I have a regex edge case that I am unable to solve. I need to grep to remove the leading period (if it exists) and the text following the last period (if it exists) from a string.
That is, given a vector:
x <- c("abc.txt", "abc.com.plist", ".abc.com")
I'd like to get the output:
[1] "abc" "abc.com" "abc"
The first two cases are solved already I obtained help in this related question. However not for the third case with leading .
I am sure it is trivial, but i'm not making the connections.
This regex does what you want:
^\.+|\.[^.]*$
Replace its matches with the empty string.
In R:
gsub("^\\.+|\\.[^.]*$", "", subject, perl=TRUE);
Explanation:
^ # Anchor the match to the start of the string
\.+ # and match one or more dots
| # OR
\. # Match a dot
[^.]* # plus any characters except dots
$ # anchored to the end of the string.
Related
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:
$pattern = '~^
(*positive_lookahead: .{0,20} (?<line_end> .* ) )
.*? \K \S+ V \b
(*positive_lookahead: .*? \g{line_end} ) ~xm';
(The alternative syntax (*positive_lookahead: ...) is available since PHP 7.3)
You can find your pattern after X chars and skip the whole string, else, match your pattern. So, if X=25:
^.{25,}\S+V.*(*SKIP)(*F)|\S+V\s*
See the regex demo. Details:
^.{25,}\S+V.*(*SKIP)(*F) - start of string, 25 or more chars other than line break chars, as many as possible, then one or more non-whitespaces and V, and then the rest of the string, the match is failed and skipped
| - or
\S+V\s* - match one or more non-whitespaces, V and zero or more whitespace chars.
Any V ending in the first 25 positions
^.{1,24}V\s
See regex
Any word ending in V in the first 25 positions
^.{1,23}[A-Z]V\s
I have a text file with the following text:
andal-4.1.0.jar
besc_2.1.0-beta
prov-3.0.jar
add4lib-1.0.jar
com_lab_2.0.jar
astrix
lis-2_0_1.jar
Is there any way i can split the name and the version using regex. I want to use the results to make two columns 'Name' and 'Version' in excel.
So i want the results from regex to look like
andal 4.1.0.jar
besc 2.1.0-beta
prov 3.0.jar
add4lib 1.0.jar
com_lab 2.0.jar
astrix
lis 2_0_1.jar
So far I have used ^(?:.*-(?=\d)|\D+) to get the Version and -\d.*$ to get the Name separately. The problem with this is that when i do it for a large text file, the results from the two regex are not in the same order. So is there any way to get the results in the way I have mentioned above?
Ctrl+H
Find what: ^(.+?)[-_](\d.*)$
Replace with: $1\t$2
check Wrap around
check Regular expression
UNCHECK . matches newline
Replace all
Explanation:
^ # beginning of line
(.+?) # group 1, 1 or more any character but newline, not greedy
[-_] # a dash or underscore
(\d.*) # group 2, a digit then 0 or more any character but newline
$ # end of line
Replacement:
$1 # content of group 1
\t # a tabulation, you may replace with what you want
$2 # content of group 2
Result for given example:
andal 4.1.0.jar
besc 2.1.0-beta
prov 3.0.jar
add4lib 1.0.jar
com_lab 2.0.jar
astrix
lis 2_0_1.jar
Not quite sure what you meant for the problem in large file, and I believe the two regex you showed are doing opposite as what you said: first one should get you the name and second one should give you version.
Anyway, here is the assumption I have to guess what may make sense to you:
"Name" may follow by - or _, followed by version string.
"Version" string is something preceded by - or _, with some digit, followed by a dot or underscore, followed by some digit, and then any string.
If these assumption make sense, you may use
^(.+?)(?:[-_](\d+[._]\d+.*))?$
as your regex. Group 1 is will be the name, Group 2 will be the Version.
Demo in regex101: https://regex101.com/r/RnwMaw/3
Explanation of regex
^ start of line
(.+?) "Name" part, using reluctant match of
at least 1 character
(?: )? Optional group of "Version String", which
consists of:
[-_] - or _
( ) Followed by the "Version" , which is
\d+ at least 1 digit,
[._] then 1 dot or underscore,
\d+ then at least 1 digit,
.* then any string
$ end of line
I'm looking find a string of unknown length that beings with abc. The strings end is defined by a space, the end of a line, the end of the file, etc.
The string may contain . characters in the middle.
Examples of what I'm trying to find include:
abc.hello.1.test.a
abc.1test.hello.b.maybe
abc.myTest.1.test.maybe
Characters after the first dot must be present, so the following would not match.
abc.
abc
Use this Pattern (abc\.\S+) Demo
( # Capturing Group (1)
abc # "abc"
\. # "."
\S # <not a whitespace character>
+ # (one or more)(greedy)
) # End of Capturing Group (1)
If you really just want abc.{any non empty string} its trivial to do ^abc\..+$ which just finds abc. at the beginning, and then matches 1 or more of anything after that
If you want abc.{any string without a space} its similar, ^abc\.[^ ]+$
the ^ and $ are called anchors, and make sure your regex is matching the whole string, instead of say, efg.abc.hij
Is there a regex that matches a string only when it starts on an odd or an even index? My use case is a hex string in which I want to replace certain "bytes".
Now, when trying to match 20 (space), 20 in "7209" would be matched as well even though it consists of the bytes 72 and 09. I am restricted to the regex implementation of Notepad++ in this case, so I'm not able to check the match index as e.g. in Java.
My sample input looks like:
324F8D8A20561205231920
I set up a testing page here, the regex should only match the first and the last occurence of 20, since the one in the middle starts on an odd index.
You can use the following regex to match 20 at even positions inside a hex string:
20(?=(?:[\da-fA-F]{2})*$)
See demo
I assume the string has no spaces in this case.
In case you have spaces between the values (or any other symbols), this could be an alternative (with $1XX-like replacement string):
((?:.{2})*?)20
See another demo
This seems to work for evens:
rx <- "^(.{2})*(20)"
strings <- c("7209","2079","9720")
grepl(rx,strings) # [1] FALSE TRUE TRUE
Not sure what Notepad++ uses for regex engine - it's been a while since I used it. This works in javascript...
/^(?:..)*?(20)/
...
/^ # start regex
(?: # non capturing group
.. # any character (two times)
)*? # close group, and repeat zero or more times, un-greedily
(20) # capture `20` in group
/ # end regex
I'm using notepad++'s regular expression search function to find all strings in a .txt document that do not contain a specific value (HIJ in the below example), where all strings begin with the same value (ABC in the below example).
How would I go about doing this?
Example
Every String starts with ABC
ABC is never used in a string other than at the beginning,
ABCABC123 would be two strings --"ABC" and "ABC123"
HIJ may appear multiple times in a string
I need to find the strings that do not contain HIJ
Input is one long file with no line breaks, but does contain special characters (*, ^, #, ~, :) and spaces
Example Input:
ABC1234HIJ56ABC7#HIJABC89ABCHIJ0ABE:HIJABC12~34HI456J
Example Input would be viewed as the following strings
ABC1234HIJ56
ABC7#HIJ
ABC89
ABCHIJ0ABE:HIJ
ABC12%34HI456J
The Third and Fifth strings both lack "HIJ" and therefore are included in the output, all others are not included in the output.
Example desired output:
ABC89
ABC12~34HI456J
I am 99% new to RegEx and will be looking more into it in the future, as my job description suddenly changed earlier this week when someone else in the company left suddenly, and therefore I have been doing this manually by searching (ABC|HIJ) and going through the search function's results looking for "ABC" appearing twice in a row. Supposedly the former employee was able to do this in an automated way, but left no documentation.
Any help would be appreciated!
This question is a repeat of a prior question I asked, but I was very very bad at formatting a question and it seems to have sunk beyond noticeable levels.
You can find the items you want with:
ABC(?:[^HA]+|H(?!IJ)|A(?!BC))*+(?=ABC|$)
Note: in this first pattern, you can replace (?=ABC|$) with (?!HIJ)
pattern details:
ABC
(?: # non-capturing group
[^HA]+ # all that is not a H or an A
| # OR
H(?!IJ) # an H not followed by IJ
|
A(?!BC) # an A not followed by BC
)*+ # repeat the group
(?=ABC|$) # followed by "ABC" or the end of the string
Note: if you want to remove all that is not the items you want you can make this search replace:
search: (?:ABC(?:[^HA]+|H(?!IJ)|A(?!BC))*+HIJ.*?(?=ABC|$))+|(?=ABC)
replace: \r\n
you could use this pattern
(ABC(?:(?!HIJ).)*?)(?=ABC|\R)
Demo
( # Capturing Group (1)
ABC # "ABC"
(?: # Non Capturing Group
(?! # Negative Look-Ahead
HIJ # "HIJ"
) # End of Negative Look-Ahead
. # Any character except line break
) # End of Non Capturing Group
*? # (zero or more)(lazy)
) # End of Capturing Group (1)
(?= # Look-Ahead
ABC # "ABC"
| # OR
\R # <line break>
) # End of Look-Ahead
You can use the following expression to match your criterion:
(^ABC(?:(?!HIJ).)*$)
This starts with ABC and looks ahead (negative) for HIJ pattern. The pattern works for the separated strings.
For a single line pattern (as provided in your question), a slight modification of this works (as follows):
(ABC(?:(?!HIJ).)*?)(?=ABC|$)