My string is like
5blog5sett5ings5[5slider5][5][5ima5ge5]
I like to match any digit into second brackets from end by regular expression.
For this case, my target digit is 5 into [5].
I like to select where before pattern like
5blog5sett5ings5[5slider5][ and after pattern like ][5ima5ge5]
I will use it for JavaScript string replace. Text can be different but the before and after patterns are like that. For better understanding see the image.
I tried something like
(?<=.+[.+?][)\d(?=][.+?])
but did not work.
I think you could just use a lookahead to check if there ] and one more [ ] ahead until end.
\d+(?=]\[[^\]]*]$)
See demo at regex101
(be aware that lookbehind you tried is not available in js regex)
I guess you can use:
\[(\d+)\][^\]]+]$
Regex Demo & Explanation
var myString = "5blog5sett5ings5[5slider5][5][5ima5ge5]";
var myRegexp = /\[(\d+)\][^\]]+]$/mg;
var match = myRegexp.exec(myString);
console.log(match[1]);
Use this:
^.*\[(\d*)\]\[[^\]]*\]$
Where:
^ is the begin of the string
.* means any character
\[ and \] matches literal squared brackets
(\d*) is what you want to match
[^\]]* is the content of the last couple of brackets
$ is the end of the string
See an example:
var str = "5blog5sett5ings5[5slider5][5][5ima5ge5]";
var res = str.match(/^.*\[(\d*)\]\[[^\]]*\]$/);
console.log(res[1])
I'm a regular expression newbie and I can't quite figure out how to write a single regular expression that would "match" any duplicate consecutive words such as:
Paris in the the spring.
Not that that is related.
Why are you laughing? Are my my regular expressions THAT bad??
Is there a single regular expression that will match ALL of the bold strings above?
Try this regular expression:
\b(\w+)\s+\1\b
Here \b is a word boundary and \1 references the captured match of the first group.
Regex101 example here
I believe this regex handles more situations:
/(\b\S+\b)\s+\b\1\b/
A good selection of test strings can be found here: http://callumacrae.github.com/regex-tuesday/challenge1.html
The below expression should work correctly to find any number of duplicated words. The matching can be case insensitive.
String regex = "\\b(\\w+)(\\s+\\1\\b)+";
Pattern p = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(input);
// Check for subsequences of input that match the compiled pattern
while (m.find()) {
input = input.replaceAll(m.group(0), m.group(1));
}
Sample Input : Goodbye goodbye GooDbYe
Sample Output : Goodbye
Explanation:
The regex expression:
\b : Start of a word boundary
\w+ : Any number of word characters
(\s+\1\b)* : Any number of space followed by word which matches the previous word and ends the word boundary. Whole thing wrapped in * helps to find more than one repetitions.
Grouping :
m.group(0) : Shall contain the matched group in above case Goodbye goodbye GooDbYe
m.group(1) : Shall contain the first word of the matched pattern in above case Goodbye
Replace method shall replace all consecutive matched words with the first instance of the word.
Try this with below RE
\b start of word word boundary
\W+ any word character
\1 same word matched already
\b end of word
()* Repeating again
public static void main(String[] args) {
String regex = "\\b(\\w+)(\\b\\W+\\b\\1\\b)*";// "/* Write a RegEx matching repeated words here. */";
Pattern p = Pattern.compile(regex, Pattern.CASE_INSENSITIVE/* Insert the correct Pattern flag here.*/);
Scanner in = new Scanner(System.in);
int numSentences = Integer.parseInt(in.nextLine());
while (numSentences-- > 0) {
String input = in.nextLine();
Matcher m = p.matcher(input);
// Check for subsequences of input that match the compiled pattern
while (m.find()) {
input = input.replaceAll(m.group(0),m.group(1));
}
// Prints the modified sentence.
System.out.println(input);
}
in.close();
}
Regex to Strip 2+ duplicate words (consecutive/non-consecutive words)
Try this regex that can catch 2 or more duplicate words and only leave behind one single word. And the duplicate words need not even be consecutive.
/\b(\w+)\b(?=.*?\b\1\b)/ig
Here, \b is used for Word Boundary, ?= is used for positive lookahead, and \1 is used for back-referencing.
Example
Source
The widely-used PCRE library can handle such situations (you won't achieve the the same with POSIX-compliant regex engines, though):
(\b\w+\b)\W+\1
Here is one that catches multiple words multiple times:
(\b\w+\b)(\s+\1)+
No. That is an irregular grammar. There may be engine-/language-specific regular expressions that you can use, but there is no universal regular expression that can do that.
This is the regex I use to remove duplicate phrases in my twitch bot:
(\S+\s*)\1{2,}
(\S+\s*) looks for any string of characters that isn't whitespace, followed whitespace.
\1{2,} then looks for more than 2 instances of that phrase in the string to match. If there are 3 phrases that are identical, it matches.
Since some developers are coming to this page in search of a solution which not only eliminates duplicate consecutive non-whitespace substrings, but triplicates and beyond, I'll show the adapted pattern.
Pattern: /(\b\S+)(?:\s+\1\b)+/ (Pattern Demo)
Replace: $1 (replaces the fullstring match with capture group #1)
This pattern greedily matches a "whole" non-whitespace substring, then requires one or more copies of the matched substring which may be delimited by one or more whitespace characters (space, tab, newline, etc).
Specifically:
\b (word boundary) characters are vital to ensure partial words are not matched.
The second parenthetical is a non-capturing group, because this variable width substring does not need to be captured -- only matched/absorbed.
the + (one or more quantifier) on the non-capturing group is more appropriate than * because * will "bother" the regex engine to capture and replace singleton occurrences -- this is wasteful pattern design.
*note if you are dealing with sentences or input strings with punctuation, then the pattern will need to be further refined.
The example in Javascript: The Good Parts can be adapted to do this:
var doubled_words = /([A-Za-z\u00C0-\u1FFF\u2800-\uFFFD]+)\s+\1(?:\s|$)/gi;
\b uses \w for word boundaries, where \w is equivalent to [0-9A-Z_a-z]. If you don't mind that limitation, the accepted answer is fine.
This expression (inspired from Mike, above) seems to catch all duplicates, triplicates, etc, including the ones at the end of the string, which most of the others don't:
/(^|\s+)(\S+)(($|\s+)\2)+/g, "$1$2")
I know the question asked to match duplicates only, but a triplicate is just 2 duplicates next to each other :)
First, I put (^|\s+) to make sure it starts with a full word, otherwise "child's steak" would go to "child'steak" (the "s"'s would match). Then, it matches all full words ((\b\S+\b)), followed by an end of string ($) or a number of spaces (\s+), the whole repeated more than once.
I tried it like this and it worked well:
var s = "here here here here is ahi-ahi ahi-ahi ahi-ahi joe's joe's joe's joe's joe's the result result result";
print( s.replace( /(\b\S+\b)(($|\s+)\1)+/g, "$1"))
--> here is ahi-ahi joe's the result
Try this regular expression it fits for all repeated words cases:
\b(\w+)\s+\1(?:\s+\1)*\b
I think another solution would be to use named capture groups and backreferences like this:
.* (?<mytoken>\w+)\s+\k<mytoken> .*/
OR
.*(?<mytoken>\w{3,}).+\k<mytoken>.*/
Kotlin:
val regex = Regex(""".* (?<myToken>\w+)\s+\k<myToken> .*""")
val input = "This is a test test data"
val result = regex.find(input)
println(result!!.groups["myToken"]!!.value)
Java:
var pattern = Pattern.compile(".* (?<myToken>\\w+)\\s+\\k<myToken> .*");
var matcher = pattern.matcher("This is a test test data");
var isFound = matcher.find();
var result = matcher.group("myToken");
System.out.println(result);
JavaScript:
const regex = /.* (?<myToken>\w+)\s+\k<myToken> .*/;
const input = "This is a test test data";
const result = regex.exec(input);
console.log(result.groups.myToken);
// OR
const regex = /.* (?<myToken>\w+)\s+\k<myToken> .*/g;
const input = "This is a test test data";
const result = [...input.matchAll(regex)];
console.log(result[0].groups.myToken);
All the above detect the test as the duplicate word.
Tested with Kotlin 1.7.0-Beta, Java 11, Chrome and Firefox 100.
You can use this pattern:
\b(\w+)(?:\W+\1\b)+
This pattern can be used to match all duplicated word groups in sentences. :)
Here is a sample util function written in java 17, which replaces all duplications with the first occurrence:
public String removeDuplicates(String input) {
var regex = "\\b(\\w+)(?:\\W+\\1\\b)+";
var pattern = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
var matcher = pattern.matcher(input);
while (matcher.find()) {
input = input.replaceAll(matcher.group(), matcher.group(1));
}
return input;
}
As far as I can see, none of these would match:
London in the
the winter (with the winter on a new line )
Although matching duplicates on the same line is fairly straightforward,
I haven't been able to come up with a solution for the situation in which they
stretch over two lines. ( with Perl )
To find duplicate words that have no leading or trailing non whitespace character(s) other than a word character(s), you can use whitespace boundaries on the left and on the right making use of lookarounds.
The pattern will have a match in:
Paris in the the spring.
Not that that is related.
The pattern will not have a match in:
This is $word word
(?<!\S)(\w+)\s+\1(?!\S)
Explanation
(?<!\S) Negative lookbehind, assert not a non whitespace char to the left of the current location
(\w+) Capture group 1, match 1 or more word characters
\s+ Match 1 or more whitespace characters (note that this can also match a newline)
\1 Backreference to match the same as in group 1
(?!\S) Negative lookahead, assert not a non whitespace char to the right of the current location
See a regex101 demo.
To find 2 or more duplicate words:
(?<!\S)(\w+)(?:\s+\1)+(?!\S)
This part of the pattern (?:\s+\1)+ uses a non capture group to repeat 1 or more times matching 1 or more whitespace characters followed by the backreference to match the same as in group 1.
See a regex101 demo.
Alternatives without using lookarounds
You could also make use of a leading and trailing alternation matching either a whitespace char or assert the start/end of the string.
Then use a capture group 1 for the value that you want to get, and use a second capture group with a backreference \2 to match the repeated word.
Matching 2 duplicate words:
(?:\s|^)((\w+)\s+\2)(?:\s|$)
See a regex101 demo.
Matching 2 or more duplicate words:
(?:\s|^)((\w+)(?:\s+\2)+)(?:\s|$)
See a regex101 demo.
Use this in case you want case-insensitive checking for duplicate words.
(?i)\\b(\\w+)\\s+\\1\\b
I have to write a regex that contains either the words “wind”, “temp”, or “press” followed by a non-digit.
So far I have:
var regex = /wind|temp|press[^0-9]{0,}/;
This doesn’t work because the [^0-9]{0,} is with “press”. How would I separate them so that all the words would be read followed by a non digit?
Just use a (non-capturing) group:
var regex = /(?:wind|temp|press)[^0-9]{0,}/;
All you need is a none capturing group to separates those words from the rest of your pattern. Use this pattern:
/(?:wind|temp|press)\D*/
By the way {0,} is the same as * in this case. Also if being a non-digit character is mandatory, you probable want to use + instead. (I mean if one of those words must be followed by at least one or more non-digit character, then use +)
Online Demo
I am trying to split a string using a regular expression for links (urls).
The regex in question is
var regex = new RegExp('(?:^(?:(?:[a-z]+:)?//)(?:\S+(?::\S*)?#)?(?:localhost|(?:(?:[a-z\u00a1-\uffff0-9]-*)*[a-z\u00a1-\uffff0-9]+)(?:\.(?:[a-z\u00a1-\uffff0-9]-*)*[a-z\u00a1-\uffff0-9]+)*(?:\.(?:[a-z\u00a1-\uffff]{2,})))(?::\d{2,5})?(?:[/?#]\S*)?$)')
If i do
regex.test("https://google.com"); // returns true
but doing -
"Go to https://google.com".split(regex);
// return ["Go to https://google.com"]
Whereas i expect it to return
["Go to ", "https://google.com"]
Any idea what's going on here?
First of all, you're using a string literal to build your regex, which means that you have to escape your backslashes (since a backslash has a special meaning in strings, used for the line feed char \n for example):
var regex = new RegExp('(?:^(?:(?:[a-z]+:)?//)(?:\\S+(?::\\S*)?#)?(?:localhost|(?:(?:[a-z\\u00a1-\\uffff0-9]-*)*[a-z\\u00a1-\\uffff0-9]+)(?:\\.(?:[a-z\\u00a1-\\uffff0-9]-*)*[a-z\\u00a1-\\uffff0-9]+)*(?:\\.(?:[a-z\\u00a1-\\uffff]{2,})))(?::\\d{2,5})?(?:[/?#]\\S*)?$)');
Another solution would be to use the regex literal, as JavaScript proposes one, but you would then have to escape the slashes:
var regex = /(?:^(?:(?:[a-z]+:)?\/\/)(?:\S+(?::\S*)?#)?(?:localhost|(?:(?:[a-z\u00a1-\uffff0-9]-*)*[a-z\u00a1-\uffff0-9]+)(?:\.(?:[a-z\u00a1-\uffff0-9]-*)*[a-z\u00a1-\uffff0-9]+)*(?:\.(?:[a-z\u00a1-\uffff]{2,})))(?::\d{2,5})?(?:[\/?#]\S*)?$)/;
Then, your regex will try to match against the entire input due to the ^ and $ anchors. So if you remove them (or better, replace them with word boundaries \b), you'll be able to find URLs in a string for example:
var regex = /(?:\b(?:(?:[a-z]+:)?\/\/)(?:\S+(?::\S*)?#)?(?:localhost|(?:(?:[a-z\u00a1-\uffff0-9]-*)*[a-z\u00a1-\uffff0-9]+)(?:\.(?:[a-z\u00a1-\uffff0-9]-*)*[a-z\u00a1-\uffff0-9]+)*(?:\.(?:[a-z\u00a1-\uffff]{2,})))(?::\d{2,5})?(?:[\/?#]\S*)?\b)/;
But, the main point is that you're misunderstanding the split concept. Given the string "hello world", if you split by space, you'll end up with ["hello", "world"]: no more space anymore since it was the char that was used to split.
That is, if you split by the URL regex, the output array won't contain the URLs anymore. It seems to me that a lookahead could suit your needs:
var regex = /(?=(?:\b(?:(?:[a-z]+:)?\/\/)(?:\S+(?::\S*)?#)?(?:localhost|(?:(?:[a-z\u00a1-\uffff0-9]-*)*[a-z\u00a1-\uffff0-9]+)(?:\.(?:[a-z\u00a1-\uffff0-9]-*)*[a-z\u00a1-\uffff0-9]+)*(?:\.(?:[a-z\u00a1-\uffff]{2,})))(?::\d{2,5})?(?:[\/?#]\S*)?\b))/;
"Go to https://google.com".split(regex) // ["Go to ", "https://google.com"]
The regex explained:
(?=(?:\b(?:(?:[a-z]+:)?//)(?:\S+(?::\S*)?#)?(?:localhost|(?:(?:[a-z\u00a1-\uffff0-9]-*)*[a-z\u00a1-\uffff0-9]+)(?:\.(?:[a-z\u00a1-\uffff0-9]-*)*[a-z\u00a1-\uffff0-9]+)*(?:\.(?:[a-z\u00a1-\uffff]{2,})))(?::\d{2,5})?(?:[/?#]\S*)?\b))
Debuggex Demo
By splitting a string with a positive lookahead (?=content_of_lookahead), you'll split by each interchar that is followed by the content of the lookahead.
Take a look at 8 Regular Expressions You Should Know.
To match an url you can use following regex :
var regex = "(https?:\/\/)?([\da-z\.-]+)\.([a-z\.]{2,6})([\/\w# \.-]*)*\/?$";
"Go to https://google.com".split(regex);
// return ["https://google.com"]
Live example.
Hope this helps.
I am sure this is really easy but how do I match
match either start of line or whitespace
match a-z
match either end of line or whitespace
I only want to return item no. 2 so for the following string
"one 1.ignore two 2ignore ignore3 three"
The expression will return
["one","two","three"]
Thanks
You would need lookbehind for a regex that matches these items, which is not supported in javascript. Either you do a manual iteration and extract matching groups (as demonstrated by #Some1.Kill.The.DJ), or you're going to split the string instead of matching:
str.split(/\s+(?:\S*?(?![a-z])\S+\s+)*/);
This expression does match all whitespaces combined with words that contain at least one character that is not [a-z]. However, this regex is complicated and not easy to maintain; also it does yield empty strings sometimes. Better, do something like
str.split(/\s+/).filter(RegExp.prototype.test.bind(/^[a-z]+$/));
Use this code:
var str = 'one 1.ignore two 2ignore ignore3 three';
str = str.replace(/\s(?=[a-z])/ig, function(text, p1) {
return p1 ? p1 : text;
});
var arr = str.match(/([a-z]+)(?=\s|$)/ig);