I have the following regular expression for capturing positive & negative time offsets.
\b(?<sign>[\-\+]?)(?<hours>2[1-3]|[01][0-9]|[1-9]):(?<minutes>[0-5]\d)\b
It matches fine but the leading sign doesn't appear in the capture group. Am I formatting it wrong?
You can see the effect here https://regex101.com/r/CQxL8q/1/
That is because of the first \b. The \b word boundary does not match between a start of the string/newline and a - or + (i.e. a non-word char).
You need to move the word boundary after the optional sign group:
(?<sign>[-+]?)\b(?<hours>2[1-3]|[01][0-9]|[1-9]):(?<minutes>[0-5][0-9])\b
^^
See the regex demo.
Now, since the char following the word boundary is a digit (a word char) the word boundary will work correctly failing all matches where the digit is preceded with another word char.
The word boundary anchor (\b) matches the transition between a word character (letter, digit or underscore) to a non-word character or vice-versa. There is no such transition in -13:21.
The word boundary anchor could stay between the sign and the hours to avoid matching it in expressions that looks similar to a time (65401:23) but you cannot prevent it match 654:01:23 or 654-01:23.
As a side note [\-\+] is just a convoluted way to write [-+]. + does not have any special meaning inside a character class, there is no need to escape it. - is a special character inside a character class but not when it is the first or the last character (i.e. [- or -]).
Another remark: you use both [0-9] and \d in your regex. They denote the same thing1 but, for readability, it's recommended to stick to only one convention. Since other character classes that contain only digits are used, I would use [0-9] and not \d.
And some bugs in the regex fragment for hours: 2[1-3]|[01][0-9]|[1-9] do not match 0 (but it matches 00) and 20.
Given all the above corrections and improvements, the regex should be:
(?<sign>[-+]?)\b(?<hours>2[0-3]|[01][0-9]|[0-9]):(?<minutes>[0-5][0-9])\b
1 \d is the same as [0-9] when the Unicode flag is not set. When Unicode is enabled, \d also matches the digits in non-Latin based alphabets.
Related
^([a-zA-Z0-9_-]+)$ matches:
BAP-78810
BAP-148080
But does not match:
B8241066 C
Q2111999 A
Q2111999 B
How can I modify regex pattern to match any space and/or special character?
For the example data, you can write the pattern as:
^[a-zA-Z0-9_-]+(?: [A-Z])?$
^ Start of string
[a-zA-Z0-9_-]+ Match 1+ chars listed in the character class
(?: [A-Z])? Optionally match a space and a char A-Z
$ End of string
Regex demo
Or a more exact match:
^[A-Z]+-?\d+(?: [A-Z])?$
^ Start of string
[A-Z]+-? Match 1+ chars A-Z and optional -
\d+(?: [A-Z])? Matchh 1+ digits and optional space and char A-Z
$ End of string
Regex demo
Whenever you want to match something that can either be a space or a special character, you would use the dot symbol .. Your regex pattern would then be modified to:
^([a-zA-Z0-9_-])+.$
This will match the empty space, or any other character. If you want to match the example provided, where strictly one alphabetical, numer character will follow the space, you could include \w such that:
^([a-zA-Z0-9_-])+.\w$
Note that \w is equivalent to [A-Za-z0-9_]
Further, be careful when you use . as it makes your pattern less specific and therefore more likely to false positives.
I suggest using this approach
^[A-Z][A-Z\d -]{6,}$
The first character must be an uppercase letter, followed by at least 6 uppercase letters, digits, spaces or -.
I removed the group because there was only one group and it was the entire regex.
You can also use \w - which includes A-Z,a-z and 0-9, as well as _ (underscore). To make it case-insensitive, without explicitly adding a-z or using \w, you can use a flag - often an i.
I thought [^0-9a-zA-Z]* excludes all alpha-numeric letters, but allows for special characters, spaces, etc.
With the search string [^0-9a-zA-Z]*ELL[^0-9A-Z]* I expect outputs such as
ELL
ELLs
The ELL
Which ELLs
However I also get following outputs
Ellis Island
Bellis
How to correct this?
You may use
(?:\b|_)ELLs?(?=\b|_)
See the regex demo.
It will find ELL or ELLs if it is surrounded with _ or non-word chars, or at the start/end of the string.
Details:
(?:\b|_) - a non-capturing alternation group matching a word boundary position (\b) or (|) a _
ELLs? - matches ELL or ELLs since s? matches 1 or 0 s chars
(?=\b|_) - a positive lookahead that requires the presence of a word boundary or _ immediately to the right of the current location.
change the * to +
a * means any amount including none. A + means one or more. What you probably want though is a word boundry:
\bELL\b
A word boundry is a position between \w and \W (non-word char), or at the beginning or end of a string if it begins or ends (respectively) with a word character ([0-9A-Za-z_]). More here about that:
What is a word boundary in regexes?
I can use \s?(\w+\s){0,2}\w*) for "up to three words" and \w{0,20} for "no more than twenty characters", but how can I combine these? Trying to merge the two via a lookahead as mentioned here seems to fail.
Some examples for clarification:
The early bird catches the worm.
should match any three words in sequence (including the worm*).
Here we have a supercalifragilisticexpialidocious sentence.
"a supercalifragilisticexpialidocious sentence" is too long a sequence and therefore should not match.
* In my actual use case I'm going for a paragraph's last three words, i.e. a (?:\r) would be at the end of the RegEx and the match "catches the worm.") Matches are then applied with a "no linebreaks" character style in Adobe InDesign in order to avoid orphans.
To match 3 words separated with whitespace(s) at the end of a line or string, you can use
\b(?!(?:\s*\w){21})\w+(?:\s+\w+){0,2}(?=$|[\r\n])
See the regex demo. Note that in the demo, I use [^\S\r\n] instead of the \s in the lookahead since the text contains newlines, use the same trick if you need that.
Regex explanation
\b - a word boundary
(?!(?:\s*\w){21}) - a lookahead check that fails the match if after the initial word boundary there are 21 word characters optionally preceded with any number of whitespace symbols
\w+ - 1 word (consisting of 1 or more word characters)
(?:\s+\w+){0,2} - zero, one or two sequences of 1+ whitespaces followed with 1+ word characters
(?=$|[\r\n]) - a positive lookahead that only allows a match to be returned if there is the end-of-string ($) or the end of a line ([\r\n]).
Now, if your words should only contain letters, use [a-zA-Z] or equivalent for your language. If the regex flavor allows, use \p{L} Unicode category/property class.
I'm writing a regular expression in Java for capturing some word without spaces.
The word can contain only letter, number, hyphens and dot.
The character set [\w+\-\\.] work well.
Now I want to edit the set for allowing a single space after the dot.
How I have to edit my regular expression?
You can add an alternation that matches this additional requirement
([\w\-.]|(?<=\.) )+
See it here on Regexr
(?<=\.) is a lookbehind assertion. It ensures that space is only matched, if it is preceded by a dot.
Other hints:
\w contains the underscore and matches per default only ASCII letters/digits. If you care about Unicode, use either the modifier UNICODE_CHARACTER_CLASS to enable Unicode for \w or use the Unicode properties \p{L} and \p{Nd} to match Unicode letters and digits.
You don't need to escape the dot in a character class.
You have \w+ in your character class, are you aware, that you just add the "+" character to the accepted characters?
In case of a dot followed by a space, I suppose this pattern should be neither the first, nor the last in the matched string? You may want to enclose it in word boundaries \b:
([0-9A-Za-z-]|\b\.( \b)?)+
I deliberately did not use \w, to exclude underscores.
For allowing ONLY a single space after the dot you can use this regex:
^(?!.*?\. {2})[\w.-]+$
You don't need to escape dot OR hyphen inside character class
(?!.*?\. {2}) is a negative lookahead that disallows 2 or more spaces after a dot
The Regular expression
/[\D\S]/
should match characters Which is not a digit or not whitespace
But When I test this expression in regexpal
It starts matching any character that's digit, whitespace
What i am doing wrong ?
\D = all characters except digits,
\S = all characters except whitespaces
[\D\S] = union (set theory) of the above character groups = all characters.
Why? Because \D contains \s and \S contains \d.
If you want to match characters which are not dights nor whitespaces you can use [^\d\s].
Your regex is invalidating itself as it goes. Putting the regex inside of [] means it has to match one of the items inside of it. These two items override each other, which end up matching everything. In theory, anything that is non digit, would match every other char. available, and any non whitespace matches any digit and any other char. as well.
You can try using [^\d\s] which says to negate the match of any digit or any space. Instead of having everything caught in the original regex, this negates the matching of both the \d and \s. You can see testing done with it here.