If I have the following example:
X-FileName: pallen (Non-Privileged).pst
Here is our forecast
Message-ID: <15464986.1075855378456.JavaMail.evans#thyme>
How can I select the text
Here is our forecast
after "X-FileName .... \n" until "Message-ID" execluded?
I read about lookahead and behind and tried this but didn't work:
(?<=X-FileName:(\n)+$).+(?=Message-ID:)
This should do it:
(?:X-FileName:[^\n]+)\n+([^\n]+)\n+(?:Message-ID:) (group #1 is the match)
Demo
Explanation:
(?:X-FileName:[^\n]+) matches X-Filename: followed by any number of characters that aren't newlines, without capturing it (?:).
\n+ matches any number of consecutive newlines.
([^\n]+) matches and captures any number of consecutive characters that aren't newlines.
\n+, again, matches any number of consecutive newlines.
(?:Message-ID:) matches Message-ID: without capturing it (?:).
Edit: as #WiktorStribiżew mentioned though, splitting your text into lines may be an easier/cleaner way to retrieve what you want.
There are two approaches here, and they depend on the broader context. If your expected substring is the second paragraph, just split with \n\n (or \r\n\r\n) and get the second item from the resulting list.
If it is a text inside some larger text, use a regex.
See a Python demo:
import re
s='''X-FileName: pallen (Non-Privileged).pst
Here is our forecast
Message-ID: <15464986.1075855378456.JavaMail.evans#thyme>'''
# Non-regex way for the string in the exact same format
print(s.split('\n\n')[1])
# Regex way to get some substring in a known context
m = re.search(r'X-FileName:.*[\r\n]+(.+)', s)
if m:
print(m.group(1))
The regex means:
X-FileName: - a literal substring
.* - any 0+ chars other than line break chars
[\r\n]+ - 1 or more CR or LF chars
(.+) - Group 1: one or more chars other than line break chars, as many as possible.
See the regex demo.
Related
I want to extract [games, games, things, things] from
the following array.
Today_games
Today_games_freq
Today_things
Today_things_freq
I have tried Today_(\w+)(?=_freq)?
Which will give me the extra "freq"
And some other combinations, but I couldn't figure out how to get just after the first hyphen.
You can use
Today_(\w+?)(?:_freq)?$
See the regex demo. This matches Today_, then captures any one or more word chars (as few as possible) into Group 1 (with (\w+?)), and then (?:_freq)?$ matches an optional occurrence of a _freq substring and asserts the position at the end of string.
Or,
Today_([^\W_]+)
See this regex demo.
Here, Today_ is matched and the ([^\W_]+) pattern captures one or more alphanumeric chars into Group 1 (same as \w+ with _ subtracted from \w).
there are 4 strings as shown below
ABC_FIXED_20220720_VALUEABC.csv
ABC_FIXED_20220720_VALUEABCQUERY_answer.csv
ABC_FIXED_20220720_VALUEDEF.csv
ABC_FIXED_20220720_VALUEDEFQUERY_answer.csv
Two strings are considered as matched based on a matching substring value (VALUEABC, VALUEDEF in the above shown strings). Thus I am looking to match first 2 (having VALUEABC) and then next 2 (having VALUEDEF). The matched strings are identified based on the same value returned for one regex group.
What I tried so far
ABC.*[0-9]{8}_(.*[^QUERY_answer])(?:QUERY_answer)?.csv
This returns regex group-1 (from (.*[^QUERY_answer])) value "VALUEABC" for first 2 strings and "VALUEDEF" for next 2 strings and thus desired matching achieved.
But the problem with above regex is that as soon as the value ends with any of the characters of "QUERY_answer", the regex doesn't match any value for the grouping. For instance, the below 2 strings doesn't match at all as the VALUESTU ends with "U" here :
ABC_FIXED_20220720_VALUESTU.csv
ABC_FIXED_20220720_VALUESTUQUERY_answer.csv
I tried to use Negative Lookahead:
ABC.*[0-9]{8}_(.*(?!QUERY_answer))(?:QUERY_answer)?.csv
but in this case the grouping-1 value is returned as "VALUESTU" for first string and "VALUESTUQUERY_answer" for second string, thus effectively making the 2 strings unmatched.
Any way to achieve the desired matching?
With your shown samples please try following regex.
^ABC_[^_]*_[0-9]+_(.*?)(?:QUERY_answer)?\.csv$
OR to match exact 8 digits try:
^ABC_[^_]*_[0-9]{8}_(.*?)(?:QUERY_answer)?\.csv$
Here is the online demo for above regex.
Explanation: Adding detailed explanation for above regex.
^ABC_[^_]*_ ##Matching from starting of value ABC followed by _ till next occurrence of _.
[0-9]+_ ##Matching continuous occurrences of digits followed by _ here.
(.*?) ##Creating one and only capturing group using lazy match which is opposite of greedy match.
(?:QUERY_answer)? ##In a non-capturing group matching QUERY_answer and keeping it optional.
\.csv$ ##Matching dot literal csv at the end of the value.
You need
ABC.*[0-9]{8}_(.*?)(?:QUERY_answer)?\.csv
See the regex demo.
Note
.*[^QUERY_answer] matches any zero or more chars other than line break chars as many as possible, and then any one char other than Q, U, E, etc., i.e. any char in the negated character class. This is replaced with .*?, to match any zero or more chars other than line break chars as few as possible.
(?:QUERY_answer)? - the group is made non-capturing to reduce grouping complexity.
\.csv - the . is escaped to match a literal dot.
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")
PCRE Regex: Is it possible for Regex to check for a pattern match within only the first X characters of a string, ignoring other parts of the string beyond that point?
My Regex:
I have a Regex:
/\S+V\s*/
This checks the string for non-whitespace characters whoich have a trailing 'V' and then a whitespace character or the end of the string.
This works. For example:
Example A:
SEBSTI FMDE OPORV AWEN STEM students into STEM
// Match found in 'OPORV' (correct)
Example B:
ARKFE SSETE BLMI EDSF BRNT CARFR (name removed) Academy Networking Event
//Match not found (correct).
Re: The capitalised text each letter and the letters placement has a meaning in the source data. This is followed by generic info for humans to read ("Academy Networking Event", etc.)
My Issue:
It can theoretically occur that sometimes there are names that involve roman numerals such as:
Example C:
ARKFE SSETE BLME CARFR Academy IV Networking Event
//Match found (incorrect).
I would like my Regex above to only check the first X characters of the string.
Can this be done in PCRE Regex itself? I can't find any reference to length counting in Regex and I suspect this can't easily be achieved. String lengths are completely arbitary. (We have no control over the source data).
Intention:
/\S+V\s*/{check within first 25 characters only}
ARKFE SSETE BLME CARFR Academy IV Networking Event
^
\- Cut off point. Not found so far so stop.
//Match not found (correct).
Workaround:
The Regex is in PHP and my current solution is to cut the string in PHP, to only check the first X characters, typically the first 20 characters, but I was curious if there was a way of doing this within the Regex without needing to manipulate the string directly in PHP?
$valueSubstring = substr($coreRow['value'],0,20); /* first 20 characters only */
$virtualCount = preg_match_all('/\S+V\s*/',$valueSubstring);
The trick is to capture the end of the line after the first 25 characters in a lookahead and to check if it follows the eventual match of your subpattern:
$pattern = '~^(?=.{0,25}(.*)).*?\K\S+V\b(?=.*\1)~m';
demo
details:
^ # start of the line
(?= # open a lookahead assertion
.{0,25} # the twenty first chararcters
(.*) # capture the end of the line
) # close the lookahead
.*? # consume lazily the characters
\K # the match result starts here
\S+V # your pattern
\b # a word boundary (that matches between a letter and a white-space
# or the end of the string)
(?=.*\1) # check that the end of the line follows with a reference to
# the capture group 1 content.
Note that you can also write the pattern in a more readable way like this:
$pattern = '~^
(*positive_lookahead: .{0,20} (?<line_end> .* ) )
.*? \K \S+ V \b
(*positive_lookahead: .*? \g{line_end} ) ~xm';
(The alternative syntax (*positive_lookahead: ...) is available since PHP 7.3)
You can find your pattern after X chars and skip the whole string, else, match your pattern. So, if X=25:
^.{25,}\S+V.*(*SKIP)(*F)|\S+V\s*
See the regex demo. Details:
^.{25,}\S+V.*(*SKIP)(*F) - start of string, 25 or more chars other than line break chars, as many as possible, then one or more non-whitespaces and V, and then the rest of the string, the match is failed and skipped
| - or
\S+V\s* - match one or more non-whitespaces, V and zero or more whitespace chars.
Any V ending in the first 25 positions
^.{1,24}V\s
See regex
Any word ending in V in the first 25 positions
^.{1,23}[A-Z]V\s
I'm new to regex, and would appreciate some guidance/help.
Currently, I'm looking to write an expression, that derives a certain part of text from the 2nd line of the provided text.
Here is the text:
123 anywhere Avenue
Winnipeg, Manitoba R3E 0L7
Canada
Pharmacy Manager: person person
Pharmacy Licence Holder/Owner: 123456 Manitoba Ltd.
see correct formatting with code here
My goal is to derive the 'Manitoba' string from the second line, however I'd like to make it dynamic rather than writing an expression to always fetch Manitoba as a static. I used the below code to target the second line:
(.*)(?=(\n.*){3}$)
(It matches 3 lines up from the last line, thus targeting the desired line)
I noticed, that within the dataset, that the Province (Manitoba) is always in between two spaces.
Is there any addition I can make to the code, so that the expression only targets the second line, then matches the first string in-between spaces?
Perhaps using a lazy expression with a positive lookaround?
If I target all matches in between spaces, it would take both 'Manitoba' and 'R3E 0L7' which I dont want.
I want it to only match the first piece of text in between spaces on the second line.
Any help is much appreciated :-)
Thanks.
One option could be to match the first line, then capture the second word in the second lines in capturing group 1.
Then match the rest of the second line and assert what follows is 3 times a line.
^.*\r?\n\S+[^\S\r\n]+(\S+).*(?=(?:\r?\n.*){3}$)
In parts:
^ Start of string
.*\r?\n Match the whole lines and a newline
\S+ Match 1+ non whitespace char (the first "word")
[^\S\r\n]+ Match 1+ times a whitespace char except newlines
(\S+) Capture group 1 Match 1+ times a non whitespace char (the second "word')
.* Match the rest of the line
(?= Positive lookahead, assert what follows on the right is
(?:\r?\n.*){3}$ Match 3 times a newline followed by 0+ times any except a newline and assert the end of the string
) Close lookahead
Regex demo
You could also turn the lookahead in to a match instead
^.*\r?\n\S+[^\S\r\n]+(\S+).*(?:\r?\n.*){3}$
Regex demo