In regards to anyone thinking of using code contributed by zmindster at gmail dot com
Please take careful consideration of possible edge cases with that regex, in example:
$url = 'http://w3.host.tld/path/to/file/..../file.extension';
$url = 'http://w3.host.tld/path/to/file/../file.extension?malicous=....';
This would cause a infinite loop and for example be a possible entry point for a denial of service attack. A correct fix would require additional code, a quick hack would be just adding a additional check, without clarity or performance in mind:
...
$i = 0;
while (substr_count($url, '../') && ++$i < strlen($url))
...
-Chris
substr_count
(PHP 4, PHP 5)
substr_count — Conta o número de ocorrências de uma substring
Descrição
int substr_count
( string $haystack
, string $needle
[, int $offset
[, int $length
]] )
substr_count() retorna o número de vezes que a substring needle ocorre na string. Note que needle faz distinção de maiúscula e minúscula.
Nota:
Esta função não conta substrings sobrepostas. Veja o exemplo abaixo!
Parâmetros
- haystack
-
A string onde será feita a busca
- needle
-
A substring a ser procurada
- offset
-
O índice onde inicia a contagem
- length
-
O valor máximo depois do índice especificado para buscar pela substring. Mostra um aviso se o tamanho do índice for maior que o tamanho de haystack.
Valor Retornado
Esta função retorna um integer.
Histórico
| Versão | Descrição |
|---|---|
| 5.1.0 | Adiciona os parâmetros offset e length |
Exemplos
Exemplo #1 Exemplo da substr_count()
<?php
$text = 'This is a test';
echo strlen($text); // 14
echo substr_count($text, 'is'); // 2
// the string is reduced to 's is a test', so it prints 1
echo substr_count($text, 'is', 3);
// the text is reduced to 's i', so it prints 0
echo substr_count($text, 'is', 3, 3);
// generates a warning because 5+10 > 14
echo substr_count($text, 'is', 5, 10);
// prints only 1, because it doesn't count overlapped subtrings
$text2 = 'gcdgcdgcd';
echo substr_count($text2, 'gcdgcd');
?>
Veja Também
- count_chars() - Retorna informações sobre os caracteres usados numa string
- strpos() - Encontra a posição da primeira ocorrência de uma string
- substr() - Retorna uma parte de uma string
- strstr() - Encontra a primeira ocorrencia de uma string
chrisstocktonaz at gmail dot com
26-Aug-2009 10:52
jrhodes at roket-enterprises dot com
17-Jun-2009 07:37
It was suggested to use
substr_count ( implode( $haystackArray ), $needle );
instead of the function described previously, however this has one flaw. For example this array:
array (
0 => "mystringth",
1 => "atislong"
);
If you are counting "that", the implode version will return 1, but the function previously described will return 0.
zmindster at gmail dot com
23-Mar-2009 07:53
For some who seeked for an easy way to resolve URL composed of /../ like http://w3.host.tld/path/to/the/file/../../file.extension, here is a solution
<?php
$url = 'http://w3.host.tld/path/to/file/../file.extension';
while (substr_count($url, "../"))
{
$url = preg_replace('#/[^/]+/\.\.#', '', $url);
}
//outputs 'http://w3.host.tld/path/to/file.extension'
?>
and seems to work perfectly!
gigi at phpmycoder dot com
12-Jan-2009 01:17
below was suggested a function for substr_count'ing an array, yet for a simpler procedure, use the following:
<?php
substr_count ( implode( $haystackArray ), $needle );
?>
Anonymous
11-Jul-2008 12:49
It should be noted that unlike the other substr functions, the offset value cannot be a negative value.
<?php
echo substr_count('abcdefg', 'efg', 4, 3); // 1
echo substr_count('abcdefg', 'efg', -3, 3); // warning
?>
danjr33 at gmail dot com
24-Jul-2007 12:37
I ran into trouble using this function when I moved a script from a server with PHP5 to a server with only PHP4.
As the last two parameters were added with 5.1.0, I wrote a substitute function:
<?php
function substr_count5($str,$search,$offset,$len) {
return substr_count(substr($str,$offset,$len),$search);
}
?>
Use it exactly as substr_count() is used in PHP5. (This will work in PHP5 as well.)
info at fat-fish dot co dot il
05-May-2007 10:07
a simple version for an array needle (multiply sub-strings):
<?php
function substr_count_array( $haystack, $needle ) {
$count = 0;
foreach ($needle as $substring) {
$count += substr_count( $haystack, $substring);
}
return $count;
}
?>
flobi at flobi dot com
25-Oct-2006 01:07
Making this case insensitive is easy for anyone who needs this. Simply convert the haystack and the needle to the same case (upper or lower).
substr_count(strtoupper($haystack), strtoupper($needle))
XinfoX X at X XkarlX X-X XphilippX X dot X XdeX
21-Dec-2003 06:27
Yet another reference to the "cgcgcgcgcgcgc" example posted by "chris at pecoraro dot net":
Your request can be fulfilled with the Perl compatible regular expressions and their lookahead and lookbehind features.
The example
$number_of_full_pattern = preg_match_all('/(cgc)/', "cgcgcgcgcgcgcg", $chunks);
works like the substr_count function. The variable $number_of_full_pattern has the value 3, because the default behavior of Perl compatible regular expressions is to consume the characters of the string subject that were matched by the (sub)pattern. That is, the pointer will be moved to the end of the matched substring.
But we can use the lookahead feature that disables the moving of the pointer:
$number_of_full_pattern = preg_match_all('/(cg(?=c))/', "cgcgcgcgcgcgcg", $chunks);
In this case the variable $number_of_full_pattern has the value 6.
Firstly a string "cg" will be matched and the pointer will be moved to the end of this string. Then the regular expression looks ahead whether a 'c' can be matched. Despite of the occurence of the character 'c' the pointer is not moved.
