-2001 03:07
18-Sep-2001 03:07
If you want to check, if a string will be converted into the exactly same number, you have to test the following:
if ( (string)(int)$val === (string)$val ) ...
This is not the same as
if ( is_numeric($val) ) ...
cause
<?php
$a="0000001";
if ( is_numeric($a) ) echo "a is numeric";
if ( (string)(float)$a !== (string)$a) echo "a cannot be converted 100%-correctly";
?>
will output:
a is numeric
a cannot be converted 100%-correctly
Ok, ok, I know. This is a hysteric type-check. It depends on the type of application, if you use it. The aim should always be, that the user could be warned, if the value cannot be stored in the way, the user types it into.
Why? Therefore is the following example:
If you set
$a="1E+17";
the above script says that it is all right.
But then we want to convert to INT, not FLOAT. is_numeric() doesn't distinct between INT and FLOAT, so a is_numeric('1E+17') is true, but the Integer of it returns 1, which isn't perhaps, what the user expects.
03-Apr-2001 02:20
Seems that this function can only verify numbers up to 16 digits long. Eg:
1111111111111111111
A 19-digit number will return false.