ctype_cntrl
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"
ctype_digit
(PHP 4 >= 4.0.4, PHP 5)
ctype_digit — Auf Ziffern überprüfen
Beschreibung
bool ctype_digit
( string $text
)
Prüft ob der übergebene String nur aus Ziffern besteht.
Parameter-Liste
- text
-
Der zu prüfende String.
Rückgabewerte
Liefert TRUE wenn jedes Zeichen in text eine Ziffer ist, ansonsten FALSE.
Beispiele
Beispiel #1 ctype_digit() Beispiel
<?php
$strings = array('1820.20', '10002', 'wsl!12');
foreach ($strings as $testcase) {
if (ctype_digit($testcase)) {
echo "Der String $testcase besteht aus Ziffern.\n";
} else {
echo "Der String $testcase enthält nicht nur Ziffern.\n";
}
}
?>
Das oben gezeigte Beispiel erzeugt folgende Ausgabe:
Der String 1820.20 enthält nicht nur Ziffern. Der String 10002 besteht aus Ziffern. Der String wsl!12 enthält nicht nur Ziffern.
add a note
User Contributed Notesctype_digit
per dot zut at gmail dot com
12-Sep-2008 12:09
12-Sep-2008 12:09
minterior at gmail dot com
10-Sep-2007 04:43
10-Sep-2007 04:43
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 12:44
19-Aug-2007 12:44
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
16-Aug-2007 06:46
16-Aug-2007 06:46
@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 10:39
12-Jul-2007 10:39
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 11:06
08-Jul-2007 11:06
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 11:46
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' :
<?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 09:31
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
07-May-2007 12:09
07-May-2007 12:09
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.