Use String for Pattern but Exclude it from Being Removed - php

Question

i'm pretty new on regex, i have learned something by the way, but is still pour knowledge!
so i want ask you for clarification on how it work!
assuming i have the following strings, as you can see they can be formatted little different way one from another but they are very similar!
DTSTART;TZID="America/Chicago":20030819T000000
DTEND;TZID="America/Chicago":20030819T010000
DTSTART;TZID=US/Pacific
DTSTART;VALUE=DATE
now i want replace everything between the first A-Z block and the colon so for example i would keep
DTSTART:20030819T000000
DTEND:20030819T010000
DTSTART
DTSTART
so on my very noobs knowledge i have worked out this shitty regex! :-(
preg_replace( '/^[A-Z](?!;[A-Z]=[\w\W]+):$/m' , '' , $data );
but why i'm sure this regex will not work!? :-)
Pls help me!
PS: the title of question is pretty explaned, i want also know how for example use a well know string block for match another...
preg_replace( '/^[DTSTART](?!;[A-Z]=[\w\W]+):$/m' , '' , $data );
..without delete DTSTART
Thanks for the time!
Regards
Luca Filosofi

Answer 1

You could use a relatively simple regex like the following.
$subject = 'DTSTART;TZID="America/Chicago":20030819T000000
DTEND;TZID="America/Chicago":20030819T010000
DTSTART;TZID=US/Pacific
DTSTART;VALUE=DATE';
echo preg_replace('/^[A-Z]+\K[^:\n]*/m', '', $subject) . PHP_EOL;
It looks for a series of capital letters at the start of a line, resets the match starting point (that's what \K does) to the end of those and matches anything not a colon or newline (i.e. the parts you want to remove). Those matched parts are then replaced with an empty string.
The output from the above would be
DTSTART:20030819T000000
DTEND:20030819T010000
DTSTART
DTSTART
If the lines that you are interested in will only ever start with DTSTART or DTEND then we could be more precise about what to match (e.g. ^DT(?:START|END)) but [A-Z] obviously covers both of those.

Answer 2

If you want to retain part of the matched pattern in a substitution, you put parentheses around it and then refer to it by $1 (or whichever grouping it is).
For example:
s/^(this is a sentence) to edit/$1/
gives "this is a sentence"

Answer 3

You can check out this example work similarly as your problem
\w+): (?P\d+)/', $str, $matches);
/* This also works in PHP 5.2.2 (PCRE 7.0) and later, however
* the above form is recommended for backwards compatibility */
// preg_match('/(?\w+): (?\d+)/', $str, $matches);
print_r($matches);
?>
The above example will output:
Array
(
[0] => foobar: 2008
[name] => foobar
[1] => foobar
[digit] => 2008
[2] => 2008
)
so if u need only digit u need to print $matches[digit]

Answer 4

You want to remove everything between a semicolon and either a colon or the end of the line, right? So use that as your expression. You're overcomplicating things.
preg_replace('/(?:;.+?:)|(?:;.+?$)/m','',$data);
It's a pretty simple expression. Either match (?:;.+?:) or (?:;.+?$), which differ only by their terminator (the first one matches up to a colon, the second one matches up to the end of the line).
Each is a non-capturing group that starts with a semicolon, reluctantly reads in all characters, then stops at the terminator. Everything matched by this is removable according to your description.