I'm trying to replace parts within my code that looks like this
Class.Method<>("SomeKeyHere"]
Notice the square bracket at the ending - that's what I want to replace with the correct bracket.
My RegEx to find it looks like this:
Class\.Method\<\>\("{[^"]+}"\]
This RegEx seems to find the occurences pretty well.
My RegEx I want to use to replace (with the correct bracket at the end) is this:
Class\.Method\<\>\("(\1)"\)
However, VS is finding everything using Quickfind or Quickreplace's Find button but it won't replace it, telling me it hasn't found any occurences
The Problem was, that Visual Studio seems not to understand, that the "Replace with"-Part contained RegEx.
The Regex that worked was:
Class.Method<>(\"\1\")
Simpler, than I thought. Seems like VisualStudio only understands the \1, \2,.. Parts within the "Replace with" field.
Related
I am dealing with a bunch of xml files that contain one-line-comments like this: // Some comment.
I am pretty sure that xml comments look like this: <!-- Some comment -->
I would like to use a regular expression in the Atom editor to find and replace all wrong comment syntax.
according to this question, the comment can be found with (?<=\s)//([^\n\r]*) and replaced with something like <!--$1-->. There must be an error somewhere since clicking replace button leaves the comment as is, instaed of replacing it. Actually I can't even replace it with a simple character.
The find and replace works with a different regex in the "Find" field:
Find: name.*
Replace: baloon
Is there anything I can write in the "Find" and "Replace" field to achieve this transformation?
Atom editor search and replace currently does not support lookbehind constructs, like (?<=\s). To "imitate" it, you may use a capturing group with an alternation between start of string, ^, and a whitespace, \s.
So, you may use
Find What: (^|\s)//([^\n\r]+)
Replace With: $1<!--$2-->
See the regex demo. NOTE \s may match newlines, so you may probably want to use (^|[^\S\r\n])//([^\n\r]+) to avoid matching across line breaks.
If you do not need to check for a whitespace, just remove that first capturing group and use a mere:
Find What: //([^\n\r]+)
Replace With: <!--$1-->
See another 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.
In Notepad++, for example if the search regex is
(?<latex>\$[^\$]*\$)(?=[\x{4e00}-\x{9fa5}])
and the replace is
~\g{latex}~
then the replacement is working properly.
But if the search regex contains look behind expression like
(?<=[\x{4e00}-\x{9fa5}])(?<latex>\$[^\$]*\$)(?=[\x{4e00}-\x{9fa5}])
then replace
~\g{latex}~
doesn't work in Notepad++, why???
Actually, I found out that your named backreference is the actual problem. As per the documentation, you need to use the syntax $+{name} for a named capture reference in the replace. So that one should work:
(?<=[\x{4e00}-\x{9fa5}])(?<latex>\$[^\$]*\$)(?=[\x{4e00}-\x{9fa5}])
And replace with:
~$+{latex}~
So the first regex you had should not be working properly, but replacing with literal ~g<latex>~. Even so, I can't really be sure here since I'm using an older version of N++ and the docs could be out of date.
Though I think that the simplest would be that you don't use the capture group. The below should work fine:
(?<=[\x{4e00}-\x{9fa5}])\$[^$]*\$(?=[\x{4e00}-\x{9fa5}])
And replace with ~$0~.
I got a string like this:
PREFIX-('STRING WITH SPACES TO REPLACE')
and i need this:
PREFIX-('STRING_WITH_SPACES_TO_REPLACE')
I'm using Notepad++ for the Regex Search and Replace, but i'm shure every other Editor capable of regex replacements can do it to.
I'm using:
PREFIX-\('(.*)(\s)(.*)'\)
for search and
PREFIX-('\1_\3')
for replace
but that replaces only one space from the string.
The regex search feature in Notepad++ is very, very weak. The only way I can see to do this in NPP is to manually select the part of the text you want to work on, then do a standard find/replace with the In selection box checked.
Alternatively, you can run the document through an external script, or you can get a better editor. EditPad Pro has the best regex support I've ever seen in an editor. It's not free, but it's worth paying for. In EPP all I had to do was this:
search: ((?:PREFIX-\('|\G)[^\s']+)\s+
replace: $1_
EDIT: \G matches the position where the previous match ended, or the beginning of the input if there was no previous match. In other words, the first time you apply the regex, \G acts like \A. You can prevent that by adding a negative lookahead, like so:
((?:PREFIX-\('|(?!\A)\G)[^\s']+)\s+
If you want to prevent a match at the very beginning of the text no matter what it starts with, you can move the lookahead outside the group:
(?!\A)((?:PREFIX-\('|\G)[^\s']+)\s+
And, just in case you were wondering, a lookbehind will work just as well as a lookahead:
((?:PREFIX-\('|(?<!\A)\G)[^\s']+)\s+
You have to keep matching from the beggining of the string untill you can match no more.
find /(PREFIX-\('[^\s']*)\s([^']*'\))/
replace $1_$2
like: while (/(PREFIX-\('[^\s']*)\s([^']*'\))/$1_$2/) {}
How about using Replace all for about 20 times? Or until you're sure no string contains more spaces
Due to nature of regex, it's not possible to do this in one step by normal regular expression.
But if I be in your place, I do such replaces in several steps:
find such patterns and mark them with special character
(Like replacing STRING WITH SPACES TO REPLACE with #STRING WITH SPACES TO REPLACE#
Replace #([^#\s]*)\s to #\1_ server times.
Remove markers!
I studied a little the regex tool in Notepad++ because I didn't know their possibilities.
I conclude that they aren't powerful enough to do what you want.
Your are obliged to learn and use a programming language having a real regex capability. There are a number of them. Personnaly, I use Python. It would take 1 mn to do what you want with it
You'd have to run the replace several times for each space but this regex will work
/(?<=PREFIX-\(')([^\s]+)\s+/g
Replace with
\1_ or $1_
See it working at http://refiddle.com/10z
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.