Matching all of a certain character after a Positive Lookbehind

Question

I have been trying to get the regex right for this all morning long and I have hit the wall. In the following string I wan't to match every forward slash which follows .com/<first_word> with the exception of any / after the URL.

$string = "http://example.com/foo/12/jacket Input/Output";
    match------------------------^--^

The length of the words between slashes should not matter.

Regex: (?<=.com\/\w)(\/) results:

$string = "http://example.com/foo/12/jacket Input/Output"; // no match
$string = "http://example.com/f/12/jacket Input/Output";   
    matches--------------------^

Regex: (?<=\/\w)(\/) results:

$string = "http://example.com/foo/20/jacket Input/O/utput"; // misses the /'s in the URL
    matches----------------------------------------^
$string = "http://example.com/f/2/jacket Input/O/utput"; // don't want the match between Input/Output
    matches--------------------^-^--------------^

Because the lookbehind can have no modifiers and needs to be a zero length assertion I am wondering if I have just tripped down the wrong path and should seek another regex combination.

Is the positive lookbehind the right way to do this? Or am I missing something other than copious amounts of coffee?

NOTE: tagged with PHP because the regex should work in any of the preg_* functions.

The`preg_match` function returns one match. You say you need to match all characters there are after some pattern. You should use `preg_match_all`. — Wiktor Stribiżew, Feb 11 '16 at 17:58
I still have an impression it is an XY problem. What are you trying to achieve? Why match those slashes? You could url_parse the URL, and then do whatever you please. Explode, e.g. — Wiktor Stribiżew, Feb 11 '16 at 18:29
No, it isn't an XY problem @WiktorStribiżew as the regex should work in *any* of the `preg_*` functions. — Jay Blanchard, Feb 11 '16 at 18:42

vks · Answer 1 · 2016-02-11T18:04:26.133

3

Use \K here along with \G.grab the groups.

^.*?\.com\/\w+\K|\G(\/)\w+\K

See demo.

https://regex101.com/r/aT3kG2/6

$re = "/^.*?\\.com\\/\\w+\\K|\\G(\\/)\\w+\\K/m"; 
$str = "http://example.com/foo/12/jacket Input/Output"; 

preg_match_all($re, $str, $matches);

Replace

$re = "/^.*?\\.com\\/\\w+\\K|\\G(\\/)\\w+\\K/m"; 
$str = "http://example.com/foo/12/jacket Input/Output"; 
$subst = "|"; 

$result = preg_replace($re, $subst, $str);

edited Feb 11 '16 at 18:04

answered Feb 11 '16 at 17:44

vks

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For whatever reason this is not working in the context of `preg_match()` – Jay Blanchard Feb 11 '16 at 17:52
OK - `preg_match_all()` works but other `preg....` functions fail - like `preg_replace()`. – Jay Blanchard Feb 11 '16 at 18:03
@JayBlanchard the only problem with replace is one extra replacement will happend at the end....guess that will have to be dealt separately – vks Feb 11 '16 at 18:06
I got this as the return `http://example.com/foo|/12/jacket Input/Output`using your code. Note that none of the slashes have been removed, but a pipe has been added. Even in your substitution example the slashes are left in place. I *really* appreciate your help with this, it seems that we're both beating our heads against the wall. One of the +1's is mine. – Jay Blanchard Feb 11 '16 at 18:08
@JayBlanchard we can do it like this ...u just need to remove an extra `|` later. https://regex101.com/r/aT3kG2/8 – vks Feb 11 '16 at 18:16
Can you also copy those regex's into your answer. They may not stay on regex101 for ever! Which would make your answers useless to others – RiggsFolly Feb 11 '16 at 18:37
That last example you provded added spaces and an extra pipe @vks `http://example.com/foo||12|jacket|wow` – Jay Blanchard Feb 11 '16 at 18:46

anubhava · Accepted Answer · 2016-02-11T18:49:54.353

3

If you want to use preg_replace then this regex should work:

$re = '~(?:^.*?\.com/|(?<!^)\G)[^/\h]*\K/~';
$str = "http://example.com/foo/12/jacket Input/Output";
echo preg_replace($re, '|', $str);
//=> http://example.com/foo|12|jacket Input/Output

Thus replacing each / by a | after first / that appears after starting .com.

Negative Lookbehind (?<!^) is needed to avoid replacing a string without starting .com like /foo/bar/baz/abcd.

RegEx Demo

edited Feb 11 '16 at 18:49

answered Feb 11 '16 at 18:15

anubhava

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The problem with this is that you're replacing the the `/` in the portion of the string with `Input/Output` – Jay Blanchard Feb 11 '16 at 18:40
It should only be the slashes *after* foo, but not in anything not in the URL – Jay Blanchard Feb 11 '16 at 18:44
There is a URL in the string, as well as some other data. The slashes in the other data should not be affected. – Jay Blanchard Feb 11 '16 at 18:48

score 1 · Answer 3 · edited May 23 '17 at 10:28

1

Another \G and \K based idea.

$re = '~(?:^\S+\.com/\w|\G(?!^))\w*+\K/~';

The (: non capture group to set entry point ^\S+\.com/\w or glue matches \G(?!^) to it.
\w*+\K/ possessively matches any amount of word characters until a slash. \K resets match.

See demo at regex101

edited May 23 '17 at 10:28

Community

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answered Feb 11 '16 at 21:33

bobble bubble

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Matching all of a certain character after a Positive Lookbehind

3 Answers3