How do I search and replace patterns like [4:5] [23:1] [1:22] etc in my file using search and replace and regex . In vim when I just do a search /[d\+\:\d\+] the pattern is highlighted but when I use %s/[d\+\:\d\+]/ /gc it says trailing characters.
though you answered your own question, I don't think that answer is correct.
your answer: /[d\+\:\d\+]
this won't match [23:1] pattern, you have to escape the [ also, the d\+ is not correct, it should be \d\+
this will replace all those matches with space: (this will replace the [ and ] as well)
:%s/\[\d\+:\d\+]/ /g
ok I figured it out this is the way
:%s/\[\(\d\+\:\d\+\)\]/ /gc
Related
My sample text would be something like this list
alpha123_4rf_Joe
45beta_Frank
Red5Great_Sam_Fun
and I would like to be left with (with a notepad++ regex find and replace)
alpha123_4rf
45beta
Red5Great_Sam
I am just looking for the Regex as I understand notepad++ :-)
Try this one:
_[^_]+$
Replace with an empty string.
To explain it:
_ match a single _ character
[^_]+ match any character that is not a _ character, one or more times (+)
$ match the end of the string
You can use:
/^(.*)(_.*?)$/\1/
Demo
If you are attempting to do this for PromQL label_replace to use for grouping as I was, this will be helpful to you:
[^.*]+_(.*?)$
I know this post is about notepad++ - however this post kept coming up in my searches and I found that the given solutions did not work with PromQLs variant of regex. Adding this answer for future folks landing here for similar reasons.
I want to use Notepadd++ replace option with regular expression to accomplish this:
From: IntegrationName
To: Integration_Name
How can do this ?
My RegEx to search is: .[A-Z]
this finds: "oN"
But I don't know what to put in the replace box so it will only add "_" between the "o" and the "N"...
Another solution using lookaround assertions would be:
(?<=[a-z])(?=[A-Z])
and replace with
_
Note: The "Match case" option needs to be active, otherwise Notepad++ will find a match between every two letters.
This regex will find every position where a lowercase is on the left and an uppercase is on the right.
You can make use of capture groups. If I have to take your current attempt and edit it as little as possible, you would get:
(.)([A-Z])
This will store the match of . into $1 and the uppercase letter in $2, thus, you can use the following in the replace entry box:
$1_$2
I know you've accepted an answer, but when I ran it, I got From: _Integration_Name
Here's my idea;
(:\s)(.{1})([a-z]*)([A-Z]{1})
And use the following replace
$1$2$3_$4
I finaly did it like this:
Find: ([a-z])([A-Z])
Replace with: $1_$2
I have the following SDDL:
O:BAG:BAD:(A;;CCDCLCSWRP;;;BA)(A;;CCDCSW;;;WD)(A;;CCDCLCSWRP;;;S-1-5-32-562)(A;;CCDCLCSWRP;;;LU)(A;;CCLCRP;;;S-1-5-21-4217728705-3687557540-3107027809-1003)
Unfortunately I keep getting this:
(A;;CCDCLCSWRP;;;BA)(A;;CCDCSW;;;WD)
And what I want is just (A;;CCDCSW;;;WD).
My regex is: (\(A;.+;WD\)) : find "(A;" some characters ending in ";WD)"
I've tried making the match lazy and I've tried excluding the ")(" pair of characters based on a search of the stackoverflow regex tag looking for examples where others have answered similar questions.
I'm really confused why the exclusion of the parens isn't working:
(\(A;.+[^\(\)]*.+;WD\)) : find "(A;" followed by some characters where none of them are ")('' followed by other characters ending in ";WD)"
And this was my guess at using negative look around:
(\(A;.+^((?!\)\().).+;WD\))
which didn't match anything.
I'm also doing this in PowerShell v3.0 with the following code:
$RegExPattern = [regex]"(\($ACE_Type;.*;$ACE_SID\))+?"
if ($SDDL -match $RegExPattern) {
$MatchingACE = $Matches[0]
Where in this instance $ACE_Type = "A" and $ACE_SID = "WD".
You almost had the solution with your second regex pattern. The problem was that you included too many . wildcards. This should be all you need:
A;[^()]+;WD
And of course if you just want to capture the string in between A; and ;WD:
A;([^()]+);WD
Then just replace with \1.
I simplified this a lot and then added lookarounds so that you only matched the intended string (in between A;...;WD). This looks behind for A;, then matches 1+ non-parenthesis characters, while looking ahead for ;WD.
(?<=A;)[^()]+(?=;WD)
Regex101
I have to make changes to URL's in a couple of notepad files. I was hoping if this could be done using regular expressions.
The URL's are in the following structure,
/web/20120730114452im_/hxxp://mysite1.com
/web/20120730114453im_/hxxp://mysite2.com
/web/20120730114454im_/hxxp://mysite3.com
/web/20120730114454im_/hxxp://mysite4.com
I have to remove the part before the hxxp so what remains after the search and replace is,
hxxp://mysite1.com
hxxp://mysite2.com
hxxp://mysite3.com
hxxp://mysite4.com
What is the regular expression I need to use to get the desired result ?
Thanks for your help.
Okay, as per your confirmation, a proper regex that won't match too much would be this:
/web/[0-9]+im_/
Where [0-9]+ will match any amount of numbers.
regex101 demo.
Don't forget to check the 'regular expression' checkbox in the Find/Replace dialog box.
USE THIS,
FIND: [ a-z 0-9 _ / ]+/hxxp
REPLACE: hxxp
Need some help in Notepad++
Example how it looks at the moment
http://www.test.com/doc/rat.rar">rat.rar
http://www.test.com/down/ung.rar">ung.rar
http://www.test.com/read/add.rar">add.rar
......
How I want it (just remove after ">....rar)
http://www.test.com/doc/rat.rar
http://www.test.com/down/ung.rar
http://www.test.com/read/add.rar
Its a list about 1000 lines. So help would be nice
Use the following expression:
">[^.]+\.rar
Explanation:
"> # literal `"` followed by literal `>`
[^.]+ # any character that is not a `.`, repeated at least once
\. # literal `.` character
rar # literal string `rar`
Note: a couple of other answers pointed out that just ">.* will work. This is true, because Notepad++ doesn't appear to support multi-line regular expressions, even with [\s\S]+. Either way will work so it's personal preference. The regex I gave in this answer is very verbose and would reduce the likelihood of false positives. ">.*, on the other hand, is shorter.
In regexp mode , replace pattern ">.* with empty string.
">.*
Search for this and replace with nothing.
Your search string should be ">.+\.rar, and you can just blank out the replace box. This should do the job.
Also, check that you've got regex selected at the bottom of the replace box ;)
If you put this in find ".* and nothing in replace, that should do what you're looking for.
Remember to check that you've got regex selected at the bottom of the replace box.
Flick the "regular expression" radio button and then use this for your FIND:
">[a-z]+\.[a-z]+
Then just put empty space for your REPLACE and replace all.
Use -
Find What : (.*)">(.*)
Replace With : \1
And check Regular expression option at the bottom of the dialog.