I'm getting insane over this, it's so simple, yet I can't figure out the right regex. I need a regex that will match blacklisted words, ie "ass".
For example, in this string:
<span class="bob">Blacklisted word was here</span>bass
I tried that regex:
((?!class)ass)
That matches the "ass" in the word "bass" bot NOT "class".
This regex flags "ass" in both occurences. I checked multiple negative lookaheads on google and none works.
NOTE: This is for a CMS, for moderators to easily find potentially bad words, I know you cannot rely on a computer to do the filtering.
If you have lookbehind available (which, IIRC, JavaScript does not and that seems likely what you're using this for) (just noticed the PHP tag; you probably have lookbehind available), this is very trivial:
(?<!cl)(ass)
Without lookbehind, you probably need to do something like this:
(?:(?!cl)..|^.?)(ass)
That's ass, with any two characters before as long as they are not cl, or ass that's zero or one characters after the beginning of the line.
Note that this is probably not the best way to implement a blacklist, though. You probably want this:
\bass\b
Which will match the word ass but not any word that includes ass in it (like association or bass or whatever else).
It seems to me that you're actually trying to use two lists here: one for words that should be excluded (even if one is a part of some other word), and another for words that should not be changed at all - even though they have the words from the first list as substrings.
The trick here is to know where to use the lookbehind:
/ass(?<!class)/
In other words, the good word negative lookbehind should follow the bad word pattern, not precede it. Then it would work correctly.
You can even get some of them in a row:
/ass(?<!class)(?<!pass)(?<!bass)/
This, though, will match both passhole and pass. ) To make it even more bullet-proof, we can add checking the word boundaries:
/ass(?<!\bclass\b)(?<!\bpass\b)(?<!\bbass\b)/
UPDATE: of course, it's more efficient to check for parts of the string, with (?<!cl)(?<!b) etc. But my point was that you can still use the whole words from whitelist in the regex.
Then again, perhaps it'd be wise to prepare the whitelists accordingly (so shorter patterns will have to be checked).
Is this one is what you want ? (?<!class)(\w+ass)
Related
I must detect the presence of some words (even polyrematic, like in "bag of words") in a user-submitted string.
I need to find the exact word, not part of it, so the strstr/strpos/stripos family is not an option for me.
My current approach (PHP/PCRE regex) is the following:
\b(first word|second word|many other words)\b
Is there any other better approach? Am I missing something important?
Words are about 1500.
Any help is appreciated
A regular expression the way you're demonstrating will work. It may be challenging to maintain if the list of words grows long or changes.
The method you're using will work in the event that you need to look for phrases with spaces and the list doesn't grow much.
If there are no spaces in the words you're looking for, you could split the input string on space characters (\s+, see https://www.php.net/manual/en/function.preg-split.php ), then check to see if any of those words are in a Set (https://www.php.net/manual/en/class.ds-set.php) made up of the words you're looking for. This will be a bit more code, but less regex maintenance, so ymmv based on your application.
If the set has spaces, consider instead using Trie. Wiktor Stribiżew suggests: https://github.com/sters/php-regexp-trie
I have a regex that will be used to match #users tags.
I use lokarround assertions, letting punctuation and white space characters surround the tags.
There is an added complication, there are a type of bbcodes that represent html.
I have two types of bbcodes, inline (^B bold ^b) and blocks (^C center ^c).
The inline ones have to be passed thru to reach for the previous or next character.
And the blocks are allowed to surround a tag, just like punctuation.
I made a regex that does work. What I want to do now is to lower the number of steps that it does in every character that’s not going to be a match.
At first I thought I could do a regex that would just look for #, and when found, it would start looking at the lookarrounds, that worked without the inline bbcodes, but since lookbehind cannot be quantifiable, it’s more difficult since I cannot add ((\^[BIUbiu])++)* inside, producing much more steps.
How could I do my regex more efficient with fewer steps?
Here is a simplified version of it, in the Regex101 link there is the full regex.
(?<=[,\.:=\^ ]|\^[CJLcjl])((\^[BIUbiu])++)*#([A-Za-z0-9\-_]{2,25})((\^[BIUbiu])++)*(?=[,\.:=\^ ]|\^[CJLcjl])
https://regex101.com/r/lTPUOf/4/
A rule of thumb:
Do not let engine make an attempt on matching each single one character if
there are some boundaries.
The quote originally comes from this answer. Following regular expression reduces steps in a significant manner because of the left side of the outermost alternation, from ~20000 to ~900:
(?:[^#^]++|[#^]{2,}+)(*SKIP)(*F)
|
(?<=([HUGE-CHARACTER-CLASS])|\^[cjleqrd])
(\^[34biu78])*+#([a-z\d][\w-.]{0,25}[a-z\d])(\^[34biu78])*+(?=(?1))
Actually I don't care much about the number of steps being reported by regex101 because that wouldn't be true within your own environment and it is not obvious if some steps are real or not or what steps are missed. But in this case since the logic of regex is clear and the difference is a lot it makes sense.
What is the logic?
We first try to match what probably is not desired at all, throw it away and look for parts that may match our pattern. [^#^]++ matches up to a # or ^ symbols (desired characters) and [#^]{2,}+ prevents engine to take extra steps before finding out it's going nowhere. So we make it to fail as soon as possible.
You can use i flag instead of defining uppercase forms of letters (this may have a little impact however).
See live demo here
I needed a regex to validate wether first and last name were provided corectly or not. Well This is what i came up with:
preg_match('/^[\p{L}]{4,25}[\s][\p{L}]{4,25}$/u', Form::post('name'))
This one works if string contains:
word (4-25 chars long and utf8 chars allowed)
space
word (4-25 chars long and utf8 chars allowed)
which rather is fine, but it seems too much complex for my script
is there a way to convert that regex so it will meet same conditions but has kind of "global" characters range instead, something like this:
(word space word){8,50}
also optionaly it could have second space and third word in case that some foreign person would want to use my site
any help will be appriciated:)
Aside from the fact that name validation is a bad idea in and of itself (see Falsehoods programmers believe about names), and that your regex can be simplified syntactically to
/^\pL{4,25}\s\pL{4,25}$/u
yes, it is possible, but ugly. You would need to use a positive lookahead assertion to make sure that there is only one space, and that it's neither at the end nor at the start of the string:
/^(?=\S+\s\S+$)[\pL\s]{8,50}$/u
If you want to allow more spaces/words, you can use
/^(?=\S+(?:\s\S+)+$)[\pL\s]{8,50}$/u
I know that I'd likely hear "Don't parse HTML with regex", so let me say that this question is just academic at this point because I actually solved my problem using the DOM, but on my road to a solution, I ran across this pattern that works on the gskinner website, but I can't figure out how to make it work in PHP preg_match().
(?<=href\=")[^]+?(?=")
I think that the [^] is causing the problem, but I'm not certain what to do about it.
What it is intended to do is pull the substring from between the quotes of an href. (One would expect it to be a web-address or at least part of one.)
[^] is a difficult construct. Basically it is an empty negated character class. But what should it match? That depends on the implementation. Some languages are interpreting it as negation of nothing, so it will match every character, that is what gskinner (means ActionScript 3) seems to be doing.
I would never use this, because it is ambiguous.
The most readable way is to use ., the meta character that matches every character (without newlines), if newlines are also wanted, just add the modifier s that enables the dotall mode, this would be exactly what you wanted to achieve with [^].
A workaround that is sometimes used is to use a character class something like this [\s\S] or [\w\W]. Those will also match every character (including newlines), because they are matching some predefined character class and their negation.
I have a couple of matching groups one after another in a long Regex pattern. Around the middle I have
...(?<number>(?:/(?:digit|num))?\d+|)...
which should match something like /num9, /digit9 or 9 or blank (because I need the named group to appear in the resulting associative array even if it's empty).
The pattern works, but is it possible to discard the / character if the one of first two cases is matched? I tried a positive lookahead, but it seems that you can't use those if you have expressions before the lookahead.
Is what I'm trying to accomplish possible using Regex?
Based on your input, I think that you need to capture / anyway at some point, otherwise your whole regex fails. At the same time you want to ignore it, so it cannot be a part of you named group. Therefore by putting it outside it and making it optional, while ensuring that a digit is not preceded directly by a / you come up with the desired results :
^/?(?<number>(?:(?:digit|num))?(?<!/)\d+|)$
However given your lack of a more complete input and regex, I am not 100% sure this will work for all your cases.