Replace all certain character except first occurrence in PHP - php

Question

Example
Input = 1.1.0.1
Expected output = 1.101

Answer 1

You could make use substr() and str_replace() fairly easily:
$str = '1.1.0.1';
$pos = strpos($str,'.');
if ($pos !== false) {
$str = substr($str,0,$pos+1) . str_replace('.','',substr($str,$pos+1));
}
echo $str;

Answer 2

$s = preg_replace('/((?<=\.)[^.]*)\./', '$1', $s);
Matches zero or more non-dot characters followed by a dot, but only if the match was preceded by a dot. This prevents a match on the initial digit(s). Replaces the match with only the non-dot characters (the digits), which were captured in group #1.

Answer 3

$input="1.1.1.1";
$s = explode(".",$input ) ;
$t=array_slice($s, 1);
print implode(".",array($s[0] , implode("",$t)) );
or
$input="1.1.1.1";
$s = explode(".",$input ,2) ;
$s[1]=str_replace(".","",$s[1]);
print implode(".",array($s[0] ,$s[1] ) );

Answer 4

Match&Release the first occurring literal dot
Replace all subsequent literal dots
Code: (Demo)
echo preg_replace('~^[^.]*\.(*SKIP)(*FAIL)|\.~', '', $string);
// 1.101
Or with the "continue" character (\G), consume and forget the first literal dot, then replace all subsequent literal dots.
Code: (Demo)
echo preg_replace('~(?:^[^.]*\.|\G(?!^))[^.]*\K\.~', '', $string);
// 1.101
Or simply check that a literal dot has a literal dot occurring earlier in the string.
Code: (Demo)
echo preg_replace('~(?<=\.)[^.]*\K\.~', '', $string);
// 1.101

Answer 5

I though substr_replace() would work here, but sadly no... Here is a regex approach:
$str = preg_replace('~(\d+\.)(\d+)\.(\d+)\.(\d+)~', '$1$2$3$4', $str);

Answer 6

You could also try the below regex with s switch,
<?php
$string = '1.1.0.1';
$pattern = "/(?s)((?<=\.).*?)(\.)/i";
$replacement = "$1";
echo preg_replace($pattern, $replacement, $string);
?>
Output:
1.101

Answer 7

Using regex matches can be clearer by depicting the desired result and avoids the error-prone approach of calls to substr and strpos. Here I assume that no text is required before the first dot, i.e., that an input may begin with a dot that must be preserved. The difference is whether a quantifier of * or + is appropriate in the patterns below.
If your inputs will always be short, a straightforward approach is to replace trailing dots until none remain:
$count = 0;
$output = $input;
do {
$output = preg_replace('/^(.*\.)(.+)\./', '$1$2', $output, -1, $count);
} while ($count != 0);
echo $output;
To do it with a single regex match, use preg_replace_callback to apply a function (str_replace in this case) to the backreference variable $2.
$output = preg_replace_callback(
'/^([^.]*\.)(.+)$/',
function ($m) { return $m[1] . str_replace('.', '', $m[2]); },
$input);
Sample results:
1.1.0.1 - 1.101
.1.0.1 - .101
111 - 111
1.1 - 1.1
1. - 1.
.1 - .1
.1. - .1
.... - .
You may want to experiment with the code and test cases at Try It Online!