I need to find all emails aren't info#example.com to config openwebmail with regex
With result:
abbb#example.com -- true
76312783dd#example.com -- true
8734289347#test.it -- true
info#example.com -- false
This would be my answer:
^(?!info#example\.com$)((\w+|\.)+#\w+(\.\w+)+)$
This will match everything in your examples except info#example.com, and a few more that would be useful, such as test.example#mail.co.uk, as shown in the example here:
Click here for the example on Regex101
Edit
You could also change the regex to:
^(?!info#example\.com$)((\w+\.?)+#\w+(\.\w+)+)$
To disallow the emails: info..example#mail.com, but as you are simply searching through emails, it is very unlikely that you will come across an email with the format info..example#mail.com
Breakdown
^ Start of string
(?!) Negative lookahead
info#example\.com$ Literal match of info.example.com, and it being until the end of the string.
(\w+|\.)+ Capture group with letters and numbers 1 or more times, or a dot, 1 or more times Will match example..test#mail.com
(\w\.?)+ Capture group with letters 1 or more times, with a following optional dot, 1 or more times. Will not match example..test#mail.com
#\w+ Matches # literally, and any letters or numbers after.
(.\w+)+ Matches a dot and any letters coming after it 1 or more times. Dot not optional because an email like: example#test.com. is invalid.
$ End of string
Use this great trick!:
^info#example\.com$|^(\w+#\w+\.\w+)$
Demo: https://regex101.com/r/zJ5nY5/1
Explanation:
a|b|(c|d) and take the group 1 value. This is one of the most popular regex trick ever. This way, if you take group 1 value, a and b will be automatically ignored and you only approve c and d.
\w is any character that we normally find inside a word.
Related
Have used an online regex learning site (regexr) and created something that works but with my very limited experience with regex creation, I could do with some help/advice.
In IIS10 logs, there is a list for time, date... but I am only interested in the cs(User-Agent) field.
My Regex:
(scan\-\d+)(?:\w)+\.shadowserver\.org
which matches these:
scan-02.shadowserver.org
scan-15n.shadowserver.org
scan-42o.shadowserver.org
scan-42j.shadowserver.org
scan-42b.shadowserver.org
scan-47m.shadowserver.org
scan-47a.shadowserver.org
scan-47c.shadowserver.org
scan-42a.shadowserver.org
scan-42n.shadowserver.org
scan-42o.shadowserver.org
but what I would like it to do is:
Match a single number with the option of capturing more than one: scan-2 or scan-02 with an optional letter: scan-2j or scan-02f
Append the rest of the User Agent: .shadowserver.org to the regex.
I will then add it to an existing URL Rewrite rule (as a condition) to abort the request.
Any advice/help would be very much appreciated
Tried:
To write a regex for IIS10 to block requests from a certain user-agent
Expected:
It to work on single numbers as well as double/triple numbers with or without a letter.
(scan\-\d+)(?:\w)+\.shadowserver\.org
Input Text:
scan-2.shadowserver.org
scan-02.shadowserver.org
scan-2j.shadowserver.org
scan-02j.shadowserver.org
scan-17w.shadowserver.org
scan-101p.shadowserver.org
UPDATE:
I eventually came up with this:
scan\-[0-9]+[a-z]{0,1}\.shadowserver\.org
This is explanation of your regex pattern if you only want the solution, then go directly to the end.
(scan\-\d+)(?:\w)+
(scan\-\d+) Group1: match the word scan followed by a literal -, you escaped the hyphen with a \, but if you keep it without escaping it also means a literal - in this case, so you don't have to escape it here, the - followed by \d+ which means one more digit from 0-9 there must be at least one digit, then the value inside the group will be saved inside the first capturing group.
(?:\w)+ non-capturing group, \w one character which is equal to [A-Za-z0-9_], but the the plus + sign after the non-capturing group (?:\w)+, means match the whole group one or more times, the group contains only \w which means it will match one or more word character, note the non-capturing group here is redundant and we can use \w+ directly in this case.
Taking two examples:
The first example: scan-02.shadowserver.org
(scan\-\d+)(?:\w)+
scan will match the word scan in scan-02 and the \- will match the hyphen after scan scan-, the \d+ which means match one or more digit at first it will match the 02 after scan- and the value would be scan-02, then the (?:\w)+ part, the plus + means match one or more word character, at least match one, it will try to match the period . but it will fail, because the period . is not a word character, at this point, do you think it is over ? No , the regex engine will return back to the previous \d+, and this time it will only match the 0 in scan-02, and the value scan-0 will be saved inside the first capturing group, then the (?:\w)+ part will match the 2 in scan-02, but why the engine returns back to \d+ ? this is because you used the + sign after \d+, (?:\w)+ which means match at least one digit, and one word character respectively, so it will try to do what it is asked to do literally.
The second example: scan-2.shadowserver.org
(scan\-\d+)(?:\w)+
(scan\-\d+) will match scan-2, (?:\w)+ will try to match the period after scan-2 but it fails and this is the important point here, then it will go back to the beginning of the string scan-2.shadowserver.org and try to match (scan\-\d+) again but starting from the character c in the string , so s in (scan\-\d+) faild to match c, and it will continue trying, at the end it will fail.
Simple solution:
(scan-\d+[a-z]?)\.shadowserver\.org
Explanation
(scan-\d+[a-z]?), Group1: will capture the word scan, followed by a literal -, followed by \d+ one or more digits, followed by an optional small letter [a-z]? the ? make the [a-z] part optional, if not used, then the [a-z] means that there must be only one small letter.
See regex demo
I want to expect some characters only if a prior regex matched. If not, no characters (empty string) is expected.
For instance, if after the first four characters appears a string out of the group (A10, B32, C56, D65) (kind of enumeration) then a "_" followed by a 3-digit number like 123 is expected. If no element of the mentioned group appears, no other string is expected.
My first attempt was this but the ELSE branch does not work:
^XXX_(?<DT>A12|B43|D14)(?(DT)(_\d{1,3})|)\.ZZZ$
XXX_A12_123.ZZZ --> match
XXX_A11.ZZZ --> match
XXX_A12_abc.ZZZ --> no match
XXX_A23_123.ZZZ --> no match
These are examples of filenames. If the filename contains a string of the mentioned group like A12 or C56, then I expect that this element if followed by an underscore followed by 1 to 3 digits. If the filename does not contain a string of that group (no character or a character sequence different from the strings in the group) then I don't want to see the underscore followed by 1 to 3 digits.
For instance, I could extend the regex to
^XXX_(?<DT>A12|B43|D14)_\d{5}(?(DT)(_\d{1,3})|)_someMoreChars\.ZZZ$
...and then I want these filenames to be valid:
XXX_A12_12345_123_wellDone.ZZZ
XXX_Q21_00000_wellDone.ZZZ
XXX_Q21_00000_456_wellDone.ZZZ
...but this is invalid:
XXX_A12_12345_wellDone.ZZZ
How can I make the ELSE branch of the conditional statement work?
In the end I intend to have two groups like
Group A: (A11, B32, D76, R33)
Group B: (A23, C56, H78, T99)
If an element of group A occurs in the filename then I expect to find _\d{1,3} in the filename.
If an element of group B occurs ion the filename then the _\d{1,3} shall be optional (it may or may not occur in the filename).
I ended up in this regex:
^XXX_(?:(?A12|B43|D14))?(?(DT)(_\d{5}_\d{1,3})|(?!(?&DT))(?!.*_\d{3}(?!\d))).*\.ZZZ$
^XXX_(?:(?<DT>A12|B43|D14))?_\d{5}(?(DT)(_\d{1,3})|(?!(?&DT))(?!.*_\d{3}(?!\d))).+\.ZZZ$
Since I have to use this regex in the OpenApi #Pattern annotation I have the problem that I get the error:
Conditionals are not supported in this regex dialect.
As #The fourth bird suggested alternation seems to do the trick:
XXX_((((A12|B43|D14)_\d{5}_\d{1,3}))|((?:(A10|B10|C20)((?:_\d{5}_\d{3})|(?:_\d{3}))))).*\.ZZZ$
The else branch is the part after the |, but if you also want to match the 2nd example, the if clause would not work as you have already matched one of A12|B43|D14
The named capture group is not optional, so the if clause will always be true.
What you can do instead is use an alternation to match either the numeration part followed by an underscore and 3 digits, or match an uppercase char and 2 digits.
^XXX_(?:(?<DT>A12|B43|D14)_\d{1,3}|[A-Z]\d{2})\.ZZZ$
Regex demo
If you want to make use of the if/else clause, you can make the named capture group optional, and then check if group 1 exists.
^XXX_(?<DT>A12|B43|D14)?(?(DT)_\d{1,3}|[A-Z]\d{2})\.ZZZ$
Regex demo
For the updated question:
^XXX_(?<DT>A12|B43|D14)?(?(DT)(?:_\d{5})?_\d{3}(?!\d)|(?!A12|B43|D14|[A-Z]\d{2}_\d{3}(?!\d))).*\.ZZZ$
The pattern matches:
^ Start of string
XXX_ Match literally
(?<DT>A12|B43|D14)?
(?(DT) If we have group DT
(?:_\d{5})? Optionally match _ and 5 digits
_\d{3}(?!\d) Match _ and 3 digits
| Or
(?! Negative lookahead, assert not to the right
A12|B43|D14| Match one of the alternatives, or
[A-Z]\d{2}_\d{3}(?!\d) Match 1 char A-Z, 2 digits _ 3 digits not followed by a digit
) Close lookahead
) Close if clause
.* Match the rest of the line
\.ZZZ Match . and ZZZ
$ End of string
Regex demo
Intro
I have a string containing diagnosis codes (ICD-10), not separated by any character. I would like to extract all valid diagnosis codes. Valid diagnosis codes are of the form
[Letter][between 2 and 4 numbers][optional letter that is not the next match starting letter]
The regex for this pattern is (I believe)
\w\d{2,4}\w?
Example
Here is an example
mystring='F328AG560F33'
In this example there are three codes:
'F328A' 'G560' 'F33'
I would like to extract these codes with a function like str_extract_all in R (preferably but not exclusively)
My solution so far
So far, I managed to come up with an expression like:
str_extract_all(mystring,pattern='\\w\\d{2,4}\\w?(?!(\\w\\d{2,4}\\w?))')
However when applied to the example above it returns
"F328" "G560F"
Basically it misses the letter A in the first code, and misses altogether the last code "F33" by mistakenly assigning F to the preceding code.
Question
What am I doing wrong? I only want to extract values that end with a letter that is not the start of the next match, and if it is, the match should not include the letter.
Application
This question is of great relevance for example when mining patient Electronic Health Records that have not been validated.
You have a letter, two-to-four numbers then an optional letter. That optional letter, if it's there, will only ever be followed by another letter; or, put another way, never followed by a number. You can write a negative lookahead to capture this:
\w\d{2,4}(?:\w(?!\d))?
This at least works with PCRE. I don't know about how R will handle it.
Your matches are overlapping. In this case, you might use str_match_all that allows easy access to capturing groups and use a pattern with a positive lookahead containing a capturing group inside:
(?i)(?=([A-Z]\d{2,4}(?:[A-Z](?!\d{2,4}))?))
See the regex demo
Details
(?= - a positive lookahead start (it will be run at every location before each char and at the end of the string
( - Group 1 start
[A-Z] - a letter (if you use a case insensitive modifier (?i), it will be case insensitive)
\d{2,4} - 2 to 4 digit
(?: - an optional non-capturing group start:
[A-Z] - a letter
(?!\d{2,4}) - not followed with 2 to 4 digits
)? - the optional non-capturing group end
) - Group 1 end
) - Lookahead end.
R demo:
> library(stringr)
> res <- str_match_all("F328AG560F33", "(?i)(?=([A-Z]\\d{2,4}(?:[A-Z](?!\\d{2,4}))?))")
> res[[1]][,2]
[1] "F328A" "G560" "F33"
I'm trying to come up with some regex to match against 1 hyphen per any number of digit groups. No characters ([a-z][A-Z]).
123-356-129811231235123-1235612346123451235
/[^\d-]/g
The one above will match the string below, but it will let the following go through:
1223--1235---123123-------
I was looking at the following post How to match hyphens with Regular Expression? for an answer, but I didn't find anything close.
#Konrad Rudolph gave a good example.
Regular expression to match 7-12 digits; may contain space or hyphen
This tool is useful for me http://www.gskinner.com/RegExr/
Assuming it can't ever start with a hyphen:
^\d(-\d|\d)*$
broken down:
^ # match beginning of line
\d # match single digit
(-\d|\d)+ # match hyphen & digit or just a digit (0 or more times)
$ # match end of line
That makes every hyphen have to have a digit immediately following it. Keep in mind though, that the following are examples of legal patterns:
213-123-12314-234234
1-2-3-4-5-6-7
12234234234
gskinner example
Alternatively:
^(\d+-)+(\d+)$
So it's one or more group(s) of digits followed by hyphen + final group of digits.
Nothing very fancy, but in my tests it matched only when there were hyphen(s) with digits on both sides.
How to write regular expression to find between one and three digits separated by periods without returning the last period? For example, find the string
1.1.
and it would also need to match
1.1
or simply
1
Likewise, it needs to support between one and three digits, so
11.11.11
and
111.111.111
need to work as well.
So..the string won't always end with a period, but it may. Further, if it does end with a period, don't return the last period (so, using a positive lookahead). So, 1.1. if matched would return 1.1
Here is what I have so far, but I am struggling to find a way to NOT return the last period:
(\d{1,3}\.?)+(?=(\Z|\s|\-|\;|\:|\?|\!|\.|\,|\)))
It is returning
6.6.
but I want it to return
6.6
You require: match d.d.d.d. or d.d.dxxx, and regardless of whether it ends with a "." or not, always stop at the last d (never the dot).
What's wrong with just: (\d(\.\d)*)
If you want your dotted-digit string to be terminated by a set of characters, put a look-ahead after it, as you have in your question:
(\d(\.\d)*)(?=(\Z|\s|\-|\;|\:|\?|\!|\.|\,|\)))
If you also want it to match a stand-alone string (with or without the terminator), add a ? after the look-ahead:
(\d(\.\d)*)(?=(\Z|\s|\-|\;|\:|\?|\!|\.|\,|\)))?
For more than one digits, just replace \d with \d{1,3} etc.
The regex (\d{1,3}(?:\.\d{1,3})*)\.{0,1} should work.
In the Group 1 (if taken Group 0 as the entire match) will be stored the string you want to keep, without the . at the end, in case it contains it.
It basically does:
Start matching 1-3 digits
Then match strings like .d, .dd, or .ddd
If the match ends with a ., it won't take it because it isn't inside the group.
Do your tests and let us know if it works with all your examples.
Edit:
Replace + with *
/\d{1,3}(\.\d{1,3})*/
Quick explanation:
\d{1,3} # Match 1-3 digits
( # Start Capture Group 1
\. # Match '.'
\d{1,3} # Match 1-3 digits
)* # End Capture Group 1 - match 0 or more times
You can write your own Regular expression and test with dummy data on following Site:
http://myregexp.com/