PHP Velho Oeste 2024

strripos

(PHP 5, PHP 7, PHP 8)

strripos计算指定字符串在目标字符串中最后一次出现的位置(不区分大小写)

说明

strripos(string $haystack, string $needle, int $offset = 0): int|false

以不区分大小写的方式查找指定字符串在目标字符串中最后一次出现的位置。与 strrpos() 不同,strripos() 不区分大小写。

参数

haystack

在此字符串中进行查找。

needle

要搜索的字符串。

Prior to PHP 8.0.0, if needle is not a string, it is converted to an integer and applied as the ordinal value of a character. This behavior is deprecated as of PHP 7.3.0, and relying on it is highly discouraged. Depending on the intended behavior, the needle should either be explicitly cast to string, or an explicit call to chr() should be performed.

offset

如果为 0 或正数,则从左到右搜索,跳过 haystack 的开头 offset 个字节。

如果为负数,则从右向左执行搜索,跳过 haystack 的最后 offset 个字节并搜索首次出现的 needle

注意:

这实际是在最后 offset 个字节之前寻找最后出现的 needle

返回值

返回 needle 相对于 haystack 字符串的位置(和搜索的方向和偏移量无关)。

注意: 字符串位置从 0 开始,而不是 1。

如果未找到 needle,则返回 false

警告

此函数可能返回布尔值 false,但也可能返回等同于 false 的非布尔值。请阅读 布尔类型章节以获取更多信息。应使用 === 运算符来测试此函数的返回值。

更新日志

版本 说明
8.0.0 needle 现在接受空字符串。
8.2.0 大小写转换不在依赖于使用 setlocale() 设置的区域。只会进行 ASCII 大小写转换。非 ASCII 字节值将通过它们的字节值进行比较。
8.0.0 不再支持将 int 传递给 needle
7.3.0 弃用将 int 传递给 needle

示例

示例 #1 strripos() 简单示例

<?php
$haystack
= 'ababcd';
$needle = 'aB';

$pos = strripos($haystack, $needle);

if (
$pos === false) {
echo
"Sorry, we did not find ($needle) in ($haystack)";
} else {
echo
"Congratulations!\n";
echo
"We found the last ($needle) in ($haystack) at position ($pos)";
}
?>

以上示例会输出:

Congratulations!
   We found the last (aB) in (ababcd) at position (2)

参见

  • strpos() - 查找字符串首次出现的位置
  • stripos() - 查找字符串首次出现的位置(不区分大小写)
  • strrchr() - 查找指定字符在字符串中的最后一次出现
  • substr() - 返回字符串的子串
  • stristr() - strstr 函数的忽略大小写版本
  • strstr() - 查找字符串的首次出现

add a note

User Contributed Notes 7 notes

up
4
Yanik Lupien
16 years ago
Simple way to implement this function in PHP 4

<?php
if (function_exists('strripos') == false) {
function
strripos($haystack, $needle) {
return
strlen($haystack) - strpos(strrev($haystack), $needle);
}
}

?>
up
2
dimmav at in dot gr
15 years ago
Suppose you just need a stripos function working backwards expecting that strripos does this job, you better use the following code of a custom function named strbipos:

<?php
function strbipos($haystack="", $needle="", $offset=0) {
// Search backwards in $haystack for $needle starting from $offset and return the position found or false

$len = strlen($haystack);
$pos = stripos(strrev($haystack), strrev($needle), $len - $offset - 1);
return ( (
$pos === false) ? false : $len - strlen($needle) - $pos );
}

// Test
$body = "01234Xy7890XYz456xy90";
$str = "xY";
$len = strlen($body);
echo
"TEST POSITIVE offset VALUES IN strbipos<br>";
for (
$i = 0; $i < $len; $i++) {
echo
"Search in [$body] for [$str] starting from offset [$i]: [" . strbipos($body, $str, $i) . "]<br>";
}
?>

Note that this function does exactly what it says and its results are different comparing to PHP 5 strripos function.
up
1
peev[dot]alexander at gmail dot com
16 years ago
I think you shouldn't underestimate the length of $needle in the search of THE FIRST POSITION of it's last occurrence in the string. I improved the posted function, with added support for offset. I think this is an exact copy of the real function:

<?php
if(!function_exists("strripos")){
function
strripos($haystack, $needle, $offset=0) {
if(
$offset<0){
$temp_cut = strrev( substr( $haystack, 0, abs($offset) ) );
}
else{
$temp_cut = strrev( substr( $haystack, $offset ) );
}
$pos = strlen($haystack) - (strpos($temp_cut, strrev($needle)) + $offset + strlen($needle));
if (
$pos == strlen($haystack)) { $pos = 0; }
return
$pos;
}
/* endfunction strripos*/
}/* endfunction exists strripos*/
?>
up
-1
Anonymous
13 years ago
Generally speaking, linear searches are from start to end, not end to start - which makes sense from a human perspective. If you need to find strings in a string backwards, reverse your haystack and needle rather than manually chopping it up.
up
-1
peev[dot]alexander at gmail dot com
15 years ago
OK, I guess this will be the final function implementation for PHP 4.x versions ( my previous posts are invalid )

<?php

if(!function_exists("stripos")){
function
stripos( $str, $needle, $offset = 0 ){
return
strpos( strtolower( $str ), strtolower( $needle ), $offset );
}
/* endfunction stripos */
}/* endfunction exists stripos */

if(!function_exists("strripos")){
function
strripos( $haystack, $needle, $offset = 0 ) {
if( !
is_string( $needle ) )$needle = chr( intval( $needle ) );
if(
$offset < 0 ){
$temp_cut = strrev( substr( $haystack, 0, abs($offset) ) );
}
else{
$temp_cut = strrev( substr( $haystack, 0, max( ( strlen($haystack) - $offset ), 0 ) ) );
}
if( (
$found = stripos( $temp_cut, strrev($needle) ) ) === FALSE )return FALSE;
$pos = ( strlen( $haystack ) - ( $found + $offset + strlen( $needle ) ) );
return
$pos;
}
/* endfunction strripos */
}/* endfunction exists strripos */
?>
up
-3
ElectroFox
16 years ago
Sorry, I made that last post a bit prematurely. One more thing wrong with the simple php4 version is that it breaks if the string is not found. It should actually look like this:

<?php
if (function_exists('strripos') == false) {
function
strripos($haystack, $needle) {
$pos = strlen($haystack) - strpos(strrev($haystack), strrev($needle));
if (
$pos == strlen($haystack)) { $pos = 0; }
return
$pos;
}
}
?>

Note, we now check to see if the $needle was found, and if it isn't, we return 0.
up
-7
admin at e-xxi dot net
13 years ago
strripos() has very strange behaviour when you provide search position. For some reason it searches forward from the given position, instead of searching backward, that is more logical.

For example if you want to find instanse of $what, previous to the last, strripos($where, $what, $last_what_pos-1) will not wark as expected. It will return $last_what_pos again and again. And that has no sence at all.

To prevent this, I just used $prev_last_what_pos = strripos(substr($where,0,$last_what_pos), $what);
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