substr_compare
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.
substr_count
(PHP 4, PHP 5)
substr_count — Ermittelt, wie oft eine Zeichenkette in einem String vorkommt
Beschreibung
int substr_count
( string
$haystack
, string $needle
[, int $offset = 0
[, int $length
]] )
Die Funktion substr_count() ermittelt, wie oft
needle in dem String haystack
vorkommt, und gibt die Anzahl der Vorkommen zurück. Beachten Sie, dass der
Parameter needle case sensitive ist.
Hinweis:
Die Funktion zählt einander überlappende Substrings nicht mit. Beachten Sie das untenstehende Beispiel!
Parameter-Liste
-
haystack -
Der String, in dem gesucht werden soll
-
needle -
Der Substring, nach dem gesucht werden soll
-
offset -
Die Zeichenposition, an der die Zählung begonnen werden soll
-
length -
Die maximale Länge nach dem angegebenen Offset, in der nach dem Substring gesucht werden soll. Es wird eine Warnung ausgegeben, wenn Offset plus Länge größer als die Länge von
haystacksind.
Rückgabewerte
Die Funktion gibt einen Wert vom Typ integer zurück.
Changelog
| Version | Beschreibung |
|---|---|
| 5.1.0 |
Hinzufügen der Parameter offset und
length
|
Beispiele
Beispiel #1 Ein substr_count() Beispiel
<?php
$text = 'Dies ist ein Test';
echo strlen($text); // 17
echo substr_count($text, 'es'); // 2
// wird der String auf 's ist ein Test' reduziert,
// lautet das ausgegebene Ergebnis 1
echo substr_count($text, 'es', 3);
// wird der String auf 's i' reduziert,
// lautet das Ergebnis 0
echo substr_count($text, 'es', 3, 3);
// generiert eine Warnung, da 5+13 > 17
echo substr_count($text, 'es', 5, 13);
// gibt 1 aus, da überlappende Substrings nicht gezählt werden
$text2 = 'gcdgcdgcd';
echo substr_count($text2, 'gcdgcd');
?>
Siehe auch
- count_chars() - Gibt Informationen über die in einem String enthaltenen Zeichen zurück
- strpos() - Sucht das erste Vorkommen des Suchstrings
- substr() - Gibt einen Teil eines Strings zurück
- strstr() - Findet das erste Vorkommen eines Strings
add a note
User Contributed Notes
substr_count - [10 notes]
jrhodes at roket-enterprises dot com ¶
3 years ago
info at fat-fish dot co dot il ¶
6 years ago
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;
}
?>
XinfoX X at X XkarlX X-X XphilippX X dot X XdeX ¶
9 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.
chrisstocktonaz at gmail dot com ¶
3 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 = '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
Douglas Dennis ¶
23 days ago
It was previously noted that if you use "substr_count(implode(array))" with this array:
array (
0 => "mystringth",
1 => "atislong"
);
The following would result:
"If you are counting "that", the implode version will return 1, but the function previously described will return 0."
You can fix this by using substr_count(implode(' ', array), 'that') which will result in the following string: "mystringth atislong".
gigi at phpmycoder dot com ¶
4 years ago
below was suggested a function for substr_count'ing an array, yet for a simpler procedure, use the following:
<?php
substr_count ( implode( $haystackArray ), $needle );
?>
danjr33 at gmail dot com ¶
5 years ago
I ran into trouble using this function when I moved a script from a server with PHP5 to a server with only PHP4.
As the last two parameters were added with 5.1.0, I wrote a substitute function:
<?php
function substr_count5($str,$search,$offset,$len) {
return substr_count(substr($str,$offset,$len),$search);
}
?>
Use it exactly as substr_count() is used in PHP5. (This will work in PHP5 as well.)
flobi at flobi dot com ¶
6 years ago
Making this case insensitive is easy for anyone who needs this. Simply convert the haystack and the needle to the same case (upper or lower).
substr_count(strtoupper($haystack), strtoupper($needle))
Anonymous ¶
4 years ago
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
?>
zmindster at gmail dot com ¶
4 years ago
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!