Goal
The goal is matching a string in JavaScript without certain delimiters, i.e. a string between two characters (the characters can be included in the match).
For example, this string should be fully matched: $ test string $. This can appear anywhere in a string. That would be trivial, however, we want to allow escaping the syntax, e.g. The price is 5\$ to 10\$.
Summarized:
Match any string that is enclosed by two $ signs.
Do not match it if the dollar signs are escaped using \$.
Solution using negative lookbehind
A solution that achieves this goal perfectly is: (?<!\\)\$(.*?)(?<!\\)\$.
Problem
This solution uses negative lookbehind, which is not supported on Safari. How can the same matches be achieved without using negative lookbehind (i.e. on Safari)?
A solution that partially works is (?<!\\)\$(.*?)(?<!\\)\$. However, this will also match the character in front of the $ sign if it is not a \.
You might rule out what you don't want by matching it, and capture what you want to keep in group 1
\\\$.*?\$|\$.*?\\\$|(\$.*?\$)
Regex demo
You may use this regex and grab your inner text using capture group #1 as you are already doing in your current regex using lookbehind:
(?:^|[^\\])\$((?:\\.|[^$])*)\$
RegEx Demo
RegEx Details:
(?:^|[^\\]): Match start position or a non-backslash character in a non-capturing group
\$: Match starting $
(: Start capturing group
(?:\\.|[^$])*: Match any escaped character or a non-$ character. Repeat this group 0 or more times
): End capturing group
\$: Match closing $
PS: This regex will give same matches as your current regex: (?<!\\)\$(.*?)(?<!\\)\$
Related
I'm trying to make this regex:
([?=section\/en\/]*)([^\/#]*)
For these examples:
https://www.test.com/en/string-to-get#cid=4949
https://www.test.com/en/section/string-to-get/page&2#cid=4949
https://www.test.com/en/section/string-to-get#cid=4949
current regex
You need to use
(?<=section\/)([^\/#]*)
Or, just
section\/([^\/#]*)
and grab Group 1 value.
Here,
(?<=section\/) - a positive lookbehind that matches a location immediately preceded with section/ substring
([^\/#]*) - Capturing group 1: zero or more chars other than / and #.
See the regex demo #1 and regex demo #2.
Depending on whether or not regex delimiters are required and if they are not /s you may use an unescaped /, (?<=section/)([^/#]*) and section/([^/#]*).
I'm building on a regular expression I found that works well for my use case. The purpose is to check for what I consider valid hashtags (I know there's a ton of hashtag regex posts on SO but this question is specific).
Here's the regex I'm using
/(^|\B)#(?![0-9_]+\b)([a-zA-Z0-9_]{1,20})(\b|\r)/g
The only problem I'm having is I can't figure out how to check if the second character is a-z (the first character would be the hashtag). I only want the first character after the hashtag to be a-z or A-Z. No numbers or non-alphanumeric.
Any help much appreciated, I'm very novice when it comes to regular expressions.
As I mentioned in the comments, you can replace [a-zA-Z0-9_]{1,20} with [a-zA-Z][a-zA-Z0-9_]{0,19} so that the first character is guaranteed to be a letter and then followed by 0 to 19 word characters (alphanumeric or underscore).
However, there are other unnecessary parts in your pattern. It appears that all you need is something like this:
/(?:^|\B)#[a-zA-Z][a-zA-Z0-9_]{0,19}\b/g
Demo.
Breakdown of (?:^|\B):
(?: # Start of a non-capturing group (don't use a capturing group unless needed).
^ # Beginning of the string/line.
| # Alternation (OR).
\B # The opposite of `\b`. In other words, it makes sure that
# the `#` is not preceded by a word character.
) # End of the non-capturing group.
Note: You may also replace [a-zA-Z0-9_] with \w.
References:
Word Boundaries.
Difference between \b and \B in regex.
The below should work.
(^|\B)#(?![0-9_]+\b)([a-zA-Z][a-zA-Z0-9_]{0,19})(\b|\r)
If you only want to accept two or more letter hashtags then change {0,19} with {1,19}.
You can test it here
In your pattern you use (?![0-9_]+\b) which asserts that what is directly on the right is not a digit or an underscore and can match a lot of other characters as well besides an upper or lower case a-z.
If you want you can use this part [a-zA-Z0-9_]{1,20} but then you have to use a positive lookahead instead (?=[a-zA-Z]) to assert what is directly to the right is an upper or lower case a-z.
(?:^|\B)#(?=[a-zA-Z])[a-zA-Z0-9_]{1,20}\b
Regex demo
I need to write a little RegEx matcher which will match any occurrence of strings in the form of
[a-zA-Z]+(_[a-zA-Z0-9]+)?
If I use the regex above it does match the sections needed but would also match onto the abc part of 4_abc which is not intended. I tried to exclude it with:
(?:[^a-zA-Z0-9_]|^)([a-zA-Z]+(_[a-zA-Z0-9]+)?)(?:[^a-zA-Z0-9_]|$)
The problem is that the 'not' matches at the beginning and end are not really working like I hoped they would. If I use them on the example
a_d Dd_da 4_d d_4
they would block matching the second Dd_da because the space was used in the first match.Sadly I can't use lookarounds because I am using JS.
So the input:
a_d Dd_da 4_d d_4
should match: a_d, Dd_da and d_4
but matches: a_d (there is a space at the end)
Is there another way to match the needed sections, or to not consume the 'anchor' matches?
I really appreciate your help.
You can make use of \b:
\b[a-zA-Z]+(_[a-zA-Z0-9]+)?\b
\b matches the (zero-width) point where either the preceding character or following character is a letter, digit or underscore, but not both. It also matches with the start/end of the string if the first/last character is a letter, digit or underscore.
How do I retrieve an entire word that has a specific portion of it that matches a regex?
For example, I have the below text.
Using ^.[\.\?\!:;,]{2,} , I match the first 3, but not the last. The last should be matched as well, but $ doesn't seem to produce anything.
a!!!!!!
n.......
c..,;,;,,
huhuhu..
I want to get all strings that have an occurrence of certain characters equal to or more than twice. I produced the aforementioned regex, but on Rubular it only matches the characters themselves, not the entire string. Using ^ and $
I've read a few stackoverflow posts similar, but not quite what I'm looking for.
Change your regex to:
/^.*[.?!:;,]{2,}/gm
i.e. match 0 more character before 2 of those special characters.
RegEx Demo
If I understand well you are trying to match an entire string that contains at least the same punctuation character two times:
^.*?([.?!:;,])\1.*
Note: if your string has newline characters, change .* to [\s\S]*
The trick is here:
([.?!:;,]) # captures the punct character in group 1
\1 # refers to the character captured in group 1
I have looked through previous questions and answers, however they do not solve the following:
https://stackoverflow.com/questions/ask#notHashTag
The closest I got to is this: (^#|(?:\s)#)(\w+), which finds the hashtag in half the necessary cases and also includes the leading space in the returned text. Here are all the cases that need to be matched:
#hashtag
a #hashtag
a #hashtag world
cool.#hashtag
##hashtag, but only until the comma and starting at second hash
#hashtag#hashtag two separate matches
And these should be skipped:
https://stackoverflow.com/questions/ask#notHashTag
Word#notHashTag
#ab is too short to be a hashtag, 3 characters minimum
This should work for everything but #hashtag#duplicates, and because JS doesn't support lookbehind, that's probably not possible to match that by itself.
\B#\w{3,}
\B is designed to match only between two word characters or two non-word characters. Since # is a non-word character, this forces the match to be preceded by a space or punctuation, or the beginning of the string.
Try this regex:
(?:^|[\s.])(#+\w{3,})(#+\w{3,})?
Online Demo: http://regex101.com/r/kG1nD5