I need to parse some text that has data enclosed within parentheses at the end of line.
Snarky Spark was at the stage with his team (Jerry Mander/Kodi Player/Bella Bella)
I need to extract the text within parantheses and separated by forward slash in capture groups.
Jerry Mander
Kodi Player
Bella Bella
I have tried the following to split by /
(?:[^\/])+
But not able to split it within parenthesis or as end of line criteria.
Appreciate help
You may use this regex with \G:
(?:\(|(?!^)\G/)([^/)]+)(?=[^()]*\))
RegEx Demo
\G asserts position at the end of the previous match or the start of the string for the first match.
Related
I have a pipe-delimited csv. Each row should have just three pipes to separate the columns. I need to find any lines that do NOT have three pipes - more OR less should match.
I'm learning regex and I came up with this (kind of hacked together finding parts I thought would work...)
^(?:[^|\r\n]*\|){3,}.*$
However, it's just matching all rows, regardless of the number of pipes in the row.
What's the correct syntax for what I want to do?
[UPDATE]
As #anubhava pointed out, I should provide an example.
This is example data in my file:
John Doe|1hgds234|Some comment|
Mary Jane|5df678|This column is the end of this record|Harry Jones|3456|Harry's record should be on the next line|
Sue Anderson|037dsf533|Another comment|
Harry Jones' record should start on a new line, starting at "Harry". Each line ends in a pipe and CRLF.
So I need a find/replace with a regex that would match on that second line and put a CRLF after the third pipe in the second line.
Assuming you don't have escaped | or | inside quoted cell value, uou can match using this regex:
^((?:[^|\n]*\|){3})(?![\r\n])
And replace this with:
$1\n
RegEx Demo
RegEx Details:
^: Start
(: Start capture group #1
(?:[^|\n]*\|){3}:
): End capture group
(?![\r\n]): Negative lookahead to assert that we don't have \r or \n ahead of the current position
The natural thing would be to put a check for 2 pipes plus some data in a positive lookbehind, but Notepad++ doesn't do variable length lookbehind. Instead we can put the lead-up into a capture group and include that capture group in the result.
^(((([^|]*?)|("[^"]*?"))\|){2}([^|]*?|"[^"]*?"))\|(?!$)
This allows for quoting between the pipes. Your replacement string should be $1\n to restore what is captured in group 1. I took the liberty of allowing a naked pipe character at line end using a negative lookahead.
Try this short code too, working as expected,
Find what: ^(.*?\K\|){3}(?=.)
Replace with: |\n
i have a source code that i need to capture. the whole file is of one line but i am not able to capture the data that i require.
allow=ok&secret=4326dwsaddsafsd286435dsfs754
now i need to capture this data 4326dwsaddsafsd286435dsfs754 which changes everytime. it contains mixed a-z and 0-9, total lenght 40 letters
i tried using Left and Right selectors by using "secret=" on left but since the source ends at the end of the value, i dont have any thing to put in right selector.
so i need to know how can i capture this data? is there any regex cmd that can let me.?
thanks
Try this:
\&secret=(.*)
and then capture the first group with $1
Use this regex:
[a-z0-9]*$
It searches for the longest sequence of alpha-numeric characters ([a-z0-9]*) at the end of the string ($).
Test here.
I have the following string and would like to extract the first few characters until the end of the word or until "Response"
<ns2:GetJobStatus
<ns10:JobIDResponse
<ns2:JobStatusResponse
<ns3:GetJobId
I would like the regular expression such that I could extract either GetJobStatus and GetJobID from all the above lines. I would like to drop "Response" from the result, such that I would get 2 of each in the above example. This is in splunk so I can't use awk or sed or any other unix /linux commands.
Here's what I have done so far
<ns\d+:(?P<ws_name>.+?)(?:Response)
with the above I am able to extract only where there is "Response"
With lookbehind and lookahead, you should be able to get the result you want with the pattern
(?<=:)(\w+?)(?=Response|\b|$)
You would be interested in the capture group (\w+?) because it'll come after the ":" character and be before the word "Response". The "\b|$" sets a word boundary or end of line.
Tested at Regex101
You're off to a good start. What you need to find after your ws_name group is either the word Response or a word boundary. Therefore, all you need to do is add |\b in your non-capturing group:
<ns\d+:(?P<ws_name>.+?)(?:Response|\b)
Here's a demo.
References:
Alternation in Regular Expressions.
I need a regex matcher to find the pattern for a list consisting of a bunch of records
all of which end with a comma.
I want to, at the first occurrence of the comma insert beginning and end h1 tags.
I tried using (.*),
This should capture everything on a line up until and including the comma:
[^,]*?,
You can use this regex:
^([^,]*),
This will locate the string before the first comma in a line. There is also capturing group that captures the text before the first comma for reference in replacement.
Try using either (.*?), or ([^,]+),. The former is preferred, but Notepad++ may not support it.
I am looking for a regular expression pattern that is able to handle the following problem:
(to make someone) happy (adj.)
I only want to get the word "happy" and the regular expression pattern should also handle lines if only one part is in brackets e.g.:
(to make someone) happy
happy (adj.)
I've tried the following: "\s*\(.*\)"
But I am somehow wrong with my idea!
This one will get you the right word in the first capturing group in all three options:
(?:\([^)]*\)\s*)?(\w+)(?:\s*\([^)]*\))?
You can adjust and be more permissive in case you'd like to get a couple of words or to allow special characters:
(?:\([^)]*\)\s*)?([^()\n]+)(?:\s*\([^)]*\))?
A regex for finding the text between two parenthesized groups is
/(?:^|\([^)]*\))([^(]*)(?:$|\([^)]*\))/m
The breakdown is a follows:
Start with some text in parentheses or the beginning of a line: (?:^|\([^)]*\)). This matches from an open paren to the first closed paren
Then match the text outside of the parentheses, and put it in a group ([^(]*). This matches up to the next open paren.
Then match more text in parentheses or the end of a line: (?:$|\([^)]*\))
I used multiline mode (m) so that ^ and $ would match line breaks as well as the start and end of the string
Try regex (?:^|\))\s*([^\(\)]+?)\s*(?:\(|$)