REGEX - php preg_replace : - php

Question

How to replace part of a string based on a "." (period) character
only if it appears after/before/between a word(s),
not when it is before/between any number(s).
Example:
This is a text string.
-Should be able to replace the "string." with "string ."
(note the Space between the end of the word and the period)
Example 2
This is another text string. 2.0 times longer.
-Should be able to replace "string." with "string ."
(note the Space between the end of the word and the period)
-Should Not replace "2.0" with "2 . 0"
It should only do the replacement if the "." appears at the end/start of a word.
Yes - I've tried various bits of regex.
But everything I do results in either nothing happening,
or the numbers are fine, but I take the last letter from the word preceeding the "."
(thus instead of "string." I end up with "strin g.")
Yes - I've looked through numerous posts here - I have seen Nothing that deals with the desire, nor the "strange" problem of grabbing the char before the ".".

Answer 1

You can use a lookbehind (?<=REXP)
preg_replace("/(?<=[a-z])\./i", "XXX", "2.0 test. abc") // <- "2.0 testXXX abc"
which will only match if the text before matches the corresponding regex (in this case [a-z]). You may use a lookahead (?=REXP) in the same way to test text after the match.
Note: There is also a negative lookbehind (?<!REXP) and a negative lookahead (?!REXP) available which will reject matches if the REXP does not match before or after.

Answer 2

$input = "This is another text string. 2.0 times longer.";
echo preg_replace("/(^\.|\s\.|\.\s|\.$)/", " $1", $input);
http://ideone.com/xJQzQ

Answer 3

I'm not too good with the regex, but this is what I would do to accomplish the task with basic PHP. Basically explode the entire string by it's . values, look at each variable to see if the last character is a letter or number and add a space if it's a number, then puts the variable back together.
<?
$string = "This is another text string. 2.0 times longer.";
print("$string<br>");
$string = explode(".",$string);
$stringcount = count($string);
for($i=0;$i<$stringcount;$i++){
if(is_numeric(substr($string[$i],-1))){
$string[$i] = $string[$i] . " ";
}
}
$newstring = implode('.',$string);
print("$newstring<br>");
?>

Answer 4

It should only do the replacement if the "." appears at the end/start of a word.
search: /([a-z](?=\.)|\.(?=[a-z]))/
replace: "$1 "
modifiers: ig (case insensitive, globally)
Test in Perl:
use strict;
use warnings;
my $samp = '
This is another text string. 2.0 times longer.
I may get a string of "this and that.that and this.this 2.34 then that 78.01."
';
$samp =~ s/([a-z](?=\.)|\.(?=[a-z]))/$1 /ig;
print "$samp";
Input:
This is another text string. 2.0 times longer.
I may get a string of "this and that.that and this.this 2.34 then that 78.01."
Output:
This is another text string . 2.0 times longer .
I may get a string of "this and that . that and this . this 2.34 then that 78.01."