Regex question

Opteron

I would like to delete every char and non-char right after and including " over".

My strings are like:
The quick brown fox jumps over a lazy dog
The quick red fox runs over a lazy dog
The quick brown fox jumps over the speedy dog
The quick fox runs over my speedy dog

What would be be correct preg_replace to do this?
Thanks

NogDog

Are you talking about right after and including "over" from the first occurrence until the end of the entire string, or on a line-by-line basis deleting from "over" until the end of the line?

Kudose

Sounds like homework.

The easiest method would be to match the space and over, using a back reference to get the data before the match.

#^{(.)\x20over.$#}

I'll leave the rest up to you.

Opteron

@
Second one. Every line will be processed individually.

@
Your regex isnt working on php 5.2.
$string = "The quick brown fox jumps over the speedy dog";
$regex = preg_replace("#^{(.)\x20over.$#",} "", $string);
print $regex;

Opteron

To clarify: Those lines aren't part of a bigger string. Each string/line is a row on mysql.
Thanks

NogDog

If we assume you want the full word "over" to match but not "flyover" or "overwhelm":

$str = preg_replace('/\bover\b.*$/', '', $str);

If you want it to be case-insensitive, add an "i" after the second "/" in the regexp.

Kudose

Opteron;10927430 wrote:
@
Second one. Every line will be processed individually.

@
Your regex isnt working on php 5.2.
$string = "The quick brown fox jumps over the speedy dog";
$regex = preg_replace("#^{(.)\x20over.$#",} "", $string);
print $regex;

It does not work without back references, which I left to you to figure out.

If you add the back reference into the replacement parameter, it works fine.

<?php

    $string = "The quick brown fox jumps over the speedy dog";
    $regex = preg_replace("#^(.*)\x20over.*$#", "$1", $string);
    print $regex;

NogDog

I think the back-reference is overkill, however, since the same result could be obtained with:

$regex = preg_replace("#\x20over.*$#", "", $string);

Kudose

I see what you mean now and you are correct. There is no point in the back reference if none of the string beyond the match is needed to be returned.

Weedpacket

If you don't care about the rest of the string, ending the pattern with ".*$" is itself overkill.

NogDog

Weedpacket;10927494 wrote:
If you don't care about the rest of the string, ending the pattern with ".*$" is itself overkill.

Well, you care about it in that you want it to be matched with the word being searched for ("over") so that it is replaced along with the search word. The "$" is probably redundant, as by default preg_() matches are "greedy", but you still need the "." part, unless I'm completely missing something?

nrg_alpha

NogDog;10927545 wrote:
Well, you care about it in that you want it to be matched with the word being searched for ("over") so that it is replaced along with the search word. The "$" is probably redundant, as by default preg_() matches are "greedy", but you still need the "." part, unless I'm completely missing something?

If by preg_() you mean statements like preg_replace or preg_match etc, it isn't the preg statement itself that is inherently greedy, but rather the quantifiers themselves (like and +).

In this case, I agree with your pattern, sans the $. As far as I can tell, . (which is in general frowned upon as it is usually used unwisely) would be the fastest, non-BS solution to match 'everything else' in cases like this (as . and the like do have their 'occasional' uses).

Opteron

Thank you very much NogDog.

Weedpacket

Ah, I was actually thinking of just searching for "over", getting its offset, and just retaining the substring up to that offset. Of course, there's then the question of what would happen if "over" did not appear on the line.