I'm trying to add in allowed urls in a watchguard firebox webblocker list using regular expression. I'm trying to keep my list short by allowing one entry to apply to both www and non-www versions of a site including subdomains. I'm currently using the following:
(www\.)?ups\.com/*
Which works great for both versions plus subdomains, but has an issue as it allows other sites through that end their domain with ups.com such as jobs-ups.com
How can I make the regular expression know that if there is no subdomain that the url is only going to be ups.com without any other letters before the u, so it will block sites like jobs-ups.com?
You can use the caret ^ to accomplish this
^(?:www\.)?ups\.com\/
DEMO
The caret forces the check at the start of the string. This means it will not match in mid-string, which is what you are wanting.
Not familiar with firebox at all, but generally you should escape your periods and forward slashes. You would also generally use a non-capturing group as well. But if this is simple regex, you can still preserve your original formatting:
^(www.)?ups.com/*
Related
I need to add url patterns for a domain that has multiple subdomains in the Url.
For example:
https://demo.site1.mybrand.company/ Where .company is the top-level domain and mybrand is the domain.
The problem is that the demo subdomain can change based on the environment in the app, so it could be demo, or test, or anything, so I would like to make sure that any subdomain with site1.mybrand.company can access the Dynamic Links API and generate links for that Url.
What I have tried:
Firebase docs cite that these are too permission and I am not sure if Firebase supports multi-tier domains such as this.
^https://.*.company/.*$
^https://.*.mybrand.company/.*$
^https://.*.site1.mybrand.company/.*$
Has anyone experienced this situation before or know if this particular scenario is supported?
References:
https://support.google.com/firebase/answer/9021429
https://github.com/google/re2/wiki/Syntax
You might use a bit more specific pattern to match either demo or test using an alternation, and extend that to all the allowed names.
^https://(?:demo|test)\.site1\.mybrand\.company/\S*$
The pattern matches:
^ Start of string
https:// Match literally
(?:demo|test) Match either demo or test
\.site1\.mybrand\.company/ Match .site1.mybrand.company/ (note to escape the dot)
\S* Match optional non whitespace chars
$ End of string
Regex demo
I want to have a regular Expression for Google Analytic so I can match all the domain including the sub domains
say we have to match a domain name called xyz.com
So i want to match every url that have xyz.com in it.
Example
abcd.xyz.com, abc1232.xyz, www.xyz.com, www.xyz.com/abc
Can anyone help me with that.
My purpose to it to have the traafic reports excluded in Google Analytics that are coming from these sites.
In general, the regular expression to match those domains would be something like .*\.xyz\.com$. The backslashes escape the dots (which are normally wildcard characters and the dollar-sign represents the end of the string.
There are different regex implementations, so you might have to tweak this for your regex engine.
To exclude subdomains like described above you can use GA filter([Exclude] [Hostname] [Matching RegEx]) along with regular expression (xyz.com)|(.*.xyz.com).
This RegEx including both main domain and it's subdomains.
You could try this regex
(.*\.)?xyz\.com
This matches all your required formats for the URL.
I'm using CodeIgniter to write an app where a user will be allowed to register an account and is assigned a URL (URL slug) of their choosing (ex. domain.com/user-name). CodeIgniter has a URL routing feature that allows the utilization of regular expressions (link).
User's are only allowed to register URL's that contain alphanumeric characters, dashes (-), and under scores (_). This is the regex I'm using to verify the validity of the URL slug: ^[A-Za-z0-9][A-Za-z0-9_-]{2,254}$
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.
How do I redirect all the following URLs to "/" using single regular expression?
members/kaleem/
members/kaleem/activity/just-me/
members/kaleem/activity/
members/kaleem/activity/favorites/
members/kaleem/activity/groups/
members/kaleem/friends/
I am using it wordpress redirect plugin.
I'm not sure how Wordpress' redirect plugin works, but this regular expression will match all of above, as well as any other pages after members/kaleem.
members/kaleem[[\w\-\/]*
Grab word characters, dashes, and slashes that appear after members/kaleem. If there are certain pages after members/kaleem that shouldn't be matched, it get's more complicated. I was assuming that the examples you showed were part of a pattern.
If you want to only match kaleem/activity and kaleem/friends, plus any pages that are children of them, you can use this:
members/kaleem/((activity|friends)[\w\/\-]*)?
It seems members/ is the common identifier. Correct? If so, you just have to match that: ^members/. Otherwise it becomes a bit more complicated: ^members/kaleem/(?:friends|activity/(?:(?:just-me|favorites|groups)/)?). See: http://regex101.com/r/jJ4rM8
10 websites need to be cached. When caching: photos, css, js, etc are not displayed properly because the base domain isn't attached to the directory. I need a regex to add the base domain to the directory. examples below
base domain: http://www.example.com
the problem occurs when reading cached pages with img src="thumb/123.jpg" or src="/inc/123.js".
they would display correctly if it was img src="http://www.example.com/thumb/123.jpg" or src="http://www.example.com/inc/123.js".
regex something like: if (src=") isn't followed by the base domain then add the base domain
without knowing the language, you can use the (maybe most portable) substitute modifier:
s/^(src=")([^"]+")$/$1www\.example\.com\/$2/
This should do the following:
1. the string 'src="' (and capture it in variable $1)
2. one or more non-double-quote (") character followed by " (and capture it in variable $2)
3. Substitutes 'www.example.com/' in between the two capture groups.
Depending on the language, you can wrap this in a conditional that checks for the existence of the domain and substitutes if it isn't found.
to check for domain: /www\.example\.com/i should do.
EDIT: See comments:
For PHP, I would do this a bit differently. I would probably use simplexml. I don't think that will translate well, though, so here's a regex one...
$html = file_get_contents('/path/to/file.html');
$regex_match = '/(src="|href=")[^(?:www.example.com\/)]([^"]+")/gi';
$regex_substitute = '$1www.example.com/$2';
preg_replace($regex_match, $regex_substitute, $html);
Note: I haven't actually run this to debug it, it's just off the cuff. I would be concerned about 3 things. first, I am unsure how preg_replace will handle the / character. I don't think you're concerned with this, though, unless VB has a similar problem. Second, If there's a chance that line breaks would get in the way, I might change the regex. Third, I added the [^(?:www\.example\.com)] bit. This should change the match to any src or href that doesn't have www.example.com/ there, but this depends on the type of regex being used (POSIX/PCRE).
The rest of the changes should be fine (I added href=" and also made it case-insensitive (\i) and there's a requirement to make it global (\g) otherwise, it will just match once).
I hope that helps.
Matching regular expression:
(?:src|href)="(http://www\.example\.com/)?.+