In notepad ++, I want to select text up to a certain text match, including the match.
The txt file I am working with contains a lot of text with also white characters, returns and some special characters. In this text, there are characters that mark an end. Let's call these stop characters "ZZ." for now.
Using RegEx, I tried to create an expression that finds the next "ZZ." and selects everything before it. This is what it looks like:
+., \c ZZ.\n
But I seem to have gotten something wrong. As it is a similar to this
problem, I tried to use their RegEx with slight modification. Here is a picture so you can figure what I'd like to accomplish:
Find the next stop marker, selext the marker and everything before it.
In the actual file, the stop marker is "გვ."
If I want to use those, maybe I need to change the RegEx even more, as those are no ASCII characters? Like so, as stated in the RegEx Wiki?
\c+ (\x{nnnn}\x{nnnn}.)\n
Not quite sure if the \c works that way. I have seen expressions that use something like (A-Za-z)(0-9) but this is a different alphabet.
To match any text up to and including some pattern, use .*? (to match any zero or more characters, as few as possible) with the . matches newline option ON and add the გვ after it:
Related
This is a follow-up question to what was solved yesterday:
Notepad++ Regex Replace Makeshift Footnotes format With Proper Markdown format
I managed to find a Regex to remove the offending semicolons in the main text area but by only cutting out the text and pasting back the result, which can only be done one by one.
I'm not sure how this can be done, but the expert can tell me.
So I have footnote references in markdown format. Two instances of the same thing:
[^1]:
[^2]:
.
.
.
[^99]:
I might not have 99 in a document but I wanted to show I need to match two digits here again.
As I said, there are two instances of these numbered references in the text. One in the main text pointing to the footnote and the footnote at the end of the document.
What I need is deleting the semi-colons from the main text and leave the
[^3]:
[^15]:
etc.
references at the end intact.
Because the main text references come after a word or at the end of a sentence (ususally before the sentence-ending period), there is never a case a reference would start a sentence (even if they seem to appear there once or twice because of word wrap).
I provided the exact opposite of my needs here:
Click here for Regex101 website link
I put in the exact opposite of what I want because I already knew of the
^
sign to match anything that is at the front of the line.
Now I would like to negate this, if possible, so that I would delete the semi-colons in the main text, not down at the bottom.
Of course, it is likely that my approach is not good and you'll come up with a completely different approach. Especially because there doesn't seem to be a NOT operator in Regex, if I read correctly.
I repeat: the Regex101 example with the match and substitution is exactly the opposite of what I want.
I am not sure if you can play around in the substitution line to get the desired negative effect.
I could have probably asked for removing the first occurence of semi-colons but I thought the important part of tackling the problem is that those items not to be matched are always at the start of the line, not the others.
Thanks for any suggestions
In Notepad++ you might use a negative lookabehind asserting not the start of the string to the left, and use \K to clear the match buffer matching only the colon that should be replaced by an empty string.
(?<!^)\[\^\d{1,2}]\K:
Explanation
(?<!^) Negative lookbehind, assert not the start of the start directly to the left
\[\^ Match [^
\d{1,2} Match 1 or 2 digits
] Match literally
\K Forget what is matched so far
: Match a colon
Regex demo
I've got a practical application for a vim regex where I'd like to remove numbers from the end of file location links. For example, if the developer is sloppy and just adds files and doesn't reuse file locations, you'll end up with something awful like this:
PATH_TO_MY_FILES>
PATH_TO_MY_FILES1>
...
PATH_TO_MY_FILES22>
PATH_TO_MY_FILES_ELSEWHERE>
PATH_TO_MY_FILES_ELSEWHERE1>
...
So all I want to do is to S&R and replace PATH_TO_MY_FILES*\d+ with PATH_TO_MY_FILES* using regex. Obviously I am not doing it quite right, so I was hoping someone here could not spoon feed the answer necessarily, but throw a regex buzzword my way to get me on track.
Here's what I have tried:
:%s\(PATH_TO_MY_FILES\w*\)\(\d+\)>:gc
But this doesn't work, i.e. if I just do a vim search on that, it doesn't find anything. However, if I use this:
:%s\(PATH_TO_MY_FILES\w*\)\(\d\)>:gc
It will match the string, but the grouping is off, as expected. For example, the string PATH_TO_MY_FILES22 will be grouped as (PATH_TO_MY_FILES2)(2), presumably because the \d only matches the 2, and the \w match includes the first 2.
Question 1: Why doesn't \d+ work?
If I go ahead and use the second string (which is wrong), Vim appears to find a match (even though the grouping is wrong), but then does the replacement incorrectly.
For example, given that we know the \d will only match the last number in the string, I would expect PATH_TO_MY_FILES22> to get replaced with PATH_TO_MY_FILES2>. However, instead it replaces it with this:
PATH_TO_MY_FILES2PATH_TO_MY_FILES22>gt
So basically, it looks like it finds PATH_TO_MY_FILES22>, but then replaces only the & with group 1, which is PATH_TO_MY_FILES2.
I tried another regex at Regexr.com to see how it would interpret my grouping, and it looked correct, but maybe a hack around my lack of regex understanding:
(PATH_TO_\D*)(\d*)>
This correctly broke my target string into the PATH part and the entire number, so I was happy. But then when I used this in Vim, it found the match, but still replaced only the &.
Question 2: Why is Vim only replacing the &?
Answer 1:
You need to escape the + or it will be taken literally. For example \d\+ works correctly.
Answer 2:
An unescaped & in the replacement portion of a substitution means "the entire matched text". You need to escape it if you want a literal ampersand.
I'm terrible at regex and need to remove everything from a large portion of text except for a certain variable declaration that occurs numerous times, id like to remove everything except for instances of mc_gross=anyint.
Generally we'd need to use "negative lookarounds" to find everything but a specified string. But these are fairly inefficient (although that's probably of little concern to you in this instance), and lookaround is not supported by all regex engines (not sure about notepad++, and even then probably depends on the version you're using).
If you're interested in learning about that approach, refer to How to negate specific word in regex?
But regardless, since you are using notepad++, I'd recommend selecting your target, then inverting the selection.
This will select each instance, allowing for optional white space either side of the '=' sign.
mc_gross\s*=\s*\d+
The following answer over on super user explains how to use bookmarks in notepad++ to achieve the "inverse selection":
https://superuser.com/questions/290247/how-to-delete-all-line-except-lines-containing-a-word-i-need
Substitute the regex they're using over there, with the one above.
You could do a regular expression replace of ^.*\b(mc_gross\s*=\s*\d+)\b.*$ with \1. That will remove everything other than the wanted text on each line. Note that on lines where the wanted text occurs two or more times, only one occurrence will be retained. In the search the ^.*\b matches from start-of-line to a word boundary before the wanted text; the \b.*$ matches everything from a word boundary after the wanted text until end of line; the round brackets capture the wanted text for the replacement text. If text such as abcmc_gross=13def should be matched and retained as mc_gross=13 then delete the \bs from the search.
To remove unwanted lines do a regular expression search for ^mc_gross\s*=\s*\d+$ from the Mark tab, tick Bookmark line and click Mark all. Then use Menu => Search => Bookmark => Remove unmarked lines.
Find what: [\s\S]*?(mc_gross=\d+|\Z)
Replace with: \1
Position the cursor at the start of the text then Replace All.
Add word boundaries \b around mc_gross=\d+ if you think it's necessary.
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 am having issues doing a string replacement in Notepad++, and need some help.
My file:
LastName,(tab)FirstName[optional]MiddleName
Some times there is data that has a middle name, sometimes not.
Public,JohnQ.
Doe,John
Clinton,WilliamJefferson
would be:
Public(tab)John(tab)Q
Doe(tab)John
Clinton(tab)William(tab)Jefferson
I want to split it out into this:
LastName(tab)FirstName(tab)MiddleName
Thanks for adding the sample input. It helps immensely to have that around. Try this and see if it does what you want.
Find, making sure Match case is checked:
([A-Z][a-z]*),([A-Z][a-z]*)(.*)
Replace with:
\1(tab)\2(tab)\3
Of course, (tab) is actually a tab character that you have to place in the replacement string yourself.
An ugly regex like this works for me on the example you've provided:
(\w+),(\w+?)(([A-Z]\w*\.?)?)\n
replace with
\1\t\2\t\3\n
Note:
This only works if the middle name starts with a letter in the A-Z. You might be able to replace [A-Z] with [[:upper:]] if notepad++ supports it (I don't know).
I need that second bracket around the middle name part because I need to match at least an empty string when there is no middle name.