Often, I run into situations where I'd like to find and replace two sides of a string, and leave the middle intact. Usually, the first part is easy enough to identify, but the second part is too common. Here is an example:
To change the code string ActiveDocument.Sections["SectionName"] to SectionName, it's easy to find the first part, but the latter "] is way too common without its relationship to ActiveDocument.Sections[". Obviously, if SectionName was a static string, this wouldn't matter, I can just find the whole code string, inclusive, to replace.
Is there a way to match both sides, skipping the middle, or can regex only find contiguous parts? Or maybe there's a way of doing what I want by temporarily storing what's found by .*? in an expression?
I'm using UltraEdit for my find/replace operation. I can also use Javascript to execute on the code as a string ie: codestring.replace()
This is going to depend on the tool you use, but in Javascript you can use $1 to return the first capture group.
So you can use the pattern ActiveDocument.Sections\["(.*?)"\] for "find" and $1 for "replace".
I tested this with the javascript regex tool at regular-expressions.info
.*\["(.*?)"\]worked for me in Python
import re
re.match(r'.*\["(.*?)"\]', 'ActiveDocument.Sections["SectionName"]').group(1)
'SectionName'
Based on an online test, it appears to work with Javascript as well.
Related
I'm trying to search and replace some strings in my code with RegEx as follows:
replacing self.word with self.indices['word']
So for example, replace:
self.example
with
self.indices['example']
But it should work for all words, not just 'example'.
So the RegEx needed for this probably doesn't need the actual word in it.
It should actually just replace everything but the word itself.
What I tried is the following:
self.(.*?)
It works for searching all the strings that match 'self.' but when I want to change it I lose the word like 'example' so I can't change it to self.indices['example'].
What you need to do is a substitution using capturing groups. Depending on the language used some of the syntax might be slightly different but in general you would want something like this:
self\.([^\s]+)
replace with:
self.indices['\1']
https://regex101.com/r/R3xFZB/1
We are using the parans to caputre what is after self. and in the replace putting that captured data into \1 which is the first (and in this case) only captured group. For some languages you might need $1 or \\1 for the substitution variable.
As with most regexes this can be done many different ways using look arounds, etc. but I think for somebody new this is easy to read, understand and maintain. The comment #wnull made has a more proper regex leveraging a look behind that should also do the trick :)
I did some searching and found tons of questions about multiple replacements with Regex, but I'm working in EditPadPro and so need a solution that works with the regex syntax of that environment. Hoping someone has some pointers as I haven't been able to work out the solution on my own.
Additional disclaimer: I suck with regex. I mean really... it's bad. Like I barely know wtf I'm doing.So that being said, here is what I need to do and how I'm currently approaching it...
I need to replace two possible values, with their corresponding replacements. My two searches are:
(.*)-sm
(.*)-rad
Currently I run these separately and replace each with simple strings:
sm
rad
Basically I need to lop off anything that comes prior to "sm" so I just detect everything up to and including sm, and then replace it all with that string (and likewise for "rad").
But it seems like there should be a way to do this in a single search/replace operation. I can do the search part fine with:
(.*)-sm|(.*)-rad
But then how to replace each with it's matching value? That's where I'm stuck. I tried:
sm|rad
but alas, that just becomes the literal complete string that is used for replacement.
Jonathan, first off let me congratulate you for using EPP Pro for regex in your text. It's my main text editor, and the main reason I chose it, as a regex lover, is that its support of regex syntax is vastly superior to competing editors. For instance Notepad++ is known for its shoddy support of regular expressions. The reason of course is that EPP's author Jan Goyvaerts is the author of the legendary RegexBuddy.
A picture is worth a thousand words... So here is how I would do your replacement. Just hit the "replace all button". The expression in the regex box assumes that anything before the dash that is not a whitespace character can be stripped, so if this is not what you want, we need to tune it.
Search for:
(.*)-(sm|rad)
Now, when you put something in parenthesis in Regex, those matches are stored in temporary variables. So whatever matched (.*) is stored in \1 and whatever matched (sm|rad) is stored in \2. Therefore, you want to replace with:
\2
Note that the replacement variable may be different depending on what programming language you are using. In Perl, for example, I would have to use $2 instead.
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 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)).
I have an text that consists of information enclosed by a certain pattern.
The only thing I know is the pattern: "${template.start}" and ${template.end}
To keep it simple I will substitute ${template.start} and ${template.end} with "a" in the example.
So one entry in the text would be:
aINFORMATIONHEREa
I do not know how many of these entries are concatenated in the text. So the following is correct too:
aFOOOOOOaaASDADaaASDSDADa
I want to write a regular expression to extract the information enclosed by the "a"s.
My first attempt was to do:
a(.*)a
which works as long as there is only one entry in the text. As soon as there are more than one entries it failes, because of the .* matching everything. So using a(.*)a on aFOOOOOOaaASDADaaASDSDADa results in only one capturing group containing everything between the first and the last character of the text which are "a":
FOOOOOOaaASDADaaASDSDAD
What I want to get is something like
captureGroup(0): aFOOOOOOaaASDADaaASDSDADa
captureGroup(1): FOOOOOO
captureGroup(2): ASDAD
captureGroup(3): ASDSDAD
It would be great to being able to extract each entry out of the text and from each entry the information that is enclosed between the "a"s. By the way I am using the QRegExp class of Qt4.
Any hints? Thanks!
Markus
Multiple variation of this question have been seen before. Various related discussions:
Regex to replace all \n in a String, but no those inside [code] [/code] tag
Using regular expressions how do I find a pattern surrounded by two other patterns without including the surrounding strings?
Use RegExp to match a parenthetical number then increment it
Regex for splitting a string using space when not surrounded by single or double quotes
What regex will match text excluding what lies within HTML tags?
and probably others...
Simply use non-greedy expressions, namely:
a(.*?)a
You need to match something like:
a[^a]*a
You have a couple of working answers already, but I'll add a little gratuitous advice:
Using regular expressions for parsing is a road fraught with danger
Edit: To be less cryptic: for all there power, flexibility and elegance, regular expression are not sufficiently expressive to describe any but the simplest grammars. Ther are adequate for the problem asked here, but are not a suitable replacement for state machine or recursive decent parsers if the input language become more complicated.
SO, choosing to use RE for parsing input streams is a decision that should be made with care and with an eye towards the future.