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ctype_graph> <ctype_cntrl
Last updated: Sun, 25 Nov 2007

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ctype_digit

(PHP 4 >= 4.0.4, PHP 5)

ctype_digit — Check for numeric character(s)

설명

bool ctype_digit ( string $text )

Checks if all of the characters in the provided string, text , are numerical.

매개변수

text

The tested string.

반환값

Returns TRUE if every character in text is a decimal digit, FALSE otherwise.

예제

Example#1 A ctype_digit() example

<?php
$strings 
= array('1820.20''10002''wsl!12');
foreach (
$strings as $testcase) {
    if (
ctype_digit($testcase)) {
        echo 
"The string $testcase consists of all digits.\n";
    } else {
        echo 
"The string $testcase does not consist of all digits.\n";
    }
}
?>

위 예제의 출력:

The string 1820.20 does not consist of all digits.
The string 10002 consists of all digits.
The string wsl!12 does not consist of all digits.



ctype_graph> <ctype_cntrl
Last updated: Sun, 25 Nov 2007
 
add a note add a note User Contributed Notes
ctype_digit
per dot zut at gmail dot com
11-Sep-2008 03:09
reuvenab at gmail dot com
"Just be aware that
ctype_digit('') == 1"

I tested this.

<?php
var_dump
(ctype_digit(''));
var_dump(phpversion());
?>

Result:

bool(false)
string(16) "5.2.4-2ubuntu5.3"
minterior at gmail dot com
10-Sep-2007 07:43
I use ctype_digit() function as a part of this IMEI validation function.

<?php

/**
 * Check the IMEI of a mobile phone
 * @param $imei IMEI to validate
 */
function is_IMEI_valid($imei){   
    if(!
ctype_digit($imei)) return false;
   
$len = strlen($imei);
    if(
$len != 15) return false;

    for(
$ii=1, $sum=0 ; $ii < $len ; $ii++){
        if(
$ii % 2 == 0) $prod = 2;
        else
$prod = 1;
       
$num = $prod * $imei[$ii-1];
        if(
$num > 9){
         
$numstr = strval($num);
         
$sum += $numstr[0] + $numstr[1];
        }else
$sum += $num;
    }

   
$sumlast = intval(10*(($sum/10)-floor($sum/10))); //The last digit of $sum
   
$dif = (10-$sumlast);
   
$diflast = intval(10*(($dif/10)-floor($dif/10))); //The last digit of $dif
   
$CD = intval($imei[$len-1]); //check digit

   
if($diflast == $CD) return true;

    return
false;
}
?>
ipernet at gmail dot com
19-Aug-2007 03:44
Be careful !

ctype_digit(36) === false because 36 is type INT and ctype_digit take only string

ctype_digit('36') === true
JustinB at harvest dot org dot REMOVE
15-Aug-2007 09:46
@robert at mediamonks dot com & withheld at withheld dot com:
I personally believe more in solutions than problems.  If you're concerned about the possibility of an int type variable being passed to ctype_digit()--and don't want to add another conditional statement using is_int() to avoid the problem--there's another simple solution: typecast it.

<?php
// The Problem
$test_values = array(123,'456','7eight9');
foreach(
$test_values as $test) {
    if(
ctype_digit($test)) echo 'True' . "\n";
    else echo
'False' . "\n";
}
// OUTPUT: False, True, False

// An Easy Solution
$test_values = array(123,'456','7eight9');
foreach(
$test_values as $test) {
    if(
ctype_digit((string)$test)) echo 'True' . "\n";
    else echo
'False' . "\n";
}
// OUTPUT: True, True, False
?>
robert at mediamonks dot com
12-Jul-2007 01:39
withheld at withheld dot com:
it is called : 'User Contributed Notes' not 'Bugs' so I am not saying there is something wrong with the function at all.

It is used for tips, tricks and simple mistakes people could make using the function. I just noted something just like that, nothing wrong with that.
withheld at withheld dot com
08-Jul-2007 02:06
robert at mediamonks dot com writes...

04-Jun-2007 11:46
I always used this function to check user input but it failed me when I wanted to check int's used in my script. This is caused by the fact that the function only functions right when the argument is a string. Here is a solution for this 'problem' :

Could we have some accuracy in these comments please? If the spec at the top of this page is taken to be correct then Robert is incorrect as it clearly states that the argument should be a string. Skimming notes in this section suggests to the reader that there is a problem with this function when actually people just arent reading the documentation - proliferation of careless mistakes.
robert at mediamonks dot com
04-Jun-2007 02:46
I always used this function to check user input but it failed me when I wanted to check int's used in my script. This is caused by the fact that the function only functions right when the argument is a string. Here is a solution for this 'problem' :

<?php
$intCheckThis
= 99;

$blnResult = ctype_digit($intCheckThis); // returns FALSE

$blnResult = ctype_digit(strval($intCheckThis)); // returns TRUE
?>

Note that user input from form or request uri is always handled as string anyway.
15-May-2007 12:31
You have your if() statement backwards. It's returning "This does not consist of only digits." when cytype_digit($var) is true.

<?php
$var
= 5;
if (
cytype_digit($var))
{
    echo
"This does not consist of only digits.";
}
else
{
    echo
"This cosists of only digits.";
}
?>
shivanfalcon at gmail dot com
06-May-2007 03:09
This function also seems to give a false for integers (PHP 4.4.4)
<?php
$var
= 5;
if (
cytype_digit($var))
{
    echo
"This does not consist of only digits.";
}
else
{
    echo
"This cosists of only digits.";
}
?>
Returns "This does not consist of only digits.".
I can only assume that this is because of how integers are seen.
2 = 00000010 bitwise
bitwise, 00000010 happens to be some ASCII formatting character, which isn't a digit.
reuvenab at gmail dot com
05-Apr-2007 02:44
Just be aware that
ctype_digit('') == 1

ctype_graph> <ctype_cntrl
Last updated: Sun, 25 Nov 2007
 
 
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