gvim regexp for nested parentheses - regex

I have file containing abc(de+fgh(2a+2b))+xyz(). i want to write regexp(preferably vim) to get a pattern like de+fgh(2a+2b) + xyz() .
I tried in gvim regexp But while matching parenthesis, if i use greedy option it will match abc(de+fgh(2a+2b))+xyz() and for non-greedy option it will matching with abc(de+fgh(2a+2b')')+xyz() , how to match with abc(de+fgh(2a+2b)')'+xyz().
Regards
keerthan

I won't do it with regex, assume that your cursor is at BOL, just do:
%di(v0p
you will get desired output.
Translating it into english, take the stuff between first (..) group and concatenate it with whatever after the first (...) group.
You can use :normal cmd or macro to apply the operations on multiple lines.

Watch out! regex cannot handle counting arbitrary numbers of brackets. If you want to do this generally, you might need to write a parser.
That said, if you only need this to work for a specific case:
http://regexr.com/3fne0
in vim this is:
%s/[^()]*(\([^()]*([^()]*)\))\(.*\)/\1\2/g

Related

Regex to insert text BEFORE a line containing a match?

I have a bunch of artists that are named in this fashion:
Killers, The
Treatment, The
Virginmarys, The
I need them to look like
The Killers
The Treatment
The Virginmarys
I'm able to match the lines with , The ((^|\n)(.*, The) is what I've used) but the more advanced syntax is eluding me. I can use regex on the replacement syntax as well (it's for a TextPipe filter so it might as well be for Notepad++ or any other Regex text editor).
You should be able to use the following:
Find: (\S+),\s\S*
Replace: The $1
Or include the The..
Find: (\S+),\s+(\S+)
Replace: $2 $1
Depending on your editor, you may be better off using \1, \2, and so on for capture groups.
Since you need to specifically capture the title before the comma, do so:
(^|\n)(.*), The
And replace it putting the "the" in the right place:
\1The \2
Regular expressions define matches but not substitutions.
How and in which way you can perform substitutions is highly dependant on the application.
Most editors that provide regular expression support work on a line per line basis.
Some of them will allow substitutions such as
s/^(.*Banana)/INSERTED LINE\n\1/
which would then insert the specific pattern before each match. Note that others may not allow newlines in the substitution pattern at all. In VIM, you can input newlines into the command prompt using Ctrl+K Return Return. YMMV.
In Java, you would just first print the insertion text, then print the matching line.

Replacing char in a String with Regular Expression

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

How to search (using regex) for a regex literal in text?

I just stumbled on a case where I had to remove quotes surrounding a specific regex pattern in a file, and the immediate conclusion I came to was to use vim's search and replace util and just escape each special character in the original and replacement patterns.
This worked (after a little tinkering), but it left me wondering if there is a better way to do these sorts of things.
The original regex (quoted): '/^\//' to be replaced with /^\//
And the search/replace pattern I used:
s/'\/\^\\\/\/'/\/\^\\\/\//g
Thanks!
You can use almost any character as the regex delimiter. This will save you from having to escape forward slashes. You can also use groups to extract the regex and avoid re-typing it. For example, try this:
:s#'\(\\^\\//\)'#\1#
I do not know if this will work for your case, because the example you listed and the regex you gave do not match up. (The regex you listed will match '/^\//', not '\^\//'. Mine will match the latter. Adjust as necessary.)
Could you avoid using regex entirely by using a nice simple string search and replace?
Please check whether this works for you - define the line number before this substitute-expression or place the cursor onto it:
:s:'\(.*\)':\1:
I used vim 7.1 for this. Of course, you can visually mark an area before (onto which this expression shall be executed (use "v" or "V" and move the cursor accordingly)).

Extract and use a part of string with a regex in GVIM

I've got a string:
doCall(valA, val.valB);
Using a regex in GVIM I would like to change this to:
valA = doCall(valA, val.valB);
How would I go about doing this? I use %s for basic regex search and replace in GVIM, but this a bit different from my normal usages.
Thanks
You can use this:
%s/\vdoCall\(<(\w*)>,/\1 = doCall(\1,/
\v enables “more magic” in regular expressions – not strictly necessary here but I usually use it to make the expressions simpler. <…> matches word boundaries and the in-between part matches the first parameter and puts it in the first capture group. The replacement uses \1 to access that capture group and insert into the right two places.

Regular expression question

I have some text like this:
dagGeneralCodes$_ctl1$_ctl0
Some text
dagGeneralCodes$_ctl2$_ctl0
Some text
dagGeneralCodes$_ctl3$_ctl0
Some text
dagGeneralCodes$_ctl4$_ctl0
Some text
I want to create a regular expression that extracts the last occurrence of dagGeneralCodes$_ctl[number]$_ctl0 from the text above.
the result should be: dagGeneralCodes$_ctl4$_ctl0
Thanks in advance
Wael
This should do it:
.*(dagGeneralCodes\$_ctl\d\$_ctl0)
The .* at the front is greedy so initially it will grab the entire input string. It will then backtrack until it finds the last occurrence of the text you want.
Alternatively you can just find all the matches and keep the last one, which is what I'd suggest.
Also, specific advice will probably need to be given depending on what language you're doing this in. In Java, for example, you will need to use DOTALL mode to . matches newlines because ordinarily it doesn't. Other languages call this multiline mode. Javascript has a slightly different workaround for this and so on.
You can use:
[\d\D]*(dagGeneralCodes\$_ctl\d+\$_ctl0)
I'm using [\d\D] instead of . to make it match new-line as well. The * is used in a greedy way so that it will consume all but the last occurrence of dagGeneralCodes$_ctl[number]$_ctl0.
I really like using this Regular Expression Cheatsheet; it's free, a single page, and printed, fits on my cube wall.