Replace colons in a string unless they are part of a disqualifying substrings

Question

I would like to replace : with \xEF\xBC\x9A however, I do not want to do this for : that follows http, https, and string with the format of @[{numbers}:{numbers}:{some text}]. What I have doesn't really work as (?<!@\[\d) does not check what's after it. My current implementation only work for something like @[1:2:text].

$string = preg_replace(
    '#(?<!https)(?<!http)(?<!@\[\d)(?<!@\[\d:\d):#',
    "\xEF\xBC\x9A",
    $string
);

Qtax · Accepted Answer · 2011-08-05 20:48:19Z

1

Try this:

preg_replace('/(@\[\d+:\d+:[^]]*]|https?:)|:/e', '"$1"?"$1":"\xEF\xBC\x9A"', $string);

edited Aug 5, 2011 at 20:48

answered Aug 5, 2011 at 20:36

Qtax

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6 Comments

hao Over a year ago

Umm I am getting Unknown modifier 'g': [/(@[\d+:\d+:[^]]*]|https?:)|:/ge]

Zoey Over a year ago

This appears to select instances of 'http:', 'https:', and '@[/d:/d:/t*]', which is opposing the point made in the question itself, I suggest rewording the question if this is the correct answer.

hao Over a year ago

@Qtax, is there a way I can do this without /e?

Alan Moore Over a year ago

@Dessix: The idea is to capture the "ineligible" colon along with the text preceding it, and plug it back in. That "gets it out of the way", so when the other alternative matches a colon, you know it's one of the ones you want to replace.

Alan Moore Over a year ago

@hao: You can do this without /e, but I don't recommend it: demo.

|

Zoey · Accepted Answer · 2011-08-05 20:53:05Z

0

Try this regex:

(?<!@\[(?::?[\d]{0,5}){1,2})(?<!https?):

It should match the first and second instances of ':' here.

: test: http: @[1:2:text]

Usage Sample:

$string = preg_replace('/(?<!@\[(?::?[\d]{0,5}){1,2})(?<!https?):/', '\xEF\xBC\x9A', $string);

edited Aug 5, 2011 at 20:53

answered Aug 5, 2011 at 20:47

Zoey

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1 Comment

Alan Moore Over a year ago

A lookbehind has to match a fixed number of characters (in PHP, at least). That's why the OP used (?<!https)(?<!http) instead of (?<!https?). But it's impossible to "fake" the whole problem that way, as the OP has discovered.

mickmackusa · Accepted Answer · 2024-06-19 22:10:12Z

0

Match the expressions that should not be mutated then disqualify them with (*SKIP)(*FAIL) -- this is far simpler than trying to avoid matching them. Then match any non-disqualified colon and replace it with the desired string.

Code: (Demo)

$string = <<<TEXT
foo : test: http: @[1:2:text] http://example.com 3:4
TEXT;

echo preg_replace(
    '/(?:https?:|@\[\d+:\d+:[^\]]+)(*SKIP)(*FAIL)|:/',
    '\xEF\xBC\x9A',
    $string
);

Output:

foo \xEF\xBC\x9A test\xEF\xBC\x9A http: @[1:2:text] http://example.com 3\xEF\xBC\x9A4

answered Jun 19, 2024 at 22:10

mickmackusa♦

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