I am using Notepad++ to find (".*)"(.*) and replace it with \1\"\2 but it doesn't seem to work. I don't know why.
Example:
Someone said "My name is "sean""
I want it to be:
Someone said "My name is \"sean\""
Edit: In my case the closing quote is always on the end of line so will (".*)"(.*"$) work?
Edit2: Also the first quote is preceded with a comma so I will use (,".*)"(.*"$) though it may not work in some cases but I think it will work with my file.
Now there is the problem with the replace it doesn't add \" it just add some space.
It should work... you just need to do a little fixing...
The Find what regex should be ("[^"]*)("\w*)(")([^"]*")
The Replace with expression should be \1\\\2\\\3\4
Make sure you select the Search Mode to be "Regular expression"
Explanation...
This is quite tricky - I've assumed that the quoted text WITHIN quotes is just a single word. If you assume something else it becomes very hard to pin down.
You need to find a
" followed by
[^"]* - any number of characters that are NOT a " and then
("\w*)(") - a quoted word, and then finally
([^"]*") - any additional number of non-quote characters + a final quote
This is important because regular expression matching is greedy by default, and a .* would continue to match all characters, including " until the end of the string (see link )
In the replacement string you need to have \\ to represent a single \
I need to reformat a text file a bit in my Notepad++ and I have a text of this kind:
This is some example text. This is some example text. This is some example text.
- This is some example text.
-This is some example text.
- This is some example text.
- This is some example text.
So as you can see in above text there are two types of "-" preceeding text the one with the space after "-" and ones without it I need to find only the ones without sapce and add it in between "-" and the "text"
If I ran piece of code below
-[A-Za-z0-9]
it finds dash and first letter right after it, which is not useful as when I replace the text it changes this first letter which is always different (depending on what is written) so I need to find this and select only the "-" and then replace it with "- " unless there is better way.
For demonstration purposes:
Find what: -([A-Za-z0-9])(.+)
Replace with: - \1\2
The parentheses denote a capture group. In the Replace with line, you use backslash and the number of group to add it.
That said, what you really want to match for is a NOT group, like -([^\s]) (match where a dash isn't immediately followed by a whitespace).
Search for
-([^ ])
and replace with
- \1
[^ ] is a negated character class and matches everything but a space. This character is stored in \1 because of the brackets () around the pattern.
I have looked around and found good answers but none work with notepad++, most are for java and php. I have found the search strings below but obviously I'm a noob with regex as i don't know what open/close tags are proper in notepad++.
I would like to add a space before each capital letter.
Example:
StackOverflowKegger
becomes
Stack Overflow Kegger
This is what i have found.
Find: [a-z]+[A-Z]+
Replace: $1 (there is a space before the $)
Find:
(?<!^)((?<![:upper:])[:upper:]|[:upper:](?![:upper:]))
("(\\p{Ll})(\\p{Lu})","$1 $2")
(?!^)(?=[A-Z])
Any help would be appreciated.
Search string: (.)([A-Z])
Replacement: \1 \2
This doesn't insert spaces before capitals that are the first letter on their line.
In Notepad++, do a search-n-Replace (ctrl+h), in 'find what' input '([a-z])([A-Z])' without single quotes. in 'Replace with' input '\1 \2' without quotes.
Select radio button 'Regular Expression' and make sure you Check 'Match Case' checkbox. Now find next and keep replacing. it will convert camel or Pascal case strings into words with a space before every capital letter except the first.
Hope it is helpful. I just used it with one of my tasks.
Find: ^([A-Z])
Replace: \1
this will add a space to the first uppercase character in notepad++
Make sure you put the space before the \1 in the replace section.
WABET : <-from
WABET : <-to
Find what: .\K([A-Z])
Replace with: $1 a space before $1
Note!!!!!! Must to check match-case see in attached photo.
If you can live with a space before the first word, then this solution worked for me.
I used the following with the Regular Expression radio button checked.:
Find what: ([A-Z])
Replace With: \1
Note the leading space before the \1 in the replace
In eclipse, is it possible to use the matched search string as part of the replace string when performing a regular expression search and replace?
Basically, I want to replace all occurrences of
variableName.someMethod()
with:
((TypeName)variableName.someMethod())
Where variableName can be any variable name at all.
In sed I could use something like:
s/[a-zA-Z]+\.someMethod\(\)/((TypeName)&)/g
That is, & represents the matched search string. Is there something similar in Eclipse?
Thanks!
Yes, ( ) captures a group. You can use it again with $i where i is the i'th capture group.
So:
search: (\w+\.someMethod\(\))
replace: ((TypeName)$1)
Hint: Ctrl + Space in the textboxes gives you all kinds of suggestions for regular expression writing.
Using ...
search = (^.*import )(.*)(\(.*\):)
replace = $1$2
...replaces ...
from checks import checklist(_list):
...with...
from checks import checklist
Blocks in regex are delineated by parenthesis (which are not preceded by a "\")
(^.*import ) finds "from checks import " and loads it to $1 (eclipse starts counting at 1)
(.*) find the next "everything" until the next encountered "(" and loads it to $2. $2 stops at the "(" because of the next part (see next line below)
(\(.*\):) says "at the first encountered "(" after starting block $2...stop block $2 and start $3. $3 gets loaded with the "('any text'):" or, in the example, the "(_list):"
Then in the replace, just put the $1$2 to replace all three blocks with just the first two.
NomeN has answered correctly, but this answer wouldn't be of much use for beginners like me because we will have another problem to solve and we wouldn't know how to use RegEx in there. So I am adding a bit of explanation to this. The answer is
search: (\w+\\.someMethod\\(\\))
replace: ((TypeName)$1)
Here:
In search:
First and last (, ) depicts a group in regex
\w depicts words (alphanumeric + underscore)
+ depicts one or more (ie one or more of alphanumeric + underscore)
. is a special character which depicts any character (ie .+ means
one or more of any character). Because this is a special character
to depict a . we should give an escape character with it, ie \.
someMethod is given as it is to be searched.
The two parenthesis (, ) are given along with escape character
because they are special character which are used to depict a group
(we will discuss about group in next point)
In replace:
It is given ((TypeName)$1), here $1 depicts the
group. That is all the characters that are enclosed within the first
and last parenthesis (, ) in the search field
Also make sure you have checked the 'Regular expression' option in
find an replace box
At least at STS (SpringSource Tool Suite) groups are numbered starting form 0, so replace string will be
replace: ((TypeName)$0)
For someone who needs an explanation and an example of how to use a regxp in Eclipse. Here is my example illustrating the problem.
I want to rename
/download.mp4^lecture_id=271
to
/271.mp4
And there can be multiple of these.
Here is how it should be done.
Then hit find/replace button
I have a value like this:
"Foo Bar" "Another Value" something else
What regex will return the values enclosed in the quotation marks (e.g. Foo Bar and Another Value)?
In general, the following regular expression fragment is what you are looking for:
"(.*?)"
This uses the non-greedy *? operator to capture everything up to but not including the next double quote. Then, you use a language-specific mechanism to extract the matched text.
In Python, you could do:
>>> import re
>>> string = '"Foo Bar" "Another Value"'
>>> print re.findall(r'"(.*?)"', string)
['Foo Bar', 'Another Value']
I've been using the following with great success:
(["'])(?:(?=(\\?))\2.)*?\1
It supports nested quotes as well.
For those who want a deeper explanation of how this works, here's an explanation from user ephemient:
([""']) match a quote; ((?=(\\?))\2.) if backslash exists, gobble it, and whether or not that happens, match a character; *? match many times (non-greedily, as to not eat the closing quote); \1 match the same quote that was use for opening.
I would go for:
"([^"]*)"
The [^"] is regex for any character except '"'
The reason I use this over the non greedy many operator is that I have to keep looking that up just to make sure I get it correct.
Lets see two efficient ways that deal with escaped quotes. These patterns are not designed to be concise nor aesthetic, but to be efficient.
These ways use the first character discrimination to quickly find quotes in the string without the cost of an alternation. (The idea is to discard quickly characters that are not quotes without to test the two branches of the alternation.)
Content between quotes is described with an unrolled loop (instead of a repeated alternation) to be more efficient too: [^"\\]*(?:\\.[^"\\]*)*
Obviously to deal with strings that haven't balanced quotes, you can use possessive quantifiers instead: [^"\\]*+(?:\\.[^"\\]*)*+ or a workaround to emulate them, to prevent too much backtracking. You can choose too that a quoted part can be an opening quote until the next (non-escaped) quote or the end of the string. In this case there is no need to use possessive quantifiers, you only need to make the last quote optional.
Notice: sometimes quotes are not escaped with a backslash but by repeating the quote. In this case the content subpattern looks like this: [^"]*(?:""[^"]*)*
The patterns avoid the use of a capture group and a backreference (I mean something like (["']).....\1) and use a simple alternation but with ["'] at the beginning, in factor.
Perl like:
["'](?:(?<=")[^"\\]*(?s:\\.[^"\\]*)*"|(?<=')[^'\\]*(?s:\\.[^'\\]*)*')
(note that (?s:...) is a syntactic sugar to switch on the dotall/singleline mode inside the non-capturing group. If this syntax is not supported you can easily switch this mode on for all the pattern or replace the dot with [\s\S])
(The way this pattern is written is totally "hand-driven" and doesn't take account of eventual engine internal optimizations)
ECMA script:
(?=["'])(?:"[^"\\]*(?:\\[\s\S][^"\\]*)*"|'[^'\\]*(?:\\[\s\S][^'\\]*)*')
POSIX extended:
"[^"\\]*(\\(.|\n)[^"\\]*)*"|'[^'\\]*(\\(.|\n)[^'\\]*)*'
or simply:
"([^"\\]|\\.|\\\n)*"|'([^'\\]|\\.|\\\n)*'
Peculiarly, none of these answers produce a regex where the returned match is the text inside the quotes, which is what is asked for. MA-Madden tries but only gets the inside match as a captured group rather than the whole match. One way to actually do it would be :
(?<=(["']\b))(?:(?=(\\?))\2.)*?(?=\1)
Examples for this can be seen in this demo https://regex101.com/r/Hbj8aP/1
The key here is the the positive lookbehind at the start (the ?<= ) and the positive lookahead at the end (the ?=). The lookbehind is looking behind the current character to check for a quote, if found then start from there and then the lookahead is checking the character ahead for a quote and if found stop on that character. The lookbehind group (the ["']) is wrapped in brackets to create a group for whichever quote was found at the start, this is then used at the end lookahead (?=\1) to make sure it only stops when it finds the corresponding quote.
The only other complication is that because the lookahead doesn't actually consume the end quote, it will be found again by the starting lookbehind which causes text between ending and starting quotes on the same line to be matched. Putting a word boundary on the opening quote (["']\b) helps with this, though ideally I'd like to move past the lookahead but I don't think that is possible. The bit allowing escaped characters in the middle I've taken directly from Adam's answer.
The RegEx of accepted answer returns the values including their sourrounding quotation marks: "Foo Bar" and "Another Value" as matches.
Here are RegEx which return only the values between quotation marks (as the questioner was asking for):
Double quotes only (use value of capture group #1):
"(.*?[^\\])"
Single quotes only (use value of capture group #1):
'(.*?[^\\])'
Both (use value of capture group #2):
(["'])(.*?[^\\])\1
-
All support escaped and nested quotes.
I liked Eugen Mihailescu's solution to match the content between quotes whilst allowing to escape quotes. However, I discovered some problems with escaping and came up with the following regex to fix them:
(['"])(?:(?!\1|\\).|\\.)*\1
It does the trick and is still pretty simple and easy to maintain.
Demo (with some more test-cases; feel free to use it and expand on it).
PS: If you just want the content between quotes in the full match ($0), and are not afraid of the performance penalty use:
(?<=(['"])\b)(?:(?!\1|\\).|\\.)*(?=\1)
Unfortunately, without the quotes as anchors, I had to add a boundary \b which does not play well with spaces and non-word boundary characters after the starting quote.
Alternatively, modify the initial version by simply adding a group and extract the string form $2:
(['"])((?:(?!\1|\\).|\\.)*)\1
PPS: If your focus is solely on efficiency, go with Casimir et Hippolyte's solution; it's a good one.
A very late answer, but like to answer
(\"[\w\s]+\")
http://regex101.com/r/cB0kB8/1
The pattern (["'])(?:(?=(\\?))\2.)*?\1 above does the job but I am concerned of its performances (it's not bad but could be better). Mine below it's ~20% faster.
The pattern "(.*?)" is just incomplete. My advice for everyone reading this is just DON'T USE IT!!!
For instance it cannot capture many strings (if needed I can provide an exhaustive test-case) like the one below:
$string = 'How are you? I\'m fine, thank you';
The rest of them are just as "good" as the one above.
If you really care both about performance and precision then start with the one below:
/(['"])((\\\1|.)*?)\1/gm
In my tests it covered every string I met but if you find something that doesn't work I would gladly update it for you.
Check my pattern in an online regex tester.
This version
accounts for escaped quotes
controls backtracking
/(["'])((?:(?!\1)[^\\]|(?:\\\\)*\\[^\\])*)\1/
MORE ANSWERS! Here is the solution i used
\"([^\"]*?icon[^\"]*?)\"
TLDR;
replace the word icon with what your looking for in said quotes and voila!
The way this works is it looks for the keyword and doesn't care what else in between the quotes.
EG:
id="fb-icon"
id="icon-close"
id="large-icon-close"
the regex looks for a quote mark "
then it looks for any possible group of letters thats not "
until it finds icon
and any possible group of letters that is not "
it then looks for a closing "
I liked Axeman's more expansive version, but had some trouble with it (it didn't match for example
foo "string \\ string" bar
or
foo "string1" bar "string2"
correctly, so I tried to fix it:
# opening quote
(["'])
(
# repeat (non-greedy, so we don't span multiple strings)
(?:
# anything, except not the opening quote, and not
# a backslash, which are handled separately.
(?!\1)[^\\]
|
# consume any double backslash (unnecessary?)
(?:\\\\)*
|
# Allow backslash to escape characters
\\.
)*?
)
# same character as opening quote
\1
string = "\" foo bar\" \"loloo\""
print re.findall(r'"(.*?)"',string)
just try this out , works like a charm !!!
\ indicates skip character
My solution to this is below
(["']).*\1(?![^\s])
Demo link : https://regex101.com/r/jlhQhV/1
Explanation:
(["'])-> Matches to either ' or " and store it in the backreference \1 once the match found
.* -> Greedy approach to continue matching everything zero or more times until it encounters ' or " at end of the string. After encountering such state, regex engine backtrack to previous matching character and here regex is over and will move to next regex.
\1 -> Matches to the character or string that have been matched earlier with the first capture group.
(?![^\s]) -> Negative lookahead to ensure there should not any non space character after the previous match
Unlike Adam's answer, I have a simple but worked one:
(["'])(?:\\\1|.)*?\1
And just add parenthesis if you want to get content in quotes like this:
(["'])((?:\\\1|.)*?)\1
Then $1 matches quote char and $2 matches content string.
All the answer above are good.... except they DOES NOT support all the unicode characters! at ECMA Script (Javascript)
If you are a Node users, you might want the the modified version of accepted answer that support all unicode characters :
/(?<=((?<=[\s,.:;"']|^)["']))(?:(?=(\\?))\2.)*?(?=\1)/gmu
Try here.
echo 'junk "Foo Bar" not empty one "" this "but this" and this neither' | sed 's/[^\"]*\"\([^\"]*\)\"[^\"]*/>\1</g'
This will result in: >Foo Bar<><>but this<
Here I showed the result string between ><'s for clarity, also using the non-greedy version with this sed command we first throw out the junk before and after that ""'s and then replace this with the part between the ""'s and surround this by ><'s.
From Greg H. I was able to create this regex to suit my needs.
I needed to match a specific value that was qualified by being inside quotes. It must be a full match, no partial matching could should trigger a hit
e.g. "test" could not match for "test2".
reg = r"""(['"])(%s)\1"""
if re.search(reg%(needle), haystack, re.IGNORECASE):
print "winning..."
Hunter
If you're trying to find strings that only have a certain suffix, such as dot syntax, you can try this:
\"([^\"]*?[^\"]*?)\".localized
Where .localized is the suffix.
Example:
print("this is something I need to return".localized + "so is this".localized + "but this is not")
It will capture "this is something I need to return".localized and "so is this".localized but not "but this is not".
A supplementary answer for the subset of Microsoft VBA coders only one uses the library Microsoft VBScript Regular Expressions 5.5 and this gives the following code
Sub TestRegularExpression()
Dim oRE As VBScript_RegExp_55.RegExp '* Tools->References: Microsoft VBScript Regular Expressions 5.5
Set oRE = New VBScript_RegExp_55.RegExp
oRE.Pattern = """([^""]*)"""
oRE.Global = True
Dim sTest As String
sTest = """Foo Bar"" ""Another Value"" something else"
Debug.Assert oRE.test(sTest)
Dim oMatchCol As VBScript_RegExp_55.MatchCollection
Set oMatchCol = oRE.Execute(sTest)
Debug.Assert oMatchCol.Count = 2
Dim oMatch As Match
For Each oMatch In oMatchCol
Debug.Print oMatch.SubMatches(0)
Next oMatch
End Sub