I need to replace character (say) x with character (say) P in a string, but only if it is contained in a quoted substring.
An example makes it clearer:
axbx'cxdxe'fxgh'ixj'k -> axbx'cPdPe'fxgh'iPj'k
Let's assume, for the sake of simplicity, that quotes always come in pairs.
The obvious way is to just process the string one character at a time (a simple state machine approach);
however, I'm wondering if regular expressions can be used to do all the processing in one go.
My target language is C#, but I guess my question pertains to any language having builtin or library support for regular expressions.
I converted Greg Hewgill's python code to C# and it worked!
[Test]
public void ReplaceTextInQuotes()
{
Assert.AreEqual("axbx'cPdPe'fxgh'iPj'k",
Regex.Replace("axbx'cxdxe'fxgh'ixj'k",
#"x(?=[^']*'([^']|'[^']*')*$)", "P"));
}
That test passed.
I was able to do this with Python:
>>> import re
>>> re.sub(r"x(?=[^']*'([^']|'[^']*')*$)", "P", "axbx'cxdxe'fxgh'ixj'k")
"axbx'cPdPe'fxgh'iPj'k"
What this does is use the non-capturing match (?=...) to check that the character x is within a quoted string. It looks for some nonquote characters up to the next quote, then looks for a sequence of either single characters or quoted groups of characters, until the end of the string.
This relies on your assumption that the quotes are always balanced. This is also not very efficient.
A more general (and simpler) solution which allows non-paired quotes.
Find quoted string
Replace 'x' by 'P' in the string
#!/usr/bin/env python
import re
text = "axbx'cxdxe'fxgh'ixj'k"
s = re.sub("'.*?'", lambda m: re.sub("x", "P", m.group(0)), text)
print s == "axbx'cPdPe'fxgh'iPj'k", s
# -> True axbx'cPdPe'fxgh'iPj'k
The trick is to use non-capturing group to match the part of the string following the match (character x) we are searching for.
Trying to match the string up to x will only find either the first or the last occurence, depending whether non-greedy quantifiers are used.
Here's Greg's idea transposed to Tcl, with comments.
set strIn {axbx'cxdxe'fxgh'ixj'k}
set regex {(?x) # enable expanded syntax
# - allows comments, ignores whitespace
x # the actual match
(?= # non-matching group
[^']*' # match to end of current quoted substring
##
## assuming quotes are in pairs,
## make sure we actually were
## inside a quoted substring
## by making sure the rest of the string
## is what we expect it to be
##
(
[^']* # match any non-quoted substring
| # ...or...
'[^']*' # any quoted substring, including the quotes
)* # any number of times
$ # until we run out of string :)
) # end of non-matching group
}
#the same regular expression without the comments
set regexCondensed {(?x)x(?=[^']*'([^']|'[^']*')*$)}
set replRegex {P}
set nMatches [regsub -all -- $regex $strIn $replRegex strOut]
puts "$nMatches replacements. "
if {$nMatches > 0} {
puts "Original: |$strIn|"
puts "Result: |$strOut|"
}
exit
This prints:
3 replacements.
Original: |axbx'cxdxe'fxgh'ixj'k|
Result: |axbx'cPdPe'fxgh'iPj'k|
#!/usr/bin/perl -w
use strict;
# Break up the string.
# The spliting uses quotes
# as the delimiter.
# Put every broken substring
# into the #fields array.
my #fields;
while (<>) {
#fields = split /'/, $_;
}
# For every substring indexed with an odd
# number, search for x and replace it
# with P.
my $count;
my $end = $#fields;
for ($count=0; $count < $end; $count++) {
if ($count % 2 == 1) {
$fields[$count] =~ s/a/P/g;
}
}
Wouldn't this chunk do the job?
Not with plain regexp. Regular expressions have no "memory" so they cannot distinguish between being "inside" or "outside" quotes.
You need something more powerful, for example using gema it would be straighforward:
'<repl>'=$0
repl:x=P
Similar discussion about balanced text replaces: Can regular expressions be used to match nested patterns?
Although you can try this in Vim, but it works well only if the string is on one line, and there's only one pair of 's.
:%s:\('[^']*\)x\([^']*'\):\1P\2:gci
If there's one more pair or even an unbalanced ', then it could fail. That's way I included the c a.k.a. confirm flag on the ex command.
The same can be done with sed, without the interaction - or with awk so you can add some interaction.
One possible solution is to break the lines on pairs of 's then you can do with vim solution.
Pattern: (?s)\G((?:^[^']*'|(?<=.))(?:'[^']*'|[^'x]+)*+)x
Replacement: \1P
\G — Anchor each match at the end of the previous one, or the start of the string.
(?:^[^']*'|(?<=.)) — If it is at the beginning of the string, match up to the first quote.
(?:'[^']*'|[^'x]+)*+ — Match any block of unquoted characters, or any (non-quote) characters up to an 'x'.
One sweep trough the source string, except for a single character look-behind.
Sorry to break your hopes, but you need a push-down automata to do that. There is more info here:
Pushdown Automaton
In short, Regular expressions, which are finite state machines can only read and has no memory while pushdown automaton has a stack and manipulating capabilities.
Edit: spelling...
Related
https://regex101.com/r/sB9wW6/1
(?:(?<=\s)|^)#(\S+) <-- the problem in positive lookbehind
Working like this on prod: (?:\s|^)#(\S+), but I need a correct start index (without space).
Here is in JS:
var regex = new RegExp(/(?:(?<=\s)|^)#(\S+)/g);
Error parsing regular expression: Invalid regular expression:
/(?:(?<=\s)|^)#(\S+)/
What am I doing wrong?
UPDATE
Ok, no lookbehind in JS :(
But anyways, I need a regex to get the proper start and end index of my match. Without leading space.
Make sure you always select the right regex engine at regex101.com. See an issue that occurred due to using a JS-only compatible regex with [^] construct in Python.
JS regex - at the time of answering this question - did not support lookbehinds. Now, it becomes more and more adopted after its introduction in ECMAScript 2018. You do not really need it here since you can use capturing groups:
var re = /(?:\s|^)#(\S+)/g;
var str = 's #vln1\n#vln2\n';
var res = [];
while ((m = re.exec(str)) !== null) {
res.push(m[1]);
}
console.log(res);
The (?:\s|^)#(\S+) matches a whitespace or the start of string with (?:\s|^), then matches #, and then matches and captures into Group 1 one or more non-whitespace chars with (\S+).
To get the start/end indices, use
var re = /(\s|^)#\S+/g;
var str = 's #vln1\n#vln2\n';
var pos = [];
while ((m = re.exec(str)) !== null) {
pos.push([m.index+m[1].length, m.index+m[0].length]);
}
console.log(pos);
BONUS
My regex works at regex101.com, but not in...
First of all, have you checked the Code Generator link in the Tools pane on the left?
All languages - "Literal string" vs. "String literal" alert - Make sure you test against the same text used in code, literal string, at the regex tester. A common scenario is copy/pasting a string literal value directly into the test string field, with all string escape sequences like \n (line feed char), \r (carriage return), \t (tab char). See Regex_search c++, for example. Mind that they must be replaced with their literal counterparts. So, if you have in Python text = "Text\n\n abc", you must use Text, two line breaks, abc in the regex tester text field. Text.*?abc will never match it although you might think it "works". Yes, . does not always match line break chars, see How do I match any character across multiple lines in a regular expression?
All languages - Backslash alert - Make sure you correctly use a backslash in your string literal, in most languages, in regular string literals, use double backslash, i.e. \d used at regex101.com must written as \\d. In raw string literals, use a single backslash, same as at regex101. Escaping word boundary is very important, since, in many languages (C#, Python, Java, JavaScript, Ruby, etc.), "\b" is used to define a BACKSPACE char, i.e. it is a valid string escape sequence. PHP does not support \b string escape sequence, so "/\b/" = '/\b/' there.
All languages - Default flags - Global and Multiline - Note that by default m and g flags are enabled at regex101.com. So, if you use ^ and $, they will match at the start and end of lines correspondingly. If you need the same behavior in your code check how multiline mode is implemented and either use a specific flag, or - if supported - use an inline (?m) embedded (inline) modifier. The g flag enables multiple occurrence matching, it is often implemented using specific functions/methods. Check your language reference to find the appropriate one.
line-breaks - Line endings at regex101.com are LF only, you can't test strings with CRLF endings, see regex101.com VS myserver - different results. Solutions can be different for each regex library: either use \R (PCRE, Java, Ruby) or some kind of \v (Boost, PCRE), \r?\n, (?:\r\n?|\n)/(?>\r\n?|\n) (good for .NET) or [\r\n]+ in other libraries (see answers for C#, PHP). Another issue related to the fact that you test your regex against a multiline string (not a list of standalone strings/lines) is that your patterns may consume the end of line, \n, char with negated character classes, see an issue like that. \D matched the end of line char, and in order to avoid it, [^\d\n] could be used, or other alternatives.
php - You are dealing with Unicode strings, or want shorthand character classes to match Unicode characters, too (e.g. \w+ to match Стрибижев or Stribiżew, or \s+ to match hard spaces), then you need to use u modifier, see preg_match() returns 0 although regex testers work - To match all occurrences, use preg_match_all, not preg_match with /...pattern.../g, see PHP preg_match to find multiple occurrences and "Unknown modifier 'g' in..." when using preg_match in PHP?- Your regex with inline backreference like \1 refuses to work? Are you using a double quoted string literal? Use a single-quoted one, see Backreference does not work in PHP
phplaravel - Mind you need the regex delimiters around the pattern, see https://stackoverflow.com/questions/22430529
python - Note that re.search, re.match, re.fullmatch, re.findall and re.finditer accept the regex as the first argument, and the input string as the second argument. Not re.findall("test 200 300", r"\d+"), but re.findall(r"\d+", "test 200 300"). If you test at regex101.com, please check the "Code Generator" page. - You used re.match that only searches for a match at the start of the string, use re.search: Regex works fine on Pythex, but not in Python - If the regex contains capturing group(s), re.findall returns a list of captures/capture tuples. Either use non-capturing groups, or re.finditer, or remove redundant capturing groups, see re.findall behaves weird - If you used ^ in the pattern to denote start of a line, not start of the whole string, or used $ to denote the end of a line and not a string, pass re.M or re.MULTILINE flag to re method, see Using ^ to match beginning of line in Python regex
- If you try to match some text across multiple lines, and use re.DOTALL or re.S, or [\s\S]* / [\s\S]*?, and still nothing works, check if you read the file line by line, say, with for line in file:. You must pass the whole file contents as the input to the regex method, see Getting Everything Between Two Characters Across New Lines. - Having trouble adding flags to regex and trying something like pattern = r"/abc/gi"? See How to add modifers to regex in python?
c#, .net - .NET regex does not support possessive quantifiers like ++, *+, ??, {1,10}?, see .NET regex matching digits between optional text with possessive quantifer is not working - When you match against a multiline string and use RegexOptions.Multiline option (or inline (?m) modifier) with an $ anchor in the pattern to match entire lines, and get no match in code, you need to add \r? before $, see .Net regex matching $ with the end of the string and not of line, even with multiline enabled - To get multiple matches, use Regex.Matches, not Regex.Match, see RegEx Match multiple times in string - Similar case as above: splitting a string into paragraphs, by a double line break sequence - C# / Regex Pattern works in online testing, but not at runtime - You should remove regex delimiters, i.e. #"/\d+/" must actually look like #"\d+", see Simple and tested online regex containing regex delimiters does not work in C# code - If you unnecessarily used Regex.Escape to escape all characters in a regular expression (like Regex.Escape(#"\d+\.\d+")) you need to remove Regex.Escape, see Regular Expression working in regex tester, but not in c#
dartflutter - Use raw string literal, RegExp(r"\d"), or double backslashes (RegExp("\\d")) - https://stackoverflow.com/questions/59085824
javascript - Double escape backslashes in a RegExp("\\d"): Why do regex constructors need to be double escaped?
- (Negative) lookbehinds unsupported by most browsers: Regex works on browser but not in Node.js - Strings are immutable, assign the .replace result to a var - The .replace() method does change the string in place - Retrieve all matches with str.match(/pat/g) - Regex101 and Js regex search showing different results or, with RegExp#exec, RegEx to extract all matches from string using RegExp.exec- Replace all pattern matches in string: Why does javascript replace only first instance when using replace?
javascriptangular - Double the backslashes if you define a regex with a string literal, or just use a regex literal notation, see https://stackoverflow.com/questions/56097782
java - Word boundary not working? Make sure you use double backslashes, "\\b", see Regex \b word boundary not works - Getting invalid escape sequence exception? Same thing, double backslashes - Java doesn't work with regex \s, says: invalid escape sequence - No match found is bugging you? Run Matcher.find() / Matcher.matches() - Why does my regex work on RegexPlanet and regex101 but not in my code? - .matches() requires a full string match, use .find(): Java Regex pattern that matches in any online tester but doesn't in Eclipse - Access groups using matcher.group(x): Regex not working in Java while working otherwise - Inside a character class, both [ and ] must be escaped - Using square brackets inside character class in Java regex - You should not run matcher.matches() and matcher.find() consecutively, use only if (matcher.matches()) {...} to check if the pattern matches the whole string and then act accordingly, or use if (matcher.find()) to check if there is a single match or while (matcher.find()) to find multiple matches (or Matcher#results()). See Why does my regex work on RegexPlanet and regex101 but not in my code?
scala - Your regex attempts to match several lines, but you read the file line by line (e.g. use for (line <- fSource.getLines))? Read it into a single variable (see matching new line in Scala regex, when reading from file)
kotlin - You have Regex("/^\\d+$/")? Remove the outer slashes, they are regex delimiter chars that are not part of a pattern. See Find one or more word in string using Regex in Kotlin - You expect a partial string match, but .matchEntire requires a full string match? Use .find, see Regex doesn't match in Kotlin
mongodb - Do not enclose /.../ with single/double quotation marks, see mongodb regex doesn't work
c++ - regex_match requires a full string match, use regex_search to find a partial match - Regex not working as expected with C++ regex_match - regex_search finds the first match only. Use sregex_token_iterator or sregex_iterator to get all matches: see What does std::match_results::size return? - When you read a user-defined string using std::string input; std::cin >> input;, note that cin will only get to the first whitespace, to read the whole line properly, use std::getline(std::cin, input); - C++ Regex to match '+' quantifier - "\d" does not work, you need to use "\\d" or R"(\d)" (a raw string literal) - This regex doesn't work in c++ - Make sure the regex is tested against a literal text, not a string literal, see Regex_search c++
go - Double backslashes or use a raw string literal: Regular expression doesn't work in Go - Go regex does not support lookarounds, select the right option (Go) at regex101.com before testing! Regex expression negated set not working golang
groovy - Return all matches: Regex that works on regex101 does not work in Groovy
r - Double escape backslashes in the string literal: "'\w' is an unrecognized escape" in grep - Use perl=TRUE to PCRE engine ((g)sub/(g)regexpr): Why is this regex using lookbehinds invalid in R?
oracle - Greediness of all quantifiers is set by the first quantifier in the regex, see Regex101 vs Oracle Regex (then, you need to make all the quantifiers as greedy as the first one)] - \b does not work? Oracle regex does not support word boundaries at all, use workarounds as shown in Regex matching works on regex tester but not in oracle
firebase - Double escape backslashes, make sure ^ only appears at the start of the pattern and $ is located only at the end (if any), and note you cannot use more than 9 inline backreferences: Firebase Rules Regex Birthday
firebasegoogle-cloud-firestore - In Firestore security rules, the regular expression needs to be passed as a string, which also means it shouldn't be wrapped in / symbols, i.e. use allow create: if docId.matches("^\\d+$").... See https://stackoverflow.com/questions/63243300
google-data-studio - /pattern/g in REGEXP_REPLACE must contain no / regex delimiters and flags (like g) - see How to use Regex to replace square brackets from date field in Google Data Studio?
google-sheets - If you think REGEXEXTRACT does not return full matches, truncates the results, you should check if you have redundant capturing groups in your regex and remove them, or convert the capturing groups to non-capturing by add ?: after the opening (, see Extract url domain root in Google Sheet
sed - Why does my regular expression work in X but not in Y?
word-boundarypcrephp - [[:<:]] and [[:>:]] do not work in the regex tester, although they are valid constructs in PCRE, see https://stackoverflow.com/questions/48670105
snowflake-cloud-data-platform snowflake-sql - If you are writing a stored procedure, and \\d does not work, you need to double them again and use \\\\d, see REGEX conversion of VARCHAR value to DATE in Snowflake stored procedure using RLIKE not consistent.
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
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.
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
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
I am having issue with a reg ex expression and can't find the answer to my question.
I am trying to build a reg ex pattern that will pull in any matches that have # around them. for example #match# or #mt# would both come back.
This works fine for that. #.*?#
However I don't want matches on ## to show up. Basically if there is nothing between the pound signs don't match.
Hope this makes sense.
Thanks.
Please use + to match 1 or more symbols:
#+.+#+
UPDATE:
If you want to only match substrings that are enclosed with single hash symbols, use:
(?<!#)#(?!#)[^#]+#(?!#)
See regex demo
Explanation:
(?<!#)#(?!#) - a # symbol that is not preceded with a # (due to the negative lookbehind (?<!#)) and not followed by a # (due to the negative lookahead (?!#))
[^#]+ - one or more symbols other than # (due to the negated character class [^#])
#(?!#) - a # symbol not followed with another # symbol.
Instead of using * to match between zero and unlimited characters, replace it with +, which will only match if there is at least one character between the #'s. The edited regex should look like this: #.+?#. Hope this helps!
Edit
Sorry for the incorrect regex, I had not expected multiple hash signs. This should work for your sentence: #+.+?#+
Edit 2
I am pretty sure I got it. Try this: (?<!#)#[^#].*?#. It might not work as expected with triple hashes though.
Try:
[^#]?#.+#[^#]?
The [^ character_group] construction matches any single character not included in the character group. Using the ? after it will let you match at the beginning/end of a string (since it matches the preceeding character zero or more times. Check out the documentation here
I am trying to combine two Regular Expression patterns to determine if a String is either a double value or a variable. My restrictions are as follows:
The variable can only begin with an _ or alphabetical letter (A-Z, ignoring case), but it can be followed by zero or more _s, letters, or digits.
Here's what I have so far, but I can't get it to work properly.
String varPattern = #"[a-zA-Z_](?: [a-zA-Z_]|\d)*";
String doublePattern = #"(?: \d+\.\d* | \d*\.\d+ | \d+ ) (?: [eE][\+-]?\d+)?";
String pattern = String.Format("({0}) | ({1})",
varPattern, doublePattern);
Regex.IsMatch(word, varPattern, RegexOptions.IgnoreCase)
It seems that it is capturing both Regular Expression patterns, but I need it to be either/or.
For example, _A2 2 is valid using the code above, but _A2 is invalid.
Some examples of valid variables are as follows:
_X6 , _ , A , Z_2_A
And some examples of invalid variables are as follows:
2_X6 , $2 , T_2$
I guess I just need clarification on the pattern format for the Regular Expression. The format is unclear to me.
As noted, the literal whitespace you've put in your regular expressions is part of the regular expression. You won't get a match unless that same whitespace is in the text being scanned by the regular expression. If you want to use whitespace to make your regex, you'll need to specify RegexOptions.IgnorePatternWhitespace, after that, if you want to match any whitespace, you'll have to do so explicitly, either by specifying \s, \x20, etc.
It should be noted that if you do specify RegexOptions.IgnorePatternWhitespace, you can use Perl-style comments (# to end of line) to document your regular expression (as I've done below). For complex regular expressions, someone 5 years from now — who might be you! — will thank you for the kindness.
Your [presumably intended] patterns are also, I think, more complex than they need be. A regular expression to match the identifier rule you've specified is this:
[a-zA-Z_][a-zA-Z0-9_]*
Broken out into its constituent parts:
[a-zA-Z_] # match an upper- or lower-case letter or an underscore, followed by
[a-zA-Z0-9_]* # zero or more occurences of an upper- or lower-case letter, decimal digit or underscore
A regular expression to match the conventional style of a numeric/floating-point literal is this:
([+-]?[0-9]+)(\.[0-9]+)?([Ee][+-]?[0-9]+)?
Broken out into its constituent parts:
( # a mandatory group that is the integer portion of the value, consisting of
[+-]? # - an optional plus- or minus-sign, followed by
[0-9]+ # - one or more decimal digits
) # followed by
( # an optional group that is the fractional portion of the value, consisting of
\. # - a decimal point, followed by
[0-9]+ # - one or more decimal digits
)? # followed by,
( # an optional group, that is the exponent portion of the value, consisting of
[Ee] # - The upper- or lower-case letter 'E' indicating the start of the exponent, followed by
[+-]? # - an optional plus- or minus-sign, followed by
[0-9]+ # - one or more decimal digits.
)? # Easy!
Note: Some grammars differ as to whether the sign of the value is a unary operator or part
of the value and whether or not a leading + sign is allowed. Grammars also vary as to whether
something like 123245. is valid (e.g., is a decimal point with no fractional digits valid?)
To combine these two regular expression,
First, group each of them with parentheses (you might want to name the containing groups, as I've done):
(?<identifier>[a-zA-Z_][a-zA-Z0-9_]*)
(?<number>[+-]?[0-9]+)(\.[0-9]+)?([Ee][+-]?[0-9]+)?
Next, combine with the alternation operation, |:
(?<identifier>[a-zA-Z_][a-zA-Z0-9_]*)|(?<number>[+-]?[0-9]+)(\.[0-9]+)?([Ee][+-]?[0-9]+)?
Finally, enclose the whole shebang in an #"..." literal and you should be good to go.
That's about all there is to it.
Spaces are not ignored in regular expressions by default, so for each space in your current expressions it is looking for a space in that string. Add the RegexOptions.IgnorePatternWhitespace flag or remove the spaces from your expressions.
You will also want to add some beginning and end of string anchors (^ and $ respectively) so you do not match just part of a string.
You should avoid having spaces in your regular expressions unless you explicitly set IgnorePatterWhiteSpace. To make sure you get only matches on complete words you should include the beginning of line (^) and end of line ($) characters. I would also suggest you build the entire expression pattern instead of using String.Format("({0}) | ({1})", ...) as you have here.
The below should work given your examples:
string pattern = #"(?:^[a-zA-Z_][a-zA-Z_\d]*)|(?:^\d+(?:\.\d+){0,1}(?:[Ee][\+-]\d+){0,1}$)";
Broadly: how do I match a word with regex rules for a)the beginning, b)the whole word, and c)the end?
More specifically: How do I match an expression of length >= 1 that has the following rules:
It cannot have any of: ! # #
It cannot begin with a space or =
It cannot end with a space
I tried:
^[^\s=][^!##]*[^\s]$
But the ^[^\s=] matching moves past the first character in the word. Hence this also matches words that begin with '!' or '#' or '#' (eg: '#ab' or '#aa'). This also forces the word to have at least 2 characters (one beginning character that is not space or = -and- one non-space character in the end).
I got to:
^[^\s=(!##)]\1*$
for a regex matching the first two rules. But how do I match no trailing spaces in the word with allowing words of length 1?
Cameron's solution is both accurate and efficient (and should be used for any production code where speed needs to be optimized). The answer presented here is less efficient, but demonstrates a general approach for applying logic using regular expressions.
You can use multiple positive and negative lookahead regex assertions (all applied at one location in the target string - typically the beginning), to apply multiple logical constraints for a match. The commented regex below demonstrates how easy this is to do for this example case. You do need to understand how the regex engine actually matches (and doesn't match), to come up with the correct expressions, but its not hard once you get the hang of it.
foundMatch = Regex.IsMatch(subjectString, #"
# Match 'word' meeting multiple logical constraints.
^ # Anchor to start of string.
(?=[^!##]*$) # It cannot have any of: ! # #, AND
(?![ =]) # It cannot begin with a space or =, AND
(?!.*\S$) # It cannot end with a space, AND
.{1,} # length >= 1 (ok to match special 'word')
\z # Anchor to end of string.
",
RegexOptions.IgnorePatternWhitespace);
This application of "regex-logic" is frequently used for complex password validation.
Your first attempt was very close. You only need to exclude more characters for the first and last parts, and make the last two parts optional:
^[^\s=!##](?:[^!##]*[^\s!##])?$
This ensures that all three sections will not include any of !##. Then, if the word is more than one character long, it will need to end with a not-space, with only select characters filling the space in-between. This is all enforced properly because of the ^ and $ anchors.
I'm not quite sure what your second example matched, since the () should be taken as literal characters when embedded within a character class, not as a capturing group.