Validating MS Teams channel names, team name and channel address in one go!
Regex1:
^(?![\s._])([^~#%&*{}+:<>?|\n]{1,50})(?<![.])( - ).{1,256}$
Which would validate this successfully:
MonkeyChannel - MonkeyTeam
but I need to check also that it doesn't contain the channel address like so:
MonkeyChannel - MonkeyTeam <12337aab.domain.com#emea.teams.ms>
so basically I'm thinking I need to incorporate this which looks for a channel address:
Regex2:
(?<![[a-z0-9]{8}\.domain\.com#emea\.teams\.ms])
into this somehow:
Regex1:
^(?![\s._])([^~#%&*{}+:<>?|\n]{1,50})(?<![.])( - ).{1,256}$
My guess looks something like this but its not working:
Regex3:
^(?![\s._])([^~#%&*{}+:<>?|\n]{1,50})(?<![.])( - ).{1,256}(?<![[a-z0-9]{8}\.domain\.com#emea\.teams\.ms])$
Any regex wizards who can spot the error of my ways?
You could either add the negative lookbhind after the anchor and note that you have <...> and not [...]
^(?![\s._])[^~#%&*{}+:<>?|\n]{1,50}(?<![.]) - .{1,256}$(?<!<[a-z0-9]{8}\.domain\.com#emea\.teams\.ms>)
Regex demo
The other way around is using a negative lookahead after matching -
^(?![\s._])[^~#%&*{}+:<>?|\n]{1,50}(?<![.]) - (?!.*<[a-z0-9]{8}\.domain\.com#emea\.teams\.ms>).{1,256}$
Regex demo
You can add the capture group accoringly if you want to after process separate parts.
If you still want to match the first part, you can optionally match the second and first assert that it does not start with the unwanted mail address
^(?![\s._])[^~#%&*{}+:<>?|\n]{1,50}(?<![.]) - [^<>\n]*(?:<(?![a-z0-9]{8}\.domain\.com#emea\.teams\.ms>).{1,256})?
Regex demo
I would like to match and block address like foo.bar#gmail.com. But it isn't that easy, since any of following:
foobar#gmail.com
fo.o....b..a..r#gmail.com
foo.bar+goo#gmail.com
fo.ob.ar+something#gmail.com
Is alias for same email account. Is it possible to create regexp that matches all possible aliases? Or do I have to normalize (remove dots and text after +) all gmail addresses before applying filters/blacklist?
I could go with : f[.]*o[.]*o[.]*b[.]*a[.]*r[.]*(+.*) but it looks ridiculous for longer email and probably has bad performance
One possibility would be a regex such as
f\.*o\.*o\.*b\.*a\.*r(?=.*\#gmail\.com)
This pattern basically says after any letter of foobar there may be some unknown number of dots .. You can always work from here on now and expand the expression to something like this
f[\.-_]*o[\.-_]*o[\.-_]*b[\.-_]*a[\.-_]*r(?=.*\#gmail\.com)
Here we also accept unknown numbers of hyphens and underscores.
Example
Here is an example in python:
# import regex
string = 'fo.o....b..a..r#gmail.com'
pattern = r'f\.*o\.*o\.*b\.*a\.*r(?=.*\#gmail\.com)'
test = regex.search(pattern, strings[0])
print(test.group(0))
# foobar
I have a text field which stores a list of email addresses e.g: x#demo.com; a.x#demo.com. I have another text field which stores the exact value matched from the list of emails i.e. if /x#demo.com/i is in x#demo.com;a.x#demo.com then it should return x#demo.com.
The issue I am having is that if I have /a.x#demo.com/i, I will get x#demo.com instead of a.x#demo.com
I know of the regex expression /^x#demo.com$/i, but this means I can only have one email in my list of email addresses which won't help.
I have tried a couple of other regex expressions with no luck.
Any ideas on how I can achieve this?
You can use this slightly changed regex:
/(^|;)x#demo.com($|;)/i
It will match from either beginning of string or start after a semi colon and end either at end of string or at a semi colon.
Edit:
Small change, this uses look behind and look forward, then you will only get the match, you want:
(?<=^|;)x#demo.com(?=$|;)
Edit2:
To allow Spaces around the semi colon and at start and end, use this (#-quoted):
#"(?<=^\s*|;\s*)x#demo.com(?=\s*$|\s*;)"
or use double escaping:
"(?<=^\\s*|;\\s*)x#demo.com(?=\\s*$|\\s*;)"
I am learning regex and am having trouble getting google from email address
String
first.name#google.com
I just want to get google, not google.com
Regex:
[^#].+(?=\.)
Result: https://regex101.com/r/wA5eX5/1
From my understanding. It ignore # find a string after that until . (dot) using (?=\.)
What did I do wrong?
[^#] means "match one symbol that is not an # sign. That is not what you are looking for - use lookbehind (?<=#) for # and your (?=\.) lookahead for \. to extract server name in the middle:
(?<=#)[^.]+(?=\.)
The middle portion [^.]+ means "one or more non-dot characters".
Demo.
Updated answer:Use a capturing group and keep it simple :)
#(\w+)
Explanation by splitting it up
( capturing group for extraction )
\w stands for word character [A-Za-z0-9_]
+ is a quantifier for one or more occurances of \w
Regex explanation and demo on Regex101
I used the solution's regex for my task, but realized that some of the emails weren't that easy: foo#us.industries.com, foobar#tm.valves.net, andfoo#ge.test.com
To anyone who came here wanting the sub domain as well (or is being cut off by it), here's the regex:
(?<=#)[^.]*.[^.]*(?=\.)
This should be the regex:
(?<=#)[^.]+
(?<=#) - places the search right after the #
[^.]+ - take all the characters that are not dot (stops on dot)
So it extracts google from the email address.
As I was working to get the domain name of email addresses and none corresponded to what I needed:
To not catch subdomains
To match countries top domains (like .com.ar or co.jp)
For example, in test#ext.domain.com.mx I need to match domain.com.mx
So I made this one:
[^.#]*?\.\w{2,}$|[^.#]*?\.com?\.\w{2}$
Here is a link to regex101 to illustrate the regex: https://regex101.com/r/vE8rP9/59
You can get the sumdomain name (without the top-level domain ex: .com or .com.mx) by adding lookaround operators (but it will match twice in test#test.com.mx):
[^.#]*?(?=\.\w{2,}$)|[^.#]*?(?=\.com?\.\w{2}$)
Maybe not strictly a "full regex answer" but more flexible ( in case the part before the # is not "first.last") would be using cut:
cut -d # -f 2 | cut -d . -f 1
The first cut will isolate the part after # and the second one will get what you want.
This will work also for another kinds of email patterns : xxxx#server.com / xxx.yyy.zzz# server.com and so on...
Thanks everyone for your great responses, I took what you had and expanded it with labelled match-groups for easy extraction of separate parts.
Caveat : Regex.Speed = Slow
Another post mentioned how SLOW and nonperformant regexes are, and that is a fair point to remember. My particular need is targeting my own background/slow/reporting processes and therefore it doesn't matter how long it takes.
But it's good to remember whenever possible Regex should NOT be used in any sort of web page load or "needs-to-be-quick" kind of application. In that case you're much better off using substring to algorithmically strip down the inputs and throw away all the junk that I'm optionally matching/allowing/including here.
https://regex101.com/r/ZnU3OC/1
One Regex to rule them all...
Subdomain/Domain/TopLevelDomain/CountryCode extraction for Emails, domain lists, & URLs
Also handles ?Querystring=junk, Slashes/With/Paths, #anchors
Now with more broth, batteries not included
^(?<Email>.*#)?(?<Protocol>\w+:\/\/)?(?<SubDomain>(?:[\w-]{2,63}\.){0,127}?)?(?<DomainWithTLD>(?<Domain>[\w-]{2,63})\.(?<TopLevelDomain>[\w-]{2,63}?)(?:\.(?<CountryCode>[a-z]{2}))?)(?:[:](?<Port>\d+))?(?<Path>(?:[\/]\w*)+)?(?<QString>(?<QSParams>(?:[?&=][\w-]*)+)?(?:[#](?<Anchor>\w*))*)?$
not overly complicated at all... why would you even say that?
Substitution / Outputs
EXAMPLE INPUT: "https://www.stackoverflow.co.uk/path/2?q=mysearch&and=more#stuff"
EXAMPLE OUTPUT:
{
Protocol: "https://"
SubDomain: "www"
DomainWithTLD: "stackoverflow.co.uk"
Domain: "stackoverflow"
TopLevelDomain: "co"
CountryCode: "uk"
Path: "/path/2"
QString: "?q=mysearch&and=more#stuff"
}
Allowed/Compliant Domains : Should ALL MATCH
www.bankofamerica.com
bankofamerica.com.securersite.regexr.com
bankofamerica.co.uk.blahblahblah.secure.com.it
dashes-bad-for-seo.but-technically-still-allowed.not-in-front-or-end
bit.ly
is.gd
foo.biz.pl
google.com.cn
stackoverflow.co.uk
level_three.sub_domain.example.com
www.thelongestdomainnameintheworldandthensomeandthensomemoreandmore.com
https://www.stackoverflow.co.uk?q=mysearch&and=more
foo://5th.4th.3rd.example.com:8042/over/there
foo://subdomain.example.com:8042/over/there?name=ferret#nose
example.com
www.example.com
example.co.uk
trailing-slash.com/
trailing-pound.com#
trailing-question.com?
probably-not-valid.com.cn?&#
probably-not-valid.com.cn/?&#
example.com/page
example.com?key=value
* NOTE: PunyCodes (Unicode in urls) handled just fine with \w ,no extra sauce needed
xn--fsqu00a.xn--0zwm56d.com
xn--diseolatinoamericano-66b.com
Emails : Should ALL MATCH
first.name#google1.co.com
foo#us.industries.com,
foobar#tm.valves.net,
andfoo#ge.test.com
jane.doe#my-bank.no
john.doe#spam.com
jane.ann.doe#sandnes.district.gov
Non-Compliant Domains : Should NOT MATCH
either not long-enough (domain min length 2), or too long (64)
v.gd
thing.y
0123456789012345678901234567890123456789012345678901234567891234.com
its-sixty-four-instead-of-sixty-three!.com
symbols-not-allowed#.com
symbols-not-allowed#.com
symbols-not-allowed$.com
symbols-not-allowed%.com
symbols-not-allowed^.com
symbols-not-allowed&.com
symbols-not-allowed*.com
symbols-not-allowed(.com
symbols-not-allowed).com
symbols-not-allowed+.com
symbols-not-allowed=.com
TBD Not handled:
* dashes as start or ending is disallowed (dropped from Regex for readability)
-junk-.com
* is underscore allowed? i donno... (but it simplifies the regex using \w instead of [a-zA-Z0-9\-] everywhere)
symbols-not-allowed_.com
* special case localhost?
.localhost
also see:
Domain Name Rules :: Super handy ASCII Diagram of a URL
see: https://stackoverflow.com/a/66660651/738895 *
Side NOTE: lazy load '?' for subdomains{0,127}? currently needed for any of the cases with country codes... (example: stackoverflow.co.uk)
Matches these, but does NOT grab $NLevelSubdomains in a match group, can only grab 3rd level only.
This is a relatively simple regex, and it grabs everything between the # and the final domain extension (e.g. .com, .org). It allows domain names that are made up of non-word characters, which exist in real-world data.
>>> regex = re.compile(r"^.+#(.+)\.[\w]+$")
>>> regex.findall('jane.doe#my-bank.no')
['my-bank']
>>> regex.findall('john.doe#spam.com')
['spam']
>>> regex.findall('jane.ann.doe#sandnes.district.gov')
['sandnes.district']
I used this regular expression to get the complete domain name '.*#+(.*)' where .* will ignore all the character before # (by #+) and start extracting cpmlete domain name by mentioning paranthesis and complete string inside(except linebrake characters)
I'm basically not in the clue about regex but I need a regex statement that will recognise anything after the / in a URL.
Basically, i'm developing a site for someone and a page's URL (Local URL of Course) is say (http://)localhost/sweettemptations/available-sweets. This page is filled with custom post types (It's a WordPress site) which have the URL of (http://)localhost/sweettemptations/sweets/sweet-name.
What I want to do is redirect the URL (http://)localhost/sweettemptations/sweets back to (http://)localhost/sweettemptations/available-sweets which is easy to do, but I also need to redirect any type of sweet back to (http://)localhost/sweettemptations/available-sweets. So say I need to redirect (http://)localhost/sweettemptations/sweets/* back to (http://)localhost/sweettemptations/available-sweets.
If anyone could help by telling me how to write a proper regex statement to match everything after sweets/ in the URL, it would be hugely appreciated.
To do what you ask you need to use groups. In regular expression groups allow you to isolate parts of the whole match.
for example:
input string of: aaaaaaaabbbbcccc
regex: a*(b*)
The parenthesis mark a group in this case it will be group 1 since it is the first in the pattern.
Note: group 0 is implicit and is the complete match.
So the matches in my above case will be:
group 0: aaaaaaaabbbb
group 1: bbbb
In order to achieve what you want with the sweets pattern above, you just need to put a group around the end.
possible solution: /sweets/(.*)
the more precise you are with the pattern before the group the less likely you will have a possible false positive.
If what you really want is to match anything after the last / you can take another approach:
possible other solution: /([^/]*)
The pattern above will find a / with a string of characters that are NOT another / and keep it in group 1. Issue here is that you could match things that do not have sweets in the URL.
Note if you do not mind the / at the beginning then just remove the ( and ) and you do not have to worry about groups.
I like to use http://regexpal.com/ to test my regex.. It will mark in different colors the different matches.
Hope this helps.
I may have misunderstood you requirement in my original post.
if you just want to change any string that matches
(http://)localhost/sweettemptations/sweets/*
into the other one you provided (without adding the part match by your * at the end) I would use a regular expression to match the pattern in the URL but them just blind replace the whole string with the desired one:
(http://)localhost/sweettemptations/available-sweets
So if you want the URL:
http://localhost/sweettemptations/sweets/somethingmore.html
to turn into:
http://localhost/sweettemptations/available-sweets
and not into:
localhost/sweettemptations/available-sweets/somethingmore.html
Then the solution is simpler, no groups required :).
when doing this I would make sure you do not match the "localhost" part. Also I am assuming the (http://) really means an optional http:// in front as (http://) is not a valid protocol prefix.
so if that is what you want then this should match the pattern:
(http://)?[^/]+/sweettemptations/sweets/.*
This regular expression will match the http:// part optionally with a host (be it localhost, an IP or the host name). You could omit the .* at the end if you want.
If that pattern matches just replace the whole URL with the one you want to redirect to.
use this regular expression (?<=://).+