First of all I created my own regex to find all URLs in a text, because:
When I searched SO and google only found regex for specific URL constructions, like images, etc.
I found a pretty complete regex from the PHP's manual itself (see "splattermania at freenet dot de 01-Oct-2009 12:01" post at http://php.net/manual/en/function.preg-match.php) that can find almost anything that resembles a URL, as little as "bit.ly".
This pattern has a few errors and constraints, so I'm fixing and enhancing it.
Now the pattern structure seems right, but I'm not sure all valid characters are present. Please post samples of URLs to test my pattern. Might be laziness, but I don't want to read pages and pages of references to find all of them, need to focus on the development. If you have a summary of valid chars for username, password, path, query and anchor that you can share, that would be very very helpful.
Best Regards!
The pattern you linked to does indeed match a lot of URLs, both valid and invalid. It's not really a surprise since nearly everything in that regex is optional; as you wrote yourself, it even matches bit.ly, so it's easy to see how it would match lots of non-URL stuff.
It doesn't take new Unicode domain names into account, for one (e.g., http://www.müller.de).
It doesn't match valid URLs like
http://msdn.microsoft.com/en-us/library/aa752574(VS.85).aspx
It doesn't match relative paths (might not be necessary, though) like /cgi-bin/version.pl.
It doesn't match mailto: links.
It doesn't match URLs like http://1.2.3.4. Don't even ask about IPv6 :)
All in all, regular expressions are NOT the right tool to reliably match or validate URLs. This is a job for a parser. If you can live with many false positive and false negative matches, then regexes are fine.
Please read Jan Goyvaerts' excellent essay on this subject: Detecting URLs in a block of text.
Related
I have this the following expression which is working well for my needs:
(www\.)?[-a-zA-Z0-9#:%._\+~#=]{2,256}\.(com|net|info)\b([-a-zA-Z0-9#:%_\+.~#?&//=]*)
It does a pretty good job of finding URLs which are hardcoded in a large codebase. Obviously, it has some false positives for things like Using System.net, but some of the valid URLs don't include http(s):// unfortunately.
So now I want to be able to exclude certain things, including system.net (false positives) and google.com (common urls in comments).
How can I do this?
Ideally the list would just include google and system. Something like (!google|!system).
Received this answer on Reddit less than an hour after posting.
you can use a negative lookahead for this.
\b(?!google\.com|system.net)
\b is there to make sure it starts at a word boundary (like after a
space or a dot there)
https://regex101.com/r/pwj3Xn/1
Credit to u/Kompaan86/
I run a Django-based forum (the framework is probably not important to the question, but still) and it has been increasingly getting spammed with posts that link to a specific website constantly (www.solidwoodkitchen.co.uk - these people are apparently the worst).
I've implemented a string blocking system that stops them posting to the forum if the URL of the website is included in the post, but as spam bots usually do, it has figured out a way around that by breaking up the URL with other characters (eg. w_w_w.s*olid_wood*kit_ch*en._*co.*uk .). So a couple of questions:
Is it even possible to build a regex capable of finding the specific URL within a block of text even when it has been modified like that?
If it is, would this cause a performance hit?
Description
You could break the url into a string of characters, then join them together with [^a-z0-9]*?. So in this case with www.solidwoodkitchen.co.uk the resulting regex would look like:
w[^a-z0-9]*?w[^a-z0-9]*?w[^a-z0-9]*?[.][^a-z0-9]*?s[^a-z0-9]*?o[^a-z0-9]*?l[^a-z0-9]*?i[^a-z0-9]*?d[^a-z0-9]*?w[^a-z0-9]*?o[^a-z0-9]*?o[^a-z0-9]*?d[^a-z0-9]*?k[^a-z0-9]*?i[^a-z0-9]*?t[^a-z0-9]*?c[^a-z0-9]*?h[^a-z0-9]*?e[^a-z0-9]*?n[^a-z0-9]*?[.][^a-z0-9]*?c[^a-z0-9]*?o[^a-z0-9]*?[.][^a-z0-9]*?u[^a-z0-9]*?k
Edit live on Debuggex
This could would basically search for the entire string of characters seperated by zero or more non alphanumeric characters.
Or you could take the input text and strip out all punctuation then simply search for wwwsolidwoodkitchencouk.
I'm using CodeIgniter to write an app where a user will be allowed to register an account and is assigned a URL (URL slug) of their choosing (ex. domain.com/user-name). CodeIgniter has a URL routing feature that allows the utilization of regular expressions (link).
User's are only allowed to register URL's that contain alphanumeric characters, dashes (-), and under scores (_). This is the regex I'm using to verify the validity of the URL slug: ^[A-Za-z0-9][A-Za-z0-9_-]{2,254}$
I am using the url routing feature to route a few url's to features on my site (ex. /home -> /pages/index, /activity -> /user/activity) so those particular URL's obviously cannot be registered by a user.
I'm largely inexperienced with regular expressions but have attempted to write an expression that would match any URL slugs with alphanumerics/dash/underscore except if they are any of the following:
default_controller
404_override
home
activity
Here is the code I'm using to try to match the words with that specific criteria:
$route['(?!default_controller|404_override|home|activity)[A-Za-z0-9][A-Za-z0-9_-]{2,254}'] = 'view/slug/$1';
but it isn't routing properly. Can someone help? (side question: is it necessary to have ^ or $ in the regex when trying to match with URL's?)
Alright, let's pick this apart.
Ignore CodeIgniter's reserved routes.
The default_controller and 404_override portions of your route are unnecessary. Routes are compared to the requested URI to see if there's a match. It is highly unlikely that those two items will ever be in your URI, since they are special reserved routes for CodeIgniter. So let's forget about them.
$route['(?!home|activity)[A-Za-z0-9][A-Za-z0-9_-]{2,254}'] = 'view/slug/$1';
Capture everything!
With regular expressions, a group is created using parentheses (). This group can then be retrieved with a back reference - in our case, the $1, $2, etc. located in the second part of the route. You only had a group around the first set of items you were trying to exclude, so it would not properly capture the entire wild card. You found this out yourself already, and added a group around the entire item (good!).
$route['((?!home|activity)[A-Za-z0-9][A-Za-z0-9_-]{2,254})'] = 'view/slug/$1';
Look-ahead?!
On that subject, the first group around home|activity is not actually a traditional group, due to the use of ?! at the beginning. This is called a negative look-ahead, and it's a complicated regular expression feature. And it's being used incorrectly:
Negative lookahead is indispensable if you want to match something not followed by something else.
There's a LOT more I could go into with this, but basically we don't really want or need it in the first place, so I'll let you explore if you'd like.
In order to make your life easier, I'd suggest separating the home, activity, and other existing controllers in the routes. CodeIgniter will look through the list of routes from top to bottom, and once something matches, it stops checking. So if you specify your existing controllers before the wild card, they will match, and your wild card regular expression can be greatly simplified.
$route['home'] = 'pages';
$route['activity'] = 'user/activity';
$route['([A-Za-z0-9][A-Za-z0-9_-]{2,254})'] = 'view/slug/$1';
Remember to list your routes in order from most specific to least. Wild card matches are less specific than exact matches (like home and activity), so they should come after (below).
Now, that's all the complicated stuff. A little more FYI.
Remember that dashes - have a special meaning when in between [] brackets. You should escape them if you want to match a literal dash.
$route['([A-Za-z0-9][A-Za-z0-9_\-]{2,254})'] = 'view/slug/$1';
Note that your character repetition min/max {2,254} only applies to the second set of characters, so your user names must be 3 characters at minimum, and 255 at maximum. Just an FYI if you didn't realize that already.
I saw your own answer to this problem, and it's just ugly. Sorry. The ^ and $ symbols are used improperly throughout the lookahead (which still shouldn't be there in the first place). It may "work" for a few use cases that you're testing it with, but it will just give you problems and headaches in the future.
Hopefully now you know more about regular expressions and how they're matched in the routing process.
And to answer your question, no, you should not use ^ and $ at the beginning and end of your regex -- CodeIgniter will add that for you.
Use the 404, Luke...
At this point your routes are improved and should be functional. I will throw it out there, though, that you might want to consider using the controller/method defined as the 404_override to handle your wild cards. The main benefit of this is that you don't need ANY routes to direct a wild card, or to prevent your wild card from goofing up existing controllers. You only need:
$route['404_override'] = 'view/slug';
Then, your View::slug() method would check the URI, and see if it's a valid pattern, then check if it exists as a user (same as your slug method does now, no doubt). If it does, then you're good to go. If it doesn't, then you throw a 404 error.
It may not seem that graceful, but it works great. Give it a shot if it sounds better for you.
I'm not familiar with codeIgniter specifically, but most frameworks routing operate based on precedence. In other words, the default controller, 404, etc routes should be defined first. Then you can simplify your regex to only match the slugs.
Ok answering my own question
I've seem to come up with a different expression that works:
$route['(^(?!default_controller$|404_override$|home$|activity$)[A-Za-z0-9][A-Za-z0-9_-]{2,254}$)'] = 'view/slug/$1';
I added parenthesis around the whole expression (I think that's what CodeIgniter matches with $1 on the right) and added a start of line identifier: ^ and a bunch of end of line identifiers: $
Hope this helps someone who may run into this problem later.
I have a regex pattern for URL's that I use to check for links in a body of text. The only problem is that the pattern will match this link
stackoverflow.com
And this sentence
I'm a sentence.Next Sentence.
Obviously this would make sense because my pattern doesn't strong check .com, .co.uk, .com.au etc
I want it to match stackoverflow.com and not the latter.
As I'm no Regex expert, does anyone know of any good Regex patterns for checking for all types of URL's in a body text, while not matching the sentences like above?
If I have to strong check the domain extension, I suppose I'll have to settle.
Here's my pattern, but i don't think it help.
(([\w]+:)?\/\/)?(([\d\w]|%[a-fA-f\d]{2,2})+(:([\d\w]|%[a-fA-f\d]{2,2})+)?#)?([\d\w][-\d\w]{0,253}[\d\w]\.)+[\w]{2,4}(:[\d]+)?(\/([-+_~.\d\w]|%[a-fA-f\d]{2,2})*)*(\?(&?([-+_~.\d\w]|%[a-fA-f\d]{2,2})=?)*)?(#([-+_~.\d\w]|%[a-fA-f\d]{2,2})*)?
I would definitely suggest finding a working regex that someone else has made (which would probably include a strong check on the domain extension), but here is one possible way to just modify your existing regex.
It requires that you make the assumption that usually links will not mix case in the domain extension, for example you might see .COM or .com but probably not .Com, if you only match domain extensions that don't mix case then you would avoid matching most sentences.
In the middle of your regex you have [\w]{2,4}, try changing this to ([A-Z]{2,4}|[a-z]{2,4}) (or (?:[A-Z]{2,4}|[a-z]{2,4}) if you don't want a new captured group).
I have a huge list of URL's, in the format:
http://www.example.com/dest/uk/bath/
http://www.example.com/dest/aus/sydney/
http://www.example.com/dest/aus/
http://www.example.com/dest/uk/
http://www.example.com/dest/nor/
What RegEx could I use to get the last three URL's, but miss the first two, so that every URL without a city attached is given, but the ones with cities are denied?
Note: I am using Google Analytics, so I need to use RegEx's to monitor my URL's with their advanced feature. As of right now Google is rejecting each regular expression.
Generally, the best suggestion I can make for parsing URL's with a Regex is don't.
Your time is much much better spent finding a libary that exists for your language dedicated to the task of processing URLs.
It will have worked out all the edge cases, be fully RFC compliant, be bug free, secure, and have a great user interface so you can just suck out the bits you really want.
In your case, the suggested way to process it would be, using your URL library, extract the element s and then work explicitly on them.
That way, at most you'll have to deal with the path on its own, and not have to worry so much wether its
http://site.com/
https://site.com/
http://site.com:80/
http://www.site.com/
Unless you really want to.
For the "Path" you might even wish to use a splitter ( or a dedicated path parser ) to tokenise the path into elements first just to be sure.
tj111's current solution doesn't work - it matches all your urls.
Here's one that works (and I checked with your values). It also matches, no matter if there is a trailing slash or not:
http:\/\/.*dest\/\w+/?$
/http:\/\/www\.site\.com\/dest\/\w+\/?$/i
matches if they're all the same site with the "dest" there. you could also do this:
/\w+:\/\/[^/]+\/dest\/\w+\/?$/i
which will match any site with any protocal (http,ftp) and any site with the /dest/country at the end, and an optional /
Note, that this will only work with a subset of what the urls could legitimately be.
Try this regular expression:
^http://www\.example\.com/dest/[^/]+/$
This would only match the last three URLs.