(PHP 4, PHP 5)
substr_count — 计算字串出现的次数
$haystack
, string $needle
[, int $offset = 0
[, int $length
]] )
substr_count() 返回子字符串needle 在字符串 haystack 中出现的次数。注意 needle 区分大小写。
Note:
该函数不会计算重叠字符串。参见下面的例子。
haystack
在此字符串中进行搜索。
needle
要搜索的字符串。
offset
开始计数的偏移位置。
length
指定偏移位置之后的最大搜索长度。如果偏移量加上这个长度的和大于 haystack 的总长度，则打印警告信息。
该函数返回整型。
| 版本 | 说明 |
|---|---|
| 5.1.0 |
新增 offset 和 length 参数。
|
Example #1 substr_count() 范例
<?php
$text = 'This is a test';
echo strlen($text); // 14
echo substr_count($text, 'is'); // 2
// 字符串被简化为 's is a test'，因此输出 1
echo substr_count($text, 'is', 3);
// 字符串被简化为 's i'，所以输出 0
echo substr_count($text, 'is', 3, 3);
// 因为 5+10 > 14，所以生成警告
echo substr_count($text, 'is', 5, 10);
// 输出 1，因为该函数不计算重叠字符串
$text2 = 'gcdgcdgcd';
echo substr_count($text2, 'gcdgcd');
?>
Douglas Dennis (2013-04-29 19:01:53)
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".
chrisstocktonaz at gmail dot com (2009-08-26 16:52:21)
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 (2009-06-18 01:37:48)
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 (2009-03-23 13:53:19)
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 (2009-01-12 08:17:33)
below was suggested a function for substr_count'ing an array, yet for a simpler procedure, use the following:
<?php
substr_count ( implode( $haystackArray ), $needle );
?>
Anonymous (2008-07-11 06:49:34)
It should be noted that unlike the other substr functions, the offset value cannot be a negative value.
<?php
echo substr_count('abcdefg', 'efg', 4, 3); // 1
echo substr_count('abcdefg', 'efg', -3, 3); // warning
?>
danjr33 at gmail dot com (2007-07-24 06:37:07)
I ran into trouble using this function when I moved a script from a server with PHP5 to a server with only PHP4.
As the last two parameters were added with 5.1.0, I wrote a substitute function:
<?php
function substr_count5($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 (2007-05-05 16:07:11)
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 $count;
}
?>
flobi at flobi dot com (2006-10-25 19:07:11)
Making this case insensitive is easy for anyone who needs this. Simply convert the haystack and the needle to the same case (upper or lower).
substr_count(strtoupper($haystack), strtoupper($needle))
XinfoX X at X XkarlX X-X XphilippX X dot X XdeX (2003-12-21 13:27:00)
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.