I have already found how to do multiple replacements, bu replacing
(from1)|(from2).....
with
(?1to1)(?2to2)
For example, if I have:
hello all! I think saying hello to all is a nice way to introduce oneself.
and I replace
(hello)|(all)
with
(?1greetings)(?2everyone)
I get
greetings everyone! I think saying greetings to everyone is a nice way to introduce oneself.
Now, I want to add a string at the very beginning and end of file - not each line. So, in that case, my desired result is:
StartOfAllgreetings everyone! I think saying greetings to everyone is a nice way to introduce oneself.EndOfAll
Can you help me with this? Things that I have tried unsuccesfully include using $,\z,\Z to identify the end of line, and using branch reset groups like this (?|(hello)|(all))*
Use
Find What: (^)(?<!(?s:.))|(hello)|(all)|($)(?!(?s:.))
Or with . matches newline ON: (^)(?<!.)|(hello)|(all)|($)(?!.)
Replace with: (?1StartOfAll)(?2greetings)(?3everyone)(?4EndOfAll)
NOTE: In order to also handle the end of file match when another alternative also matches at the end of the file, you need to add optional groups and handle them in the replacement pattern, too:
Find What: (?s)(^)(?<!.)|(hello)(?:($)(?!.))?|(all)(?:($)(?!.))?|($)(?!.)
Replace with: (?1StartOfAll)(?2greetings)(?3EndOfAll)(?4everyone)(?5EndOfAll)(?6EndOfAll)
Now, the (?:($)(?!.))? optional non-capturing groups ensure an additional capture for end of file positions, and that is why there are additional (?nEndOfAll) in the replacement pattern.
Details
The (^)(?<!(?s:.))|(hello)|(all)|($)(?!(?s:.)) has four alternatives, the ones that you are interested are
(^)(?<!(?s:.)) - The first alternative and the start of file is matched (and captured into Group 1) with ^ that is not preceded with any char (ensured with a negative lookbehind (?<!.) - the inline modifier group is added to make sure the regex works regardless of extra regex Notepad++ settings)
($)(?!(?s:.)) - matches (and captures into Group 4) the end of line that is not followed with any char (see the (?!(?s:.)) negative lookahead).
Settings & demo:
Related
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 $1.
This will force the cursor to move forward in the string, so the ^ anchor can't match again.
Another workaround could be finding:
^.(.)?
and replacing it with:
\1
I'm sure this is a subject of a bug report but couldn't find it as of now. In N++:
Anchors are buggy
By Replace All functionality, replacements are supposed to not be a subject to re-matching. But they are, when replacement strings are invisible / zero-length characters.
Take care of them.
I have such txt file:
ххх.prontube.ru
salo.ru
bbb.antichat.ru
yyy.ru
xx.bb.prontube.ru
zzz.com
srfsf.jwbefw.com.ua
Trying to delete all subdomains with such regex:
Find: .+\.((.*?)\.(ru|ua|com\.ua|com|net|info))$
Replace with: \1
Receive:
prontube.ru
salo.ru
antichat.ru
yyy.ru
prontube.ru
zzz.com
com.ua
Why last line becomes com.ua instead of jwbefw.com.ua ?
This works without look around:
Find: [a-zA-Z0-9-.]+\.([a-zA-Z0-9-]+)\.([a-zA-Z0-9-]+)$
Replace: \1\.\2
It finds something with at least 2 periods and only letters, numbers, and dashes following the last two periods; then it replaces it with the last 2 parts. More intuitive, in my opinion.
There's something funny going on with that leading xxx. It doesn't appear to be plain ASCII. For the sake of this question, I'm going to assume that's just something funny with this site and not representative of your real data.
Incorrect
Interestingly, I previously had an incorrect answer here that accumulated a lot of upvotes. So I think I should preserve it:
Find: [a-zA-Z0-9-]+\.([a-zA-Z0-9-]+)\.(.+)$
Replace: \1\.\2
It just finds a host name with at least 2 periods in it, then replaces it with everything after the first dot.
The .+ part is matching as much as possible. Try using .+? instead, and it will capture the least possible, allowing the com.ua option to match.
.+?\.([\w-]*?\.(?:ru|ua|com\.ua|com|net|info))$
This answer still uses the specific domain names that the original question was looking at. As some TLD (top level domains) have a period in them, and you could theoretically have a list including multiple subdomains, whitelisting the TLD in the regex is a good idea if it works with your data set. Both current answers (from 2013) will not handle the difference between "xx.bb.prontube.ru" and "srfsf.jwbefw.com.ua" correctly.
Here is a quick explanation of why this psnig's original regex isn't working as intended:
The + is greedy.
.+ will zip all the way to the right at the end of the line capturing everything,
then work its way backwards (to the left) looking for a match from here:
(ru|ua|com\.ua|com|net|info)
With srfsf.jwbefw.com.ua the regex engine will first fail to match a,
then it will move the token one place to the left to look at "ua"
At that point, ua from the regex (the second option) is a match.
The engine will not keep looking to find "com.ua" because ".ua" met that requirement.
Niet the Dark Absol's answer tells the regex to be "lazy"
.+? will match any character (at least one) and then try to find the next part of the regex. If that fails, it will advance the token, .+ matching one more character and then evaluating the rest of the regex again.
The .+? will eventually consume: srfsf.jwbefw before matching the period, and then matching com.ua.
But the implimentation of ? also creates issues.
Adding in the question mark makes that first .+ lazy, but then causes group1 to match bb.prontube.ru instead of prontube.ru
This is because that first period after the bb will match, then inside group 1 (.*?) will match bb.prontube. before \.(ru|ua|com\.ua|com|net|info))$ matches .ru
To avoid this, change that third group from (.*?) to ([\w-]*?) so it won't capture . only letters and numbers, or a dash.
resulting regex:
.+?\.(([\w-])*?\.(ru|ua|com\.ua|com|net|info))$
Note that you don't need to capture any groups other than the first. Adding ?: makes the TLD options non-capturing.
last change:
.+?\.([\w-]*?\.(?:ru|ua|com\.ua|com|net|info))$
Search what: .+?\.(\w+\.(?:ru|com|com\.au))
Replace with: $1
Look in the picture above, what regex capture referring
It's color the way you will not need a regex explaination anymore ....
I'm creating some reports in Google Analytics.
I am trying to write a RegEx that will match
www.website.com/
www.website.com/?_string_begins_with_question_mark
But will not match
www.website.com/string_doesnt_begin_with_question_mark
Using Reggy (with POSIX Extended), I tried to create an optional group that would match a string beginning with a question mark, followed by any number of characters. I thought
(\?.+)?
would do the trick, but it ignores the question mark requirement, and matches any string.
I tried some variations:
www.website.com/(\?(.+))?
www.website.com/(\?.+)?
www.website.com/(?.+)?
Et cetera.
Any help is appreciated - Sorry if this has already been asked! I'm new to RegEx.
Thank you!
Your regexp
www.website.com/(\?.+)?
will still match anything that contains www.website.com/, no matter what comes after the slash. Have you tried appending a $ (end-of-input marker)?
www\.website\.com/(?:\?.+)?$
(Escape the dots for more precision; the ?: is just a way of indicating that the group is of no special meaning and does not have to be remembered -- if you omit this, you can access the contents of the group by \1, e.g., in a replace operation.)
Try this one
www.website.com/\?(.+)
I would suggest
www.website.com/($|\?.+$)
Either empty or question mark plus some string
I've been looking around and could not make this happen. I am not totally noob.
I need to get text delimited by (including) START and END that doesn't contain START. Basically I can't find a way to negate a whole word without using advanced stuff.
Example string:
abcSTARTabcSTARTabcENDabc
The expected result:
STARTabcEND
Not good:
STARTabcSTARTabcEND
I can't use backward search stuff. I am testing my regex here: www.regextester.com
Thanks for any advice.
Try this
START(?!.*START).*?END
See it here online on Regexr
(?!.*START) is a negative lookahead. It ensures that the word "START" is not following
.*? is a non greedy match of all characters till the next "END". Its needed, because the negative lookahead is just looking ahead and not capturing anything (zero length assertion)
Update:
I thought a bit more, the solution above is matching till the first "END". If this is not wanted (because you are excluding START from the content) then use the greedy version
START(?!.*START).*END
this will match till the last "END".
START(?:(?!START).)*END
will work with any number of START...END pairs. To demonstrate in Python:
>>> import re
>>> a = "abcSTARTdefENDghiSTARTjlkENDopqSTARTrstSTARTuvwENDxyz"
>>> re.findall(r"START(?:(?!START).)*END", a)
['STARTdefEND', 'STARTjlkEND', 'STARTuvwEND']
If you only care for the content between START and END, use this:
(?<=START)(?:(?!START).)*(?=END)
See it here:
>>> re.findall(r"(?<=START)(?:(?!START).)*(?=END)", a)
['def', 'jlk', 'uvw']
The really pedestrian solution would be START(([^S]|S*S[^ST]|ST[^A]|STA[^R]|STAR[^T])*(S(T(AR?)?)?)?)END. Modern regex flavors have negative assertions which do this more elegantly, but I interpret your comment about "backwards search" to perhaps mean you cannot or don't want to use this feature.
Update: Just for completeness, note that the above is greedy with respect to the end delimiter. To only capture the shortest possible string, extend the negation to also cover the end delimiter -- START(([^ES]|E*E[^ENS]|EN[^DS]|S*S[^STE]|ST[^AE]|STA[^RE]|STAR[^TE])*(S(T(AR?)?)?|EN?)?)END. This risks to exceed the torture threshold in most cultures, though.
Bug fix: A previous version of this answer had a bug, in that SSTART could be part of the match (the second S would match [^T], etc). I fixed this but by the addition of S in [^ST] and adding S* before the non-optional S to allow for arbitrary repetitions of S otherwise.
May I suggest a possible improvement on the solution of Tim Pietzcker?
It seems to me that START(?:(?!START).)*?END is better in order to only catch a START immediately followed by an END without any START or END in between. I am using .NET and Tim's solution would match also something like START END END. At least in my personal case this is not wanted.
[EDIT: I have left this post for the information on capture groups but the main solution I gave was not correct.
(?:START)((?:[^S]|S[^T]|ST[^A]|STA[^R]|STAR[^T])*)(?:END)
as pointed out in the comments would not work; I was forgetting that the ignored characters could not be dropped and thus you would need something such as ...|STA(?![^R])| to still allow that character to be part of END, thus failing on something such as STARTSTAEND; so it's clearly a better choice; the following should show the proper way to use the capture groups...]
The answer given using the 'zero-width negative lookahead' operator "?!", with capture groups, is: (?:START)((?!.*START).*)(?:END) which captures the inner text using $1 for the replace. If you want to have the START and END tags captured you could do (START)((?!.*START).*)(END) which gives $1=START $2=text and $3=END or various other permutations by adding/removing ()s or ?:s.
That way if you are using it to do search and replace, you can do, something like BEGIN$1FINISH. So, if you started with:
abcSTARTdefSTARTghiENDjkl
you would get ghi as capture group 1, and replacing with BEGIN$1FINISH would give you the following:
abcSTARTdefBEGINghiFINISHjkl
which would allow you to change your START/END tokens only when paired properly.
Each (x) is a group, but I have put (?:x) for each of the ones except the middle which marks it as a non-capturing group; the only one I left without a ?: was the middle; however, you could also conceivably capture the BEGIN/END tokens as well if you wanted to move them around or what-have-you.
See the Java regex documentation for full details on Java regexes.
I'm trying to append "/index.html" to some folder paths in a list like this:
path/one/
/another/index.html
other/file/index.html
path/number/two
this/is/the/third/path/
path/five
sixth/path/goes/here/
Obviously the text only needs to be added where it does not exist yet. I could achieve some good results with (vim command):
:%s/^\([^.]*\)$/\1\/index.html/
The only problem is that after running this command, some lines like the 1st, 5th and 7th in the previous example end up with duplicated slashes. That's easy to solve too, all I have to do is search for duplicates and replace with a single slashes.
But the question is:
Isn't there a better way to achieve the correct result at once?
I'm a Vim beginner, and not a regex master also. Any tips are really appreciated!
Thanks!
So very close :)
Just add an optional slash to the end of the regex:
\/\?
Then you need to change the rest of the pattern to a non-greedy match so that it ignores a trailing slash. The syntax for a non-greedy match in vim (replacing the *) is:
\{-}
So we end up with:
:%s/^\([^\.]\{-}\)\/\?$/\1\/index.html/
(Doesn't hurt to be safe and escape the period.)
Vim's regex supports the ability to match a bit of text foo if it does or doesn't precedes or follows some other text bar without matching bar, and this is exactly the sort of thing you're looking for. Here you want to match the end of line with an optional /, but only if the / isn't followed by index.html, and then replace it with /index.html. A quick look at Vim's help tells me \#<! is exactly what to use. It tells Vim that the preceding atom must be in the text but not in what's matched. With a little experimentation, I get
:%s;/\?\(index\.html\)\#<!$;/index.html;
I use ; to delimit the parts of the :s command so that I don't have to escape any / in the regex or replacement expression. In this particular situation, it's not a big deal though.
The / is optional, and we say so with \?.
We need to group index.html together because otherwise our special \#<! would only affect the l otherwise.