I have a file which will contain contents similar to the following example:
?[A7]DA<DA-SG
'[G7]G%SD\$DF
#[27]F:./4FFF
?[P9]W3_2SS_F
'[90]GA\\WTER
Each line ends in \r\n.
From this particular file, I need to replace the F:./4FFF part of the line #[27]F:./4FFF.
So far to start, I have this pattern in order to try and capture the part I need to replace:
\#\[27\]([\w\W]*)\r\n
The problem is that between the closing ] and the \r\n, could be any alphanumeric character or symbol.
I think the problem lies in the capturing group??? What is the correct pattern for this; I will be doing this in VBA.
You might be trying to design an expression, that would somewhat look like:
(?<=^#\[\d{2}\])\S*
DEMO 1
Or maybe just:
^(#\[\d+\])\S*
DEMO 2
Use the multiline option with these (regex.MULTILINE or something)
Two ways to do it
^#\[27\](.*)
or
^#\[27\]([^\r\n]*)
The thing is that \r\n is not needed to stop the match on the line.
It goes to the end without matching them.
This is advantageous if the line is the last one in the file.
Related
Is it possible to use a regex to generate matches until a pattern is broken?
https://regex101.com/r/bRQkWM/1
(?m)(?=.*?\*)(\d+)|\*\w*.*$
In this instance, capture the digits at the start of the line, plus the rest of the line provided the line begins with a *.
If the line does not begin with a *, do not match digits or rest of line.
Thank you in advance!
The solution should be (link):
(?m)\G(\d+)\s+\*(\w*.*)(?:[\n\r]+|$)
However... The example you provided has broken pattern right in its first line, as there is no * in such line. That leads me to a conclusion that you wish to ignore all lines before the fist match. If that is your desired specification, then the solution should be (link):
(?m)\A(?:\d+\s+[^*]\w*.*$[\n\r]*)*|\G(\d+)\s+\*(\w*.*)(?:[\n\r]+|$)
This extended regex pattern will work even if there is no broken pattern before the first match.
Please keep in mind that the first match of this solution has to be ignored, as it contains those ignored lines before the first match, or it is empty if there are no lines needed to be ignored.
The key of the above solution(s) is a use of \G, the anchor that matches at the position where the previous match ended.
When one tries to remove some characters from the start of a line and the anchored pattern can be found again after the first replace, it will be removed again.
For a very simple example given the input 012345, search pattern ^. and empty replacement, Notepad++ will remove the whole line when using replace all. This is most likely due to the case, that the cursor is still at the start of the line after the first replace and thus matches the ^ anchor again.
How can one ensure that only the actual first character is removed (in my case the expected output would be 12345)?
You can see my workaround in my answer, but maybe there is another nice trick to achieve it.
One can match the rest of the line, capture the match into a group and then use this group as replacement. The pattern in the question could be adjusted to ^.(.*) and be replaced by $1.
This will force the cursor to move forward in the string, so the ^ anchor can't match again.
Another workaround could be finding:
^.(.)?
and replacing it with:
\1
I'm sure this is a subject of a bug report but couldn't find it as of now. In N++:
Anchors are buggy
By Replace All functionality, replacements are supposed to not be a subject to re-matching. But they are, when replacement strings are invisible / zero-length characters.
Take care of them.
I have a latex file in which I want to get rid of the last \\ before a \end{quoting}.
The section of the file I'm working on looks similar to this:
\myverse{some text \\
some more text \\}%
%
\myverse{again some text \\
this is my last line \\}%
\footnote{possibly some footnotes here}%
%
\end{quoting}
over several hundred lines, covering maybe 50 quoting environments.
I tried with :%s/\\\\}%\(\_.\{-}\)\\end{quoting}/}%\1\\end{quoting}/gc but unfortunately the non-greedy quantifier \{-} is still too greedy.
It catches starting from the second line of my example until the end of the quoting environment, I guess the greedy quantifier would catch up to the last \end{quoting} in the file. Is there any possibility of doing this with search and replace, or should I write a macro for this?
EDIT: my expected output would look something like this:
this is my last line }%
\footnote{possibly some footnotes here}%
%
\end{quoting}
(I should add that I've by now solved the task by writing a small macro, still I'm curious if it could also be done by search and replace.)
I think you're trying to match from the last occurrence of \\}% prior to end{quoting}, up to the end{quoting}, in which case you don't really want any character (\_.), you want "any character that isn't \\}%" (yes I know that's not a single character, but that's basically it).
So, simply (ha!) change your pattern to use \%(\%(\\\\}%\)\#!\_.\)\{-} instead of \_.\{-}; this means that the pattern cannot contain multiple \\}% sequences, thus achieving your aims (as far as I can determine them).
This uses a negative zero-width look-ahead pattern \#! to ensure that the next match for any character, is limited to not match the specific text we want to avoid (but other than that, anything else still matches). See :help /zero-width for more of these.
I.e. your final command would be:
:%s/\\\\}%\(\%(\%(\\\\}%\)\#!\_.\)\{-}\)\\end{quoting}/}%\1\\end{quoting}/g
(I note your "expected" output does not contain the first few lines for some reason, were they just omitted or was the command supposed to remove them?)
You’re on the right track using the non-greedy multi. The Vim help files
state that,
"{-}" is the same as "*" but uses the shortest match first algorithm.
However, the very next line warns of the issue that you have encountered.
BUT: A match that starts earlier is preferred over a shorter match: "a{-}b" matches "aaab" in "xaaab".
To the best of my knowledge, your best solution would be to use the macro.
I’m working with a text file with 200.000+ lines in Notepad++. Each line has only one word. I need to strip out and remove all words which only contains one letter (e.g.: I) and words which contains only two letters (e.g.: as).
I thought I could just pas in regular regex like this [a-zA-Z]{1,2} but I does not recognize anything (I’m trying to Mark them).
I’ve done manual search and I know that there do exists words of that length so therefor can it only be my regex code that’s wrong. Anyone knows how to do this in Notepad++ ???
Cheers,
- Mestika
If you want to remove only the words but leave the lines empty, this works:
^[a-zA-Z]{1,2}$
Replace this with an empty string. ^ and $ are anchors for the beginning and the end of a line (because Notepad++'s regexes work in multi-line mode).
If you want to remove the lines completely, search for this:
^[a-zA-Z]{1,2}\r\n
And replace with an empty string. However, this won't work before Notepad++ 6, so make sure yours is up-to-date.
Note that you will have to replace \r\n with the specific line-endings of your file!
As Tim Pietzker suggested, a platform independent solution that also removes empty lines would be:
^[a-zA-Z]{1,2}[\r\n]+
A platform-independent solution that does not remove empty lines but only those with one or two letters would be:
^[a-zA-Z]{1,2}(\r\n?|\n)
I don't use Notepad++ but my guess is it could be because you have too many matches - try including word boundaries (your exp will match every set of 2 letters)
\b[a-zA-Z]{1,2}\b
The regex you specified should find 1-or-2 characters (even in Notepad++'s Find-dialog), but not in the way you'd think. You want to have the regex make sure it starts at the beginning of the line and ends at the end with ^ and $, respecitevely:
^[a-zA-Z]{1,2}$
Notepad++ version 6.0 introduced the PCRE engine, so if this doesn't work in your current version try updating to the most recent.
You seem to use the version of Notepad++ that doesn't support explicit quantifiers: that's why there's no match at all (as { and } are treated as literals, not special symbols).
The solution is to use their somewhat more lengthy replacement:
\w\w?
... but that's only part of the story, as this regex will match any symbol, and not just short words. To do that, you need something like this:
^\w\w?$
I've the following RegExp to find the URIs listed above:
"^w{3}\.[\S\-\n|\S]+[^\s.!?,():]+$"
URLs to find:
www.example.org
www.example-example.org
www.example-example.org/product
You'll find it at www.example-
example.org/product.
www.example.org
You'll find it there.
Number 1, 2 and 3 will be found, but 4. delivers "www.example-" as URI.
When there is no point at the end of 4. it would deliver it correct.
EDIT: With deleting ^ and $ only number 5 is not working.
Does anyone can help here?
Your pattern
^w{3}\.[\S\-\n|\S]+[^\s.!?,():]+$
can be simplified to
^w{3}\.[\S\n]+[^\s.!?,():]$
[\S\-\n|\S] this is a character class, no OR possible, no repetition needed, - is included in \S. So [\S\n] is doing the same.
[^\s.!?,():]+ because you match every non whitespace with the expression before this one, here the + is not needed. I assume you just want your pattern not to end with one of the characters from the class.
See your pattern on Regexr (I added \r to your first class, because the line breaks there needs it)
This is a very useful tool to test regexes
I think your problem is that you want to allow line breaks in the link. How do you want to handle this? How do you want to distinguish when the line ends with a link if the word in the next line is just a word or part of the link. I think this is not possible!
The problem is the '^\s' in the second squared bracketed part. Depending on your programming language, '\s' might match the new line. So, you are telling it to match anything that is not a whitespace and it finds a whitespace (new line).
However, this should only be one of your issues. Your regex uses the '^' and '$' characters which mean start and end of line respectively. Try this URL example:
hello from www.example.org
Did it match? I think it will not.