I would need to retrieve an email id from a email address.
(i.e. this-is-the-best.email#gmail.com => this-is-the-best.email)
The regex that I used is (.*)#.* .
Now I need truncate the string with N characters.
(i.e. N=7 => this-is N=30 =>this-is-the-best.email)
How would I add this to a existing regex?
Any other recommendations?
What about: ([^#]{1,7}).+?
this-is-the-best-email#gmail.com
short#hotmail.co.uk
Becomes:
this-is
short
I think that this is what you are looking for:
((.{1,7}).*)#.+
The first capturing group contains the full id and the second group contains up to 7 chars.
In your pattern (.*)#.* you don't need the trailing .* as it is optional, and the dot can match any character including spaces and the # itself which can match much more that just an email like address.
The thing of interest is the non whitespace chars excluding an # char before actually matching the #, and in that case you can use a capture group matching 7 non whitespace chars.
([^\s#]{7})[^\s#]*#
The pattern matches:
([^\s#]{7}) Capture group 1, match 7 non whitespace chars excluding #
[^\s#]* Optionally match any non whitespace char excluding #
# Match literally
Regex demo
Related
I only have access to a function that can match a pattern and replace it with some text:
Syntax
regexReplace('text', 'pattern', 'new text'
And I need to return only the 5 digit string from text in the following format:
CRITICAL - 192.111.6.4: rta nan, lost 100%
Created Time Tue, 5 Jul 8:45
Integration Name CheckMK Integration
Node 192.111.6.4
Metric Name POS1
Metric Value DOWN
Resource 54871
Alert Tags 54871, POS1
So from this text, I want to replace everything with "" except the "54871".
I have come up with the following:
regexReplace("{{ticket.description}}", "\w*[^\d\W]\w*", "")
Which almost works but it doesn't match the symbols. How can I change this to match any word that includes a letter or symbol, essentially.
As you can see, the pattern I have is very close, I just need to include special characters and letters, whereas currently it is only letters:
You can match the whole string but capture the 5-digit number into a capturing group and replace with the backreference to the captured group:
regexReplace("{{ticket.description}}", "^(?:[\w\W]*\s)?(\d{5})(?:\s[\w\W]*)?$", "$1")
See the regex demo.
Details:
^ - start of string
(?:[\w\W]*\s)? - an optional substring of any zero or more chars as many as possible and then a whitespace char
(\d{5}) - Group 1 ($1 contains the text captured by this group pattern): five digits
(?:\s[\w\W]*)? - an optional substring of a whitespace char and then any zero or more chars as many as possible.
$ - end of string.
The easiest regex is probably:
^(.*\D)?(\d{5})(\D.*)?$
You can then replace the string with "$2" ("\2" in other languages) to only place the contents of the second capture group (\d{5}) back.
The only issue is that . doesn't match newline characters by default. Normally you can pass a flag to change . to match ALL characters. For most regex variants this is the s (single line) flag (PCRE, Java, C#, Python). Other variants use the m (multi line) flag (Ruby). Check the documentation of the regex variant you are using for verification.
However the question suggest that you're not able to pass flags separately, in which case you could pass them as part of the regex itself.
(?s)^(.*\D)?(\d{5})(\D.*)?$
regex101 demo
(?s) - Set the s (single line) flag for the remainder of the pattern. Which enables . to match newline characters ((?m) for Ruby).
^ - Match the start of the string (\A for Ruby).
(.*\D)? - [optional] Match anything followed by a non-digit and store it in capture group 1.
(\d{5}) - Match 5 digits and store it in capture group 2.
(\D.*)? - [optional] Match a non-digit followed by anything and store it in capture group 3.
$ - Match the end of the string (\z for Ruby).
This regex will result in the last 5-digit number being stored in capture group 2. If you want to use the first 5-digit number instead, you'll have to use a lazy quantifier in (.*\D)?. Meaning that it becomes (.*?\D)?.
(?s) is supported by most regex variants, but not all. Refer to the regex variant documentation to see if it's available for you.
An example where the inline flags are not available is JavaScript. In such scenario you need to replace . with something that matches ALL characters. In JavaScript [^] can be used. For other variants this might not work and you need to use [\s\S].
With all this out of the way. Assuming a language that can use "$2" as replacement, and where you do not need to escape backslashes, and a regex variant that supports an inline (?s) flag. The answer would be:
regexReplace("{{ticket.description}}", "(?s)^(.*\D)?(\d{5})(\D.*)?$", "$2")
I'm trying to match a comma separated list with prefixed values which contains also a comma.
I finally made it to match all occurrence which doesn't have a ,.
Sample String (With NL for visualization - original string doesn't have NL):
field01=Value 1,
field02=Value 2,
field03=<xml value>,
field04=127.0.0.1,
field05=User-Agent: curl/7.28.0\r\nHost: example.org\r\nAccept: */*,
field06=Location, Resource,
field07={Item 1},{Item 2}
My actual RegEx looks like this not optimized piece ....
(?'fields'(field[0-9]{2,3})=?([\s\w\d_<>.:="*?\-\/\\(){}<>'#]+))([^,](?&fields))*
Any one has a clue how to solve this?
EDIT:
The first pattern is near to my expected result.
This is a anonymized full example of the string:
asm01=Predictable Resource Location,Information Leakage,asm02=N/A,asm04=Uncategorized,asm08=2021-02-15 09:18:16,asm09=127.0.0.1,asm10=443,asm11=N/A,asm15=,asm16=DE,asm17=User-Agent: curl/7.29.0\r\nHost: dev.example.com\r\nAccept: */*\r\nX-Forwarded-For: 127.0.0.1\r\n\r\n,asm18=/Common/_www.example.com_live_v1,asm20=127.0.0.1,asm22=,asm27=HEAD,asm34=/Common/_www.example.com_live_v1,asm35=HTTPS,asm39=blocked,asm41=0,asm42=3,asm43=0,asm44=Error,asm46=200000028,200100015,asm47=Unix hidden (dot-file) access,.htaccess access,asm48={Unix/Linux Signatures},{Apache/NCSA HTTP Server Signatures},asm50=40622,asm52=200000028,asm53=Unix hidden (dot-file) access,asm54={Unix/Linux Signatures},asm55=,asm61=,asm62=,asm63=8985143867830069446,asm64=example-waf.example.com,asm65=/.htaccess,asm67=Attack signature detected,asm68=<?xml version='1.0' encoding='UTF-8'?><BAD_MSG><violation_masks><block>13020008202d8a-f803000000000000</block><alarm>417020008202f8a-f803000000000000</alarm><learn>13000008202f8a-f800000000000000</learn><staging>200000-0</staging></violation_masks><request-violations><violation><viol_index>42</viol_index><viol_name>VIOL_ATTACK_SIGNATURE</viol_name><context>request</context><sig_data><sig_id>200000028</sig_id><blocking_mask>7</blocking_mask><kw_data><buffer>Ly5odGFjY2Vzcw==</buffer><offset>0</offset><length>2</length></kw_data></sig_data><sig_data><sig_id>200000028</sig_id><blocking_mask>4</blocking_mask><kw_data><buffer>Ly5odGFjY2Vzcw==</buffer><offset>0</offset><length>3</length></kw_data></sig_data><sig_data><sig_id>200100015</sig_id><blocking_mask>7</blocking_mask><kw_data><buffer>Ly5odGFjY2Vzcw==</buffer><offset>1</offset><length>9</length></kw_data></sig_data></violation></request-violations></BAD_MSG>,asm69=5,asm71=/Common/_dev.example.com_SSL,asm75=127.0.0.1,asm100=,asm101=HEAD /.htaccess HTTP/1.1\r\nUser-Agent: curl/7.29.0\r\nHost: dev.example.com\r\nAccept: */*\r\nX-Forwarded-For: 127.0.0.1\r\n\r\n#015
The pattern does not work as the fields group matches the string field
You are trying to repeat the named group fields but the example strings do not have the string field.
Note that [^,] matches any char except a comma, you can omit the capture group inside the named group field as it already is a group and \w also matches \d
With 2 capture groups:
\b(asm[0-9]+)=(.*?)(?=,asm[0-9]+=|$)
\b A word boundary
(asm[0-9]+) Capture group 1, match asm and 1+ digits
= Match literally
(.*?) Capture group 2, match any char as least as possible
(?= Positive lookahead, assert what is at the right is
,asm[0-9]+= Match ,asm followed by 1+ digits and =
| Or
$ Assert the end of the string
) Close lookahead
Regex demo
A simple solution would be (see regexr.com/5mg1b):
/((asm\d{2,3})=(.*?))(?=,asm|$)/g
Match groupings will be:
group #1 - asm01=Predictable Resource Location,Information Leakage
group #2 - asm01
group #3 - Predictable Resource Location,Information Leakage
Conditions:
This will match everything including empty values
The key here is to make sure that each match is delimited by either a comma and your field descriptor, or an end of string. A look ahead will be handy here: (?=,asm|$).
I'm looking to return the URL string to the right of a specific set of text using RegEx:
URL:
www.websitename/countrycode/websitename/contact/thank-you/whitepaper/countrycode/whitepapername.pdf
What I would like to just return:
/whitepapername.pdf
I've tried using ^\w+"countrycode"(\w.*) but the match won't recognize countrycode.
In Google Data Studio, I want to create a new field to remove the beginning of the URL using the REGEX_REPLACE function.
Ideally using:
REGEX_REPLACE(Page,......)
The REGEXP_REPLACE function below does the trick, capturing all (.*) the characters after the last countrycode, where Page represents the respective field:
REGEXP_REPLACE(Page, ".*(countrycode)(.*)$", "\\2")
Alternatively - Adapting the RegEx by The fourth bird to Google Data Studio:
REGEXP_REPLACE(Page, "^.*/countrycode(/[^/]+\\.\\w+)$", "\\1")
Google Data Studio Report as well as a GIF to elaborate:
You could use a capturing group and replace with group 1. You could match /countrycode literally or use the pattern to match 2 times chars a-z with an underscore in between like /[a-z]{2}_[a-z]{2}
In the replacement use group 1 \\1
^.*/countrycode(/[^/]+\.\w+)$
Regex demo
Or using a country code pattern from the comments:
^.*/[a-z]{2}_[a-z]{2}(/[^/]+\.\w+)$
Regex demo
The second pattern in parts
^ Start of string
.*/ Match until the last occurrence of a forward slash
[a-z]{2}_[a-z]{2} Match the country code part, an underscore between 2 times 2 chars a-z
( Capture group 1
/[^/]+ Match a forward slash, then match 1+ occurrences of any char except / using a negated character class
\.\w+ Match a dot and 1+ word chars
) Close group
$ End of string
What I have here is a string
sdsdsfd2312341232434545:4343523543435454
I want to get the values that are before the : char and also wants to exclude 16 chars to get
sdsdsfd
I have tried by using the expression
^.*(?=(\:).)
which matches all the strings before :. I want to exclude 16 chars before all the strings before : keyword. How do one do that? I want to return the sdsdsfd and 2312341232434545 too
You may use this regex:
.*?(?=.{16}:)
RegEx Demo
RegEx Details:
.*?: Match 0 or more characters (non-greedy)
(?=.{16}:): Lookahead to assert that there are 16 characters followed by a colon ahead
Alternative Approach
You can use avoid using lookahead by using a capture group here:
(.*?).{16}:
RegEx Demo 2
In this example I try to validate for a city name. It works if I enter San Louis Obispo but not if I enter Boulder Creek or Boulder. I thought ? was supposed to make a block optional.
if (!/^[a-zA-Z'-]+\s[a-zA-Z'-]*\s([a-zA-Z']*)?$/.test(field)){
return "Enter City only a-z A-Z .\' allowed and not over 20 characters.\n";
}
I think spaces are the problem (\s). You made second and third words optional (by using * instead of +), but not the spaces. Question mark is only being applied to the third word because of parentheses.
The issue with your regex is that, in english, it says to match a word that's required to be followed by a space that's optionally followed by another word but then is required to have another space and then optionally another word. So, a single-word would not match - however, a word followed by two spaces would. Additionally two words that have a space at the end would also match - but neither without the trailing spaces would match.
To fix your exact regex you should add another grouping (non-matching group with (?: instead of just () around the second word to the end of the sentence) and have this group as optional with ?. Also, move the \s's inside the optional groups as well.
Try this:
^[a-zA-Z'-]+(?:\s[a-zA-Z'-]+(?:\s[a-zA-Z']+)?)?$
Regex explaind:
^ # beginning of line
[a-zA-Z'-]+ # first matching word
(?: # start of second-matching word
\s[a-zA-Z'-]+ # space followed by matching word
(?: # start of third-matching word
\s[a-zA-Z']+ # space followed by matching word
)? # third-matching word is optional
)? # second-matching word is optional
$ # end of line
Alternatively, you can try the following regex:
^([a-zA-Z'-]+(?:\s[a-zA-Z'-]+){0,2})$
This will match 1 through 3 words, or "cities", in a given line with the ability to adjust the range of words without having to further-duplicate the matching set for each new word.
Regex explained:
^( # start of line & matching group
[a-zA-Z'-]+ # required first matching word
(?: # start a non-matching group (required to "match", but not returned as an individual group)
\s # sub-group required to start with a space
[a-zA-Z'-]+ # sub-group matching word
){0,2} # sub-group can match 0 -> 2 times
)$ # end of matching group & line
So, if you want to add the ability to match more than 3 words, you can change the 2 in the {0,2} range above to be the number of words you want to match minus 1 (i.e. if you want to match 4 words, you'll set it to {0,3}).