(PHP 4, PHP 5)

substr_countВозвращает число вхождений подстроки


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

substr_count() возвращает число вхождений подстроки needle в строку haystack. Заметьте, что параметр needle чувствителен к регистру.


Эта функция не подсчитывает перекрывающиеся подстроки. Смотрите пример ниже!

Список параметров


Строка, в которой ведется поиск


Искомая подстрока


Смещение начала отсчета


Максимальная длина строки в которой будет производится поиск подстроки после указанного смещения. Если сумма смещения и максимальной длины будет больше длины haystack, то будет выведено предупреждение.

Возвращаемые значения

Эта функция возвращает integer.

Список изменений

Версия Описание
5.1.0 Добавлены параметры offset и length


Пример #1 Пример использования substr_count()

'This is a test';
strlen($text); // 14

echo substr_count($text'is'); // 2

// строка уменьшается до 's is a test', поэтому вывод будет 1
echo substr_count($text'is'3);

// текст уменьшается до 's i', поэтому вывод будет 0
echo substr_count($text'is'33);

// генерирует предупреждение, так как  5+10 > 14
echo substr_count($text'is'510);

// выводит только 1, т.к. перекрывающиеся подстроки не учитываются
$text2 'gcdgcdgcd';

Смотрите также

  • count_chars() - Возвращает информацию о символах, входящих в строку
  • strpos() - Возвращает позицию первого вхождения подстроки
  • substr() - Возвращает подстроку
  • strstr() - Находит первое вхождение подстроки

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User Contributed Notes 7 notes

jrhodes at roket-enterprises dot com
4 years ago
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.
gigi at phpmycoder dot com
5 years ago
below was suggested a function for substr_count'ing an array, yet for a simpler procedure, use the following:

( implode( $haystackArray ), $needle );
info at fat-fish dot co dot il
6 years ago
a simple version for an array needle (multiply sub-strings):

function substr_count_array( $haystack, $needle ) {
$count = 0;
     foreach (
$needle as $substring) {
$count += substr_count( $haystack, $substring);
flobi at flobi dot com
7 years ago
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
10 years ago
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.
qeremy [atta] gmail [dotta] com
7 months ago
Unicode example with "case-sensitive" option;

function substr_count_unicode($str, $substr, $caseSensitive = true, $offset = 0, $length = null) {
    if (
$offset) {
$str = substr_unicode($str, $offset, $length);

$pattern = $caseSensitive
? '~(?:'. preg_quote($substr) .')~u'
: '~(?:'. preg_quote($substr) .')~ui';
preg_match_all($pattern, $str, $matches);

    return isset(
$matches[0]) ? count($matches[0]) : 0;

substr_unicode($str, $start, $length = null) {
join('', array_slice(
preg_split('~~u', $str, -1, PREG_SPLIT_NO_EMPTY), $start, $length));

$s = 'Ümit yüzüm gözüm...';
substr_count_unicode($s, 'ü');            // 3
print substr_count_unicode($s, 'ü', false);     // 4
print substr_count_unicode($s, 'ü', false, 10); // 1

print substr_count_unicode($s, 'üm');           // 2
print substr_count_unicode($s, 'üm', false);    // 3
chrisstocktonaz at gmail dot com
4 years ago
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 = '';
$url = '';

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))

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