substr_replace

substr_compare

Last updated: Fri, 13 Nov 2009

substr_count

(PHP 4, PHP 5)

substr_count — 副文字列の出現回数を数える

説明

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

substr_count() は、文字列 haystack の中での副文字列 needle の出現回数を返します。 needle は英大小文字を区別することに注意してください。

注意: この関数は重なり合う副文字列をカウントしません。以下の例を見てください !

パラメータ

haystack: 検索対象の文字列
needle: 検索する副文字列
offset: 開始位置のオフセット
length: 指定したオフセット以降に副文字列で検索する最大長。オフセットと長さの総和が haystack の長さよりも長い場合、警告が発生します。

返り値

この関数は整数を返します。

変更履歴

バージョン	説明
5.1.0	`offset` と `length` パラメータが追加されました。

例

例1 substr_count() の例


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

echo substr_count($text, 'is'); // 2

// 文字列は 's is a test' になっているので, 1 が表示される
echo substr_count($text, 'is', 3);

// テキストは 's i' になっているので, 0 が表示される
echo substr_count($text, 'is', 3, 3);

// 5+10 > 14 なので、警告が発生する
echo substr_count($text, 'is', 5, 10);


// 重なっている副文字列はカウントされないので、1 が表示される
$text2 = 'gcdgcdgcd';
echo substr_count($text2, 'gcdgcd');
?>

参考

count_chars() - 文字列で使用されている文字に関する情報を返す
strpos() - 文字列が最初に現れる場所を見つける
substr() - 文字列の一部分を返す
strstr() - 文字列が最初に現れる位置を見つける

substr_replace

substr_compare

Last updated: Fri, 13 Nov 2009

add a note User Contributed Notes
substr_count

chrisstocktonaz at gmail dot com
27-Aug-2009 08:52


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 05:37


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
24-Mar-2009 06:53


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
13-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 10:49


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 10:37


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
06-May-2007 08:07


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
26-Oct-2006 11:07


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
22-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.

add a note

substr_replace

substr_compare

Last updated: Fri, 13 Nov 2009