For example, this regex
(.*)<FooBar>
will match:
abcde<FooBar>
But how do I get it to match across multiple lines?
abcde
fghij<FooBar>
Try this:
((.|\n)*)<FooBar>
It basically says "any character or a newline" repeated zero or more times.
It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:
/(.*)<FooBar>/s
The s at the end causes the dot to match all characters including newlines.
The question is, can the . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.
A special note about lua-patterns: they are not considered regular expressions, but . matches any character there, the same as POSIX-based engines.
Another note on matlab and octave: the . matches any character by default (demo): str = "abcde\n fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcde\n fghij item).
Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).
As for oracle (it is POSIX based), use the n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual
POSIX-based engines:
A mere . already matches line breaks, so there isn't a need to use any modifiers, see bash (demo).
The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.
However, most POSIX-based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:
sed - There are multiple workarounds. The most precise, but not very safe, is sed 'H;1h;$!d;x; s/\(.*\)><Foobar>/\1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.
perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (\n\n) as the record separator.
gnu-grep - grep -Poz '(?si)abc\K.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, \K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.
pcregrep - pcregrep -Mi "(?si)abc\K.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for macOS grep users.
See demos.
Non-POSIX-based engines:
php - Use the s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)
c# - Use RegexOptions.Singleline flag (demo): - var result = Regex.Match(s, #"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;- var result = Regex.Match(s, #"(?s)(.*)<Foobar>").Groups[1].Value;
powershell - Use the (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
perl - Use the s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
python - Use the re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))
java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
kotlin - Use RegexOption.DOT_MATCHES_ALL : "(.*)<FooBar>".toRegex(RegexOption.DOT_MATCHES_ALL)
groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcde\n fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
javascript - Use [^] or workarounds [\d\D] / [\w\W] / [\s\S] (demo): s.match(/([\s\S]*)<FooBar>/)[1]
c++ (std::regex) Use [\s\S] or the JavaScript workarounds (demo): regex rex(R"(([\s\S]*)<FooBar>)");
vba vbscript - Use the same approach as in JavaScript, ([\s\S]*)<Foobar>. (NOTE: The MultiLine property of the RegExp object is sometimes erroneously thought to be the option to allow . match across line breaks, while, in fact, it only changes the ^ and $ behavior to match start/end of lines rather than strings, the same as in JavaScript regex)
behavior.)
ruby - Use the /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
rtrebase-r - Base R PCRE regexps - use (?s): regmatches(x, regexec("(?s)(.*)<FooBar>",x, perl=TRUE))[[1]][2] (demo)
ricustringrstringi - in stringr/stringi regex funtions that are powered with the ICU regex engine. Also use (?s): stringr::str_match(x, "(?s)(.*)<FooBar>")[,2] (demo)
go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
objective-c - The same as Swift. (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern options:NSRegularExpressionDotMatchesLineSeparators error:®exError];
re2, google-apps-script - Use the (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))
NOTES ON (?s):
In most non-POSIX engines, the (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.
If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those .s will be affected that are located to the right of it unless this is a pattern passed to Python's re. In Python re, regardless of the (?s) location, the whole pattern . is affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g., Delim1(?s:.*?)\nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).
POSIX note:
In non-POSIX regex engines, to match any character, [\s\S] / [\d\D] / [\w\W] constructs can be used.
In POSIX, [\s\S] is not matching any character (as in JavaScript or any non-POSIX engine), because regex escape sequences are not supported inside bracket expressions. [\s\S] is parsed as bracket expressions that match a single character, \ or s or S.
If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:
(?s).*<FooBar>
In many regex dialects, /[\S\s]*<Foobar>/ will do just what you want. Source
([\s\S]*)<FooBar>
The dot matches all except newlines (\r\n). So use \s\S, which will match ALL characters.
We can also use
(.*?\n)*?
to match everything including newline without being greedy.
This will make the new line optional
(.*?|\n)*?
In Ruby you can use the 'm' option (multiline):
/YOUR_REGEXP/m
See the Regexp documentation on ruby-doc.org for more information.
"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [\S\s].
For Eclipse, the following expression worked:
Foo
jadajada Bar"
Regular expression:
Foo[\S\s]{1,10}.*Bar*
Note that (.|\n)* can be less efficient than (for example) [\s\S]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.
Use:
/(.*)<FooBar>/s
The s causes dot (.) to match carriage returns.
Use RegexOptions.Singleline. It changes the meaning of . to include newlines.
Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);
In notepad++ you can use this
<table (.|\r\n)*</table>
It will match the entire table starting from
rows and columns
You can make it greedy, using the following, that way it will match the first, second and so forth tables and not all at once
<table (.|\r\n)*?</table>
In a Java-based regular expression, you can use [\s\S].
This works for me and is the simplest one:
(\X*)<FooBar>
Generally, . doesn't match newlines, so try ((.|\n)*)<foobar>.
In JavaScript you can use [^]* to search for zero to infinite characters, including line breaks.
$("#find_and_replace").click(function() {
var text = $("#textarea").val();
search_term = new RegExp("[^]*<Foobar>", "gi");;
replace_term = "Replacement term";
var new_text = text.replace(search_term, replace_term);
$("#textarea").val(new_text);
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<button id="find_and_replace">Find and replace</button>
<br>
<textarea ID="textarea">abcde
fghij<Foobar></textarea>
Solution:
Use pattern modifier sU will get the desired matching in PHP.
Example:
preg_match('/(.*)/sU', $content, $match);
Sources:
Pattern Modifiers
In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.
In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcde\nfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.
Line-based regular expression use is usually for command line things like egrep.
Try: .*\n*.*<FooBar> assuming you are also allowing blank newlines. As you are allowing any character including nothing before <FooBar>.
I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:
mystring = Regex.Replace(mystring, "\r\n", "")
I am manipulating HTML so line breaks don't really matter to me in this case.
I tried all of the suggestions above with no luck. I am using .NET 3.5 FYI.
I wanted to match a particular if block in Java:
...
...
if(isTrue){
doAction();
}
...
...
}
If I use the regExp
if \(isTrue(.|\n)*}
it included the closing brace for the method block, so I used
if \(!isTrue([^}.]|\n)*}
to exclude the closing brace from the wildcard match.
Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an XML element:
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>81</PercentComplete>
</TASK>
Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including \n till .PercentCompleted.. The regular expression pattern and the replace specification are:
String hw = new String("<TASK>\n <UID>21</UID>\n <Name>Architectural design</Name>\n <PercentComplete>81</PercentComplete>\n</TASK>");
String pattern = new String ("(<UID>21</UID>)((.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
String replaceSpec = new String ("$1$2$440$6");
// Note that the group (<PercentComplete>) is $4 and the group ((.|\n)*?) is $2.
String iw = hw.replaceFirst(pattern, replaceSpec);
System.out.println(iw);
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>40</PercentComplete>
</TASK>
The subgroup (.|\n) is probably the missing group $3. If we make it non-capturing by (?:.|\n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:
pattern = new String("(<UID>21</UID>)((?:.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
replaceSpec = new String("$1$2$340$5")
and the replacement works correctly as before.
Typically searching for three consecutive lines in PowerShell, it would look like:
$file = Get-Content file.txt -raw
$pattern = 'lineone\r\nlinetwo\r\nlinethree\r\n' # "Windows" text
$pattern = 'lineone\nlinetwo\nlinethree\n' # "Unix" text
$pattern = 'lineone\r?\nlinetwo\r?\nlinethree\r?\n' # Both
$file -match $pattern
# output
True
Bizarrely, this would be Unix text at the prompt, but Windows text in a file:
$pattern = 'lineone
linetwo
linethree
'
Here's a way to print out the line endings:
'lineone
linetwo
linethree
' -replace "`r",'\r' -replace "`n",'\n'
# Output
lineone\nlinetwo\nlinethree\n
Option 1
One way would be to use the s flag (just like the accepted answer):
/(.*)<FooBar>/s
Demo 1
Option 2
A second way would be to use the m (multiline) flag and any of the following patterns:
/([\s\S]*)<FooBar>/m
or
/([\d\D]*)<FooBar>/m
or
/([\w\W]*)<FooBar>/m
Demo 2
RegEx Circuit
jex.im visualizes regular expressions:
Related
My regex pattern looks something like
<xxxx location="file path/level1/level2" xxxx some="xxx">
I am only interested in the part in quotes assigned to location. Shouldn't it be as easy as below without the greedy switch?
/.*location="(.*)".*/
Does not seem to work.
You need to make your regular expression lazy/non-greedy, because by default, "(.*)" will match all of "file path/level1/level2" xxx some="xxx".
Instead you can make your dot-star non-greedy, which will make it match as few characters as possible:
/location="(.*?)"/
Adding a ? on a quantifier (?, * or +) makes it non-greedy.
Note: this is only available in regex engines which implement the Perl 5 extensions (Java, Ruby, Python, etc) but not in "traditional" regex engines (including Awk, sed, grep without -P, etc.).
location="(.*)" will match from the " after location= until the " after some="xxx unless you make it non-greedy.
So you either need .*? (i.e. make it non-greedy by adding ?) or better replace .* with [^"]*.
[^"] Matches any character except for a " <quotation-mark>
More generic: [^abc] - Matches any character except for an a, b or c
How about
.*location="([^"]*)".*
This avoids the unlimited search with .* and will match exactly to the first quote.
Use non-greedy matching, if your engine supports it. Add the ? inside the capture.
/location="(.*?)"/
Use of Lazy quantifiers ? with no global flag is the answer.
Eg,
If you had global flag /g then, it would have matched all the lowest length matches as below.
Here's another way.
Here's the one you want. This is lazy [\s\S]*?
The first item:
[\s\S]*?(?:location="[^"]*")[\s\S]* Replace with: $1
Explaination: https://regex101.com/r/ZcqcUm/2
For completeness, this gets the last one. This is greedy [\s\S]*
The last item:[\s\S]*(?:location="([^"]*)")[\s\S]*
Replace with: $1
Explaination: https://regex101.com/r/LXSPDp/3
There's only 1 difference between these two regular expressions and that is the ?
The other answers here fail to spell out a full solution for regex versions which don't support non-greedy matching. The greedy quantifiers (.*?, .+? etc) are a Perl 5 extension which isn't supported in traditional regular expressions.
If your stopping condition is a single character, the solution is easy; instead of
a(.*?)b
you can match
a[^ab]*b
i.e specify a character class which excludes the starting and ending delimiiters.
In the more general case, you can painstakingly construct an expression like
start(|[^e]|e(|[^n]|n(|[^d])))end
to capture a match between start and the first occurrence of end. Notice how the subexpression with nested parentheses spells out a number of alternatives which between them allow e only if it isn't followed by nd and so forth, and also take care to cover the empty string as one alternative which doesn't match whatever is disallowed at that particular point.
Of course, the correct approach in most cases is to use a proper parser for the format you are trying to parse, but sometimes, maybe one isn't available, or maybe the specialized tool you are using is insisting on a regular expression and nothing else.
Because you are using quantified subpattern and as descried in Perl Doc,
By default, a quantified subpattern is "greedy", that is, it will
match as many times as possible (given a particular starting location)
while still allowing the rest of the pattern to match. If you want it
to match the minimum number of times possible, follow the quantifier
with a "?" . Note that the meanings don't change, just the
"greediness":
*? //Match 0 or more times, not greedily (minimum matches)
+? //Match 1 or more times, not greedily
Thus, to allow your quantified pattern to make minimum match, follow it by ? :
/location="(.*?)"/
import regex
text = 'ask her to call Mary back when she comes back'
p = r'(?i)(?s)call(.*?)back'
for match in regex.finditer(p, str(text)):
print (match.group(1))
Output:
Mary
I need to identify matching course number that have xx.3xxxxxx.
These are some examples of the course numbers.
26.3730004
27.0210000
26.3730009
26.7114001
23.9610071
26.0A34430
23.3670005
26.0B05430
I tried many patterns one example I used is the pattern below. It did not get any match.
"[^0-9]{2}\Q.\E3[^0-9]+$"
I tried using grep and grepl. I actually need the code to return indexes.
This code shows my attempt to tag the rows that have matches.
Teacher$virtual[
which(
grepl("[^0-9]{2}\\Q.\\E3[^0-9]+$",Teacher$CourseNumber))]
<- "1"
I need to remove any row from my dataframe that have the course number with that pattern. XX.3XXXXXX
But, my code did not find any match. Can you please help me?
You should use
grepl("^[0-9]{2}\\.3", Teacher$CourseNumber)
See the regex graph:
Details:
^ - start of a string
[0-9]{2} - two digits
\\. - a dot (note that a regex escape is a literal backslash, but inside a string literal, "...", a single backslash is used to form string escape sequences, hence the backslash must be double to obtain a literal backslash char necessary for a regex escape)
3 - a 3 char.
NOTE: If you want to use in-pattern quoting with \Q and \E (in between which all chars are treated literally) you need to use PCRE regex, add perl=TRUE and use
grepl("^[0-9]{2}\\Q.\\E3", Teacher$CourseNumber, perl=TRUE)
Now, the dot is treated as a literal dot, not a . metacharacter that matches any char but a line break char (in a PCRE regex, . does not match line break chars by default).
Here, this simple expression would likely cover that:
^[0-9]{2}\.[3].+$
which has a [3] boundary right after the .. It would probably work without start and end anchors:
[0-9]{2}\.[3].+
Demo
We can add or reduce the boundaries, if it'd be necessary.
I experimented with a pattern to reject any line beginning with 0 to N spaces followed by a "#" . At this test webpage, the following pattern works fine:
^(?!(\s*[#])).*
I used as test lines of text the following:
#tbadword
#test
one two
abadwo#rds
#three
And only the "non-comment" lines are selected.
But in R, using the Windows Rgui , if I try
> history(Inf, pattern = '^(?!(\\s*[#])).*' )
I get the error message "Invalid regexp" .
Can someone point out what R is unhappy with here? Do I need to set a global "perl=TRUE" or some such thing? Or is there a simpler way to do this?
The history() command has a ... for values that will be passed to grep(), so you can use the invert= flag rather than a look-ahead to find what you need. How about
history(Inf, pattern="^\\s*#", invert=TRUE)
You may have R history parse your regex with a PCRE regex engine:
history(Inf, pattern="^(?!\\s*#)", perl=TRUE)
Now, ^(?!\s*#) will be parsed correctly as
^ - start of string
(?!\s*#) - a negative lookahead that fails the match if, immediately to the right of the current location (i.e. at the start of the string) there are 0+ whitespaces and then #.
Although the solution with invert=TRUE and an opposite regex is more natural for the current scenario, you may need the more advanced regex functionality for other cases, and perl=TRUE will help cover them.
Simple regex question. I have a string on the following format:
this is a [sample] string with [some] special words. [another one]
What is the regular expression to extract the words within the square brackets, ie.
sample
some
another one
Note: In my use case, brackets cannot be nested.
You can use the following regex globally:
\[(.*?)\]
Explanation:
\[ : [ is a meta char and needs to be escaped if you want to match it literally.
(.*?) : match everything in a non-greedy way and capture it.
\] : ] is a meta char and needs to be escaped if you want to match it literally.
(?<=\[).+?(?=\])
Will capture content without brackets
(?<=\[) - positive lookbehind for [
.*? - non greedy match for the content
(?=\]) - positive lookahead for ]
EDIT: for nested brackets the below regex should work:
(\[(?:\[??[^\[]*?\]))
This should work out ok:
\[([^]]+)\]
Can brackets be nested?
If not: \[([^]]+)\] matches one item, including square brackets. Backreference \1 will contain the item to be match. If your regex flavor supports lookaround, use
(?<=\[)[^]]+(?=\])
This will only match the item inside brackets.
To match a substring between the first [ and last ], you may use
\[.*\] # Including open/close brackets
\[(.*)\] # Excluding open/close brackets (using a capturing group)
(?<=\[).*(?=\]) # Excluding open/close brackets (using lookarounds)
See a regex demo and a regex demo #2.
Use the following expressions to match strings between the closest square brackets:
Including the brackets:
\[[^][]*] - PCRE, Python re/regex, .NET, Golang, POSIX (grep, sed, bash)
\[[^\][]*] - ECMAScript (JavaScript, C++ std::regex, VBA RegExp)
\[[^\]\[]*] - Java, ICU regex
\[[^\]\[]*\] - Onigmo (Ruby, requires escaping of brackets everywhere)
Excluding the brackets:
(?<=\[)[^][]*(?=]) - PCRE, Python re/regex, .NET (C#, etc.), JGSoft Software
\[([^][]*)] - Bash, Golang - capture the contents between the square brackets with a pair of unescaped parentheses, also see below
\[([^\][]*)] - JavaScript, C++ std::regex, VBA RegExp
(?<=\[)[^\]\[]*(?=]) - Java regex, ICU (R stringr)
(?<=\[)[^\]\[]*(?=\]) - Onigmo (Ruby, requires escaping of brackets everywhere)
NOTE: * matches 0 or more characters, use + to match 1 or more to avoid empty string matches in the resulting list/array.
Whenever both lookaround support is available, the above solutions rely on them to exclude the leading/trailing open/close bracket. Otherwise, rely on capturing groups (links to most common solutions in some languages have been provided).
If you need to match nested parentheses, you may see the solutions in the Regular expression to match balanced parentheses thread and replace the round brackets with the square ones to get the necessary functionality. You should use capturing groups to access the contents with open/close bracket excluded:
\[((?:[^][]++|(?R))*)] - PHP PCRE
\[((?>[^][]+|(?<o>)\[|(?<-o>]))*)] - .NET demo
\[(?:[^\]\[]++|(\g<0>))*\] - Onigmo (Ruby) demo
If you do not want to include the brackets in the match, here's the regex: (?<=\[).*?(?=\])
Let's break it down
The . matches any character except for line terminators. The ?= is a positive lookahead. A positive lookahead finds a string when a certain string comes after it. The ?<= is a positive lookbehind. A positive lookbehind finds a string when a certain string precedes it. To quote this,
Look ahead positive (?=)
Find expression A where expression B follows:
A(?=B)
Look behind positive (?<=)
Find expression A where expression B
precedes:
(?<=B)A
The Alternative
If your regex engine does not support lookaheads and lookbehinds, then you can use the regex \[(.*?)\] to capture the innards of the brackets in a group and then you can manipulate the group as necessary.
How does this regex work?
The parentheses capture the characters in a group. The .*? gets all of the characters between the brackets (except for line terminators, unless you have the s flag enabled) in a way that is not greedy.
Just in case, you might have had unbalanced brackets, you can likely design some expression with recursion similar to,
\[(([^\]\[]+)|(?R))*+\]
which of course, it would relate to the language or RegEx engine that you might be using.
RegEx Demo 1
Other than that,
\[([^\]\[\r\n]*)\]
RegEx Demo 2
or,
(?<=\[)[^\]\[\r\n]*(?=\])
RegEx Demo 3
are good options to explore.
If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
RegEx Circuit
jex.im visualizes regular expressions:
Test
const regex = /\[([^\]\[\r\n]*)\]/gm;
const str = `This is a [sample] string with [some] special words. [another one]
This is a [sample string with [some special words. [another one
This is a [sample[sample]] string with [[some][some]] special words. [[another one]]`;
let m;
while ((m = regex.exec(str)) !== null) {
// This is necessary to avoid infinite loops with zero-width matches
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`Found match, group ${groupIndex}: ${match}`);
});
}
Source
Regular expression to match balanced parentheses
(?<=\[).*?(?=\]) works good as per explanation given above. Here's a Python example:
import re
str = "Pagination.go('formPagination_bottom',2,'Page',true,'1',null,'2013')"
re.search('(?<=\[).*?(?=\])', str).group()
"'formPagination_bottom',2,'Page',true,'1',null,'2013'"
The #Tim Pietzcker's answer here
(?<=\[)[^]]+(?=\])
is almost the one I've been looking for. But there is one issue that some legacy browsers can fail on positive lookbehind.
So I had to made my day by myself :). I manged to write this:
/([^[]+(?=]))/g
Maybe it will help someone.
console.log("this is a [sample] string with [some] special words. [another one]".match(/([^[]+(?=]))/g));
if you want fillter only small alphabet letter between square bracket a-z
(\[[a-z]*\])
if you want small and caps letter a-zA-Z
(\[[a-zA-Z]*\])
if you want small caps and number letter a-zA-Z0-9
(\[[a-zA-Z0-9]*\])
if you want everything between square bracket
if you want text , number and symbols
(\[.*\])
This code will extract the content between square brackets and parentheses
(?:(?<=\().+?(?=\))|(?<=\[).+?(?=\]))
(?: non capturing group
(?<=\().+?(?=\)) positive lookbehind and lookahead to extract the text between parentheses
| or
(?<=\[).+?(?=\]) positive lookbehind and lookahead to extract the text between square brackets
In R, try:
x <- 'foo[bar]baz'
str_replace(x, ".*?\\[(.*?)\\].*", "\\1")
[1] "bar"
([[][a-z \s]+[]])
Above should work given the following explaination
characters within square brackets[] defines characte class which means pattern should match atleast one charcater mentioned within square brackets
\s specifies a space
+ means atleast one of the character mentioned previously to +.
I needed including newlines and including the brackets
\[[\s\S]+\]
If someone wants to match and select a string containing one or more dots inside square brackets like "[fu.bar]" use the following:
(?<=\[)(\w+\.\w+.*?)(?=\])
Regex Tester
What is the difference between below two regular expressions
(.|[\r\n]){1,1500}
^.{1,1500}$
The first matches up-to-1500 chars, and the second (assuming you haven't set certain regex options) matches a first single line of up-to-1500 chars, with no newlines.
. does not match new lines.
The second one matches the first 1500 characteres of a line IF the line contains 1500 characters or less
First expression matches some <= 1500 characters of the file(or other source).
Second expression matches a entire line with charsNumber <= 1500.
. matches any character except \n newline.
If it's for use in a RegularExpressionValidator, you probably want to use this regex:
^[\s\S]{1,1500}$
This is because the regex may be run on either the server (.NET) or the client (JavaScript). In .NET regexes you can use the RegexOptions.Singleline flag (or its inline equivalent, (?s)) to make the dot match newlines, but JavaScript has no such mechanism.
[\s\S] matches any whitespace character or anything that's not a whitespace character--in other words, anything. It's the most popular idiom for matching anything including a newline in JavaScript; it's much, much more efficient than alternation-based approaches like (.|\n).
Note that you'll still need to use a RequiredFieldValidator if you don't want the user to leave the textbox empty.