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)).
Related
In vim I need to search all strings in quotes e.g. 'foo'
Does one see the problem in this regex? E486: Pattern not found \'([^']*)'
:\/'([^']*)'
Regex Tester
First problem is that your use of find is a bit confusing. If you want
to just find, use /. The colon is not necessary (which indicates
command mode). If you're using the find as a range (basically the same
thing, / is just an empty command with a range) you can use the colon,
but either way escaping the first slash is not necessary.
The other main problem is that parenthesis by default need to be escaped
if you meant a capturing group. All of this is dependant on your
'magic' option reading the help for the /magic topic (you can do a
:h magic) is highly recommended. With "vanilla" Vim settings, the
regex you need looks live this:
/'\([^']*\)'
With very magic enable (by using the \v atom) this can be simplified
to your original design:
/\v'([^']*)'
Alternatively you can use
\v'(\a+)'
this regex performs similar than yours, except when nested quotes are encountered. In the text:
The user's first 'answer'.
The regex \v'(\a+)' will capture answer while your original regex (corrected by sidyll) \v'([^']*)' will capture 's first '.
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.
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 $1_
See it working at http://refiddle.com/10z
I have some text like this:
dagGeneralCodes$_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]$_ctl0.
I really like using this Regular Expression Cheatsheet; it's free, a single page, and printed, fits on my cube wall.
I have an text that consists of information enclosed by a certain pattern.
The only thing I know is the pattern: "${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.