best method to stop users posting urls - php

I am looking to implement a system to strip out url's from text posted by a user.
I know there is no perfect solution and users will still attempt things like:
www dot google dot com
so I know that ultimately any solution will be flawed in some way... all I am looking to do really is reduce the number of people doing it.
Any suggestions, source or approaches appriciated,
Thanks

There are number of regular expression pattern matchers here. Some of them are quite complex.
I would suggest that running multiple ones may be a good idea.

You need to define exactly what you want to strip out. The stricter the definition, the more false positives you will get. The following example will remove any string with 3 characters, followed by a period, more letters, another period and 2-4 more letters:
$text = preg_replace('/[a-z]{3}\.[a-z]+\.[a-z]{2,4}/i', '', $text);
The other end of strictness might be anything that ends on a period and 2-4 letters (like .com):
$text = preg_replace('/[a-z]+\.[a-z]{2,4}/i', '', $text);
Note that the latter will strip out the last word of a sentence, the full stop and the first word of the next sentence if someone forgets to add a space inbetween the sentences.

Related

What's the best approach to find words from a set of words in a string?

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

Regex PHP. Reduce steps: limited by fixed width Lookbehind

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

No periods at beginning or end of expression?

I want to allow alphanumeric characters and periods; however, the phrase cannot contain more two or more periods in a row, it cannot start or end with a period, and spaces are not allowed.
I am using both PHP and Javascript.
So far, I have /^(?!.*\.{2})[a-zA-Z0-9.]+$/
This works for allowing alphanumeric characters and periods, while denying spaces and consecutive periods, but I still am not sure how to check for starting and/or ending periods. How might I do this? and, is there an even better way to do what I already have?
It nearly always helps to draw a finite state machine to conceptualize what your regular expression should look like.
^(?:\w\.?)*\w$
here's a possible way
/^(?!\.)((?:[a-z\d]|(?<!\.)\.)+)(?<!\.)$/i
for more explanations and tests see here: http://www.regex101.com/r/rZ6yH4
edit: according to tyler's solution, here's him way, shortened and reduced to letters and digits
/^(?:[a-z\d]+(?:\.(?!$))?)+$/i
( http://www.regex101.com/r/dL5aG0 )
A start would be:
/^[^. ](?!.*\.{2})[a-zA-Z0-9.]+[^. ]$/
but it should be tested carefully.

Regex with negative lookahead to ignore the word "class"

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)

how to detect telephone numbers in a text (and replace them)?

I know it can be done for bad words (checking an array of preset words) but how to detect telephone numbers in a long text?
I'm building a website in PHP for a client who needs to avoid people using the description field to put their mobile phone numbers..(see craigslist etc..)
beside he's going to need some moderation but i was wondering if there is a way to block at least the obvious like nnn-nnn-nnnn, not asking to block other weird way of writing like HeiGHT*/four*/nine etc...
Welcome to the world of regular expressions. You're basically going to want to use preg_replace to look for (some pattern) and replace with a string.
Here's something to start you off:
$text = preg_replace('/\+?[0-9][0-9()\-\s+]{4,20}[0-9]/', '[blocked]', $text);
this looks for:
a plus symbol (optional), followed by a number, followed by between 4-20 numbers, brackets, dashes or spaces, followed by a number
and replaces with the string [blocked].
This catches all the obvious combinations I can think of:
012345 123123
+44 1234 123123
+44(0)123 123123
0123456789
Placename 123456 (although this one will leave 'Placename')
however it will also strip out any succession of 6+ numbers, which might not be desirable!
To do so you must use regular expressions as you may know.
I found this pattern that could be useful for your project:
<?php
preg_match("/(^(([\+]\d{1,3})?[ \.-]?[\(]?\d{3}[\)]?)?[ \.-]?\d{3}[ \.-]?\d{4}$)/", $yourText, $matches);
//matches variable will contain the array of matched strings
?>
More information about this pattern can be found here http://gskinner.com/RegExr/?2rirv where you can even test it online. It's a great tool to test regular expressions.
preg_match($pattern, $subject) will return 1 (true) if pattern is found in subject, and 0 (false) otherwise.
A pattern to match the example you give might be '/\d{3}-\d{3}\d{4}/'
However whatever you choose for your pattern will suffer from both false positives and false negatives.
You might also consider looking for words like mob, cell or tel next to the number.
The fill details of the php pattern matching can be found at http://www.php.net/manual/en/reference.pcre.pattern.syntax.php
Ian
p.s. It can't be done for bad words, as the people in Scunthorpe will tell you.
I think that use a too tight regular espression would lead to loose a great number of detections.
You should check for portions of 10 consecutive chatacters containing more than 5 digits.
So it is similar you will have an analisys routine queued to be called after any message insertion due to the computational weight.
After the 6 or more digits have been isolated replace them as you prefer, including other syblings digits.
Better in any case to preserve original data, so you can try and train your detection algorithm until it works the best way.
Then you can also study your user data to create more complex euristics, such like case insensitive numbers written as letters, mixed, dot separated, etc...
It's not about write the most perfect regex, is about approaching the problem statistically and dinamically.
And remember, after you take action, user will change their insertion habits as consequence, so stats will change and you will need to learn and update your euristics.

Categories