How to split some strings defined in a specific format:
[length namevalue field]name=value[length namevalue field]name=value[length namevalue field]name=value[length namevalue field]name=value
Is it possible with a Find/Replace regex in Notepad++ isolate the pair name=value replacing [length namevalue field] with a white space?
The main problem is related to numeric value where a simple \d{4} search doesn't work.
Eg.
INPUT:
0010name=mario0013surname=rossi0006age=180006phone=0014address=street
0013name=marianna0013surname=rossi0006age=210006phone=0015address=street1
0003name=pia0015surname=rossini0005age=30017phone=+39221122330020address=streetstreet
OUTPUT:
name=mario surname=rossi age=18 phone= address=street
name=mario surname=rossi age=18 phone= address=street
name=marianna surname=rossi age=21 phone= address=street1
name=pia surname=rossini age=3 phone=+3922112233 address=streetstreet
You can use
\d{4}(?=[[:alpha:]]\w*=)
\d{4}(?=[^\W\d]\w*=)
See the regex demo.
The patterns match
\d{4} - four digits
(?=[[:alpha:]]\w*=) - that are immediately followed with a letter and then any zero or more word chars followed with a = char immediately to the right of the current position.
(?=[^\W\d]\w*=) - that are immediately followed with a letter or an underscore and then any zero or more word chars followed with a = char immediately to the right of the current position.
In Notepad++, if you want to remove the match at the start of the line and replace with space anywhere else, you can use
^(\d{4}(?=[[:alpha:]]\w*=))|(?1)
and replace with (?1: ). The above explained pattern, \d{4}(?=[[:alpha:]]\w*=), is matched and captured into Group 1 if it is at the start of a line (^), and just matched anywhere else ((?1) recurses the Group 1 pattern, so as not to repeat it). The (?1: ) replacement means we replace with empty string if Group 1 matched, else, we replace with a space.
See the demo screenshot:
I want to validate a field of string so that it only accept string that contains words with certain format.
Example accepted string:
#key;
#key1; #key2;#key3;
Example rejected string:
key;
%key1X #key2X$key3X
My regex:
\B(\#[a-zA-Z0-9_; ]+\b)(\;)
It seems my regex still accept a string as long as it has a word with valid format, while I only want it to be accepted if whole words are in the correct format.
Current example:
%key1; %key2 #keysz;#key3; #key4;
From the above Current Example still accepted because it contains #keysz; and #key3; while I want it to be rejected because there are %key1; %key2 and #key4;.
I've do some search and the closest I can found is this question, but it returns similar result as my current regex.
What did i do wrong in my regex? What is the right regex?
Sorry if this is dumb question but I'm a newbie in regex.
The main thing needed are start ^ and end $ anchors. The rest can be simplified too:
^( *#\w+;)+$
See live demo.
Breaking it down:
^ = start
* = 0-n spaces
# = a literal hash (these don't need escaping in regex)
\w+ = one or more word characters (letters, digits and the underscore)`
$
If underscore can be in the input and must not be, then use:
^( *#[A-Za-z0-9]+;)+$
Your regex matches a full sentence because in your regex pattern(\B(\#[a-zA-Z0-9_; ]+\b)(\;)) you haven't specified where the matching process should start and end. So regex engine will try to match every position of the string on which you run the regex.match.
The way to specify where regex should try to match is done by adding anchors(^-beginning and $-end) to regex pattern.
You can edit your pattern to look like this: /(?:\s|^)(#[a-zA-Z0-9_; ]+?);(?:\s|$)/gm
Explanation:
/(?:\s|^)
- (?: means a non capture group, means dont include whatever is matched in between these () in the result. \s|^ means start matching if the beginning is a white space or beginning of a string.
(#[a-zA-Z0-9_; ]+);
- () is a regular capture group, which means that things captured in this group are included in the result.
You don't need to insert a '\' before every symbol
(?:\s|$)/
- another non capture group, specifying to match a white space or end position of a string.
gm
- global and multiline flags of javascript regex
Here is an example:
let regex_pattern = /(?:\s|^)(#[a-zA-Z0-9_; ]+);(?=\s|$)/gm
let input1 = " #key;" // string with just one word
let input2 = "#key1; #key2;#key3;" // string with one whole word and another word which will match your pattern
let input3 = "soemthing random #key;andjointstring" // a string with a word that will match the pattern but its not a whole word
console.log(input1.match(regex_pattern)) // it matches
console.log(input2.match(regex_pattern)) // it matches
console.log(input3.match(regex_pattern)) // it doesnt matches
I am trying to write a regex that finds the first word in each line that contains the character a.
For a string like:
The cat ate the dog
and the mouse
The expression should find cat and
So far, I have:
/\b\w*a\w*\b/g
However this will return every match in each line, not just the first match (cat ate and).
What is the easiest way to only return the first occurrence?
Assuming you are onluy looking for words without numbers and underscores (\w would include those), I'd advise to maybe use:
(?i)^.*?(?<!\S)([b-z]*a[a-z]*)(?!\S)
And use whatever is in the 1st capture group. See an online demo. Or, if supported:
(?i)^.*?\K(?<!\S)[b-z]*a[a-z]*(?!\S)
See an online demo.
Please note that I used lookaround to assert that the word is not inbetween anything other than whitespace characters. You may also use word-boundaries if you please and swap those lookarounds for \b. Also, depending on your application you can probably scratch the inline case-insensitive switch to a 'flag'. For example, if you happen to use JavaScript /^.*?(?<!\S)([b-z]*a[a-z]*)(?!\S)/gmi should probably be your option. See for example:
var myString = "The cat ate the dog\nand the mouse";
var myRegexp = new RegExp("^.*?(?<!\S)([b-z]*a[a-z]*)(?!\S)", "gmi");
m = myRegexp.exec(myString);
while (m != null) {
console.log(m[1])
m = myRegexp.exec(myString);
}
If you want to match a word using \w you might also use a negated character class matching any character except a or a newline.
Then match a word that consists of at least an a char with word boundaries \b
^[^a\n\r]*\b([^\Wa]*a\w*)
The pattern matches:
^ Start of string
[^a\n\r]*\b Optionally match any character except a or a newline
( Capture group 1
[^\Wa]*a\w* Optionally match a word character without a, then match a and optional word characters
) Close group 1
Regex demo
Using whitespace boundaries on the left and right:
^[^a\n\r]*(?<!\S)([^\Wa]*a\w*)(?!\S)
Regex demo
The text could be matched with the regular expression
(?=(\b[a-z]*a[a-z]*\b)).*\r?\n
with the multiline and case-indifferent flags set. For each match capture group 1 contains the first word (comprised only of letters) in a line that contains an "a". There are no matches in lines that do not contain an "a".
Demo
The expression can be broken down as follows.
(?= # begin a positive lookahead
\b # match a word boundary
([a-z]*a[a-z]*) # match a word containing an "a" and save to
# capture group 1
)
.*\r?\n # match the remainder of the line including the
# line terminator
Suppose I have an email address, 'abcdef#gmail.com'. I want to replace all the characters between 'a' and 'f' so the result would look like 'a****f#gmail.com'.
Trying to do this with a regex and replace
str.replace(/^(.*?)#/gi, '*');
But the results look like this
*gmail.com
Is there a way to do what I need?
This is not an answer to your actual question, but I'd like to challenge you that your idea is not a good one. It's best not to show how long an email address is by replacing the internal letters with the same number of *s. It's better to use a fixed number of *s.
You seem to be using javascript, which doesn't have lookbehind assertions, and capturing in this case may be simpler to understand too, so I'd do this to replace with a constant number of *s
str.replace(/^(.).*(.#)/, '$1***$2')
I'd use a replace with a callback, where the user middle part can be also replaced with *s:
var email = "abcdef#gmail.com";
document.write(email.replace(/^(.)(.*)(.#[^#]*)$/, function(m, g1, g2, g3) {
return g1 + g2.replace(/./g, "*") + g3;
}));
Here is how the "outer" /^(.)(.*)(.#[^#]*)$/ regex works:
^ - matches start of a string
(.) - Group 1: any first character
(.*) - Group 2: any characters up to the character before the last #`
(.#[^#]*) - Group 3: one character before the last #, then # and then any 0+ characters other than # up to...
$ - end of string
The .replace(/./g, "*") will just replace any character with *. And it will be done only on the Group 2.
The regex you suggested in the comment should also work.
/(?!^).(?=[^#]+#)/g matches any character but a newline that is not the first character ((?!^)) and that has 1+ characters other than # after it and a #.
var re = /(?!^).(?=[^#]+#)/g;
document.body.innerHTML = "fake#gmail.com".replace(re, "*");
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