substr_count

(PHP 4, PHP 5)

substr_count — Връща броя на появите на подниз

Описание

int substr_count ( string $haystack , string $needle [, int $offset = 0 [, int $length ]] )

substr_count() връща броя появи на низа needle в низа haystack . Моля забележете, че параметърът needle е чувствителен към регистъра.

Забележка: Функцията не брои застъпващите се поднизове. Вижте примера по-долу!

Параметри

haystack: Низът в който ще се търси
needle: Поднизът за който ще се търси
offset: Отместването от което да започне броенето
length: Максималната дължина, след указаното отместване за търсене на подниза. Ако отместването плюс дължината на низа needle е по-голямо от дължината на haystack се извежда предупреждение.

Връщани стойности

Функцията връща цяло число.

Дневник на промените

Версия	Описание
5.1.0	Добавени са параметрите `offset` и `length`

Примери

Example #1 Примери за substr_count()


<?php
$text = 'This is a test';
echo strlen($text); // 14

echo substr_count($text, 'is'); // 2

// текстът е намален до 's is a test', поради което се извежда 1
echo substr_count($text, 'is', 3);

// текстът е намален до 's i', поради което се извежда 0
echo substr_count($text, 'is', 3, 3);

// генерира се предупреждение тъй като 5+10 > 14
echo substr_count($text, 'is', 5, 10);


// извежда само 1, тъй като не се отброяват застъпващите се поднизове
$text2 = 'gcdgcdgcd';
echo substr_count($text2, 'gcdgcd');
?>

Вж. също

count_chars() - Връща информация за знаците в даден низ
strpos() - Връща позицията на първата поява на подниз
substr() - Връща част от низ
strstr() - Връща първата поява на низ

substr_replace

substr_compare

[edit] Last updated: Fri, 18 Sep 2009

add a note User Contributed Notes substr_count

chrisstocktonaz at gmail dot com 26-Aug-2009 04:52


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

jrhodes at roket-enterprises dot com 18-Jun-2009 01:37


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 01:53


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 08:17


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 06:49


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 06:37


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 04:07


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 07:07


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 01:27


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.

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