This is an example of what the regex would and wouldn't match:
# Matches
AAAA: aaaa
# Matches
ABCD: abcd
# Doesn't match
AAAA: abcd
# Doesn't match
AAAA: AaAa
How can I accomplish this?
I found this, but it doesn't work for matches because \L transforms the matches in the replace. Besides, \L seems to be only available in PHP and not in Javascript:
This works, but only when the case-insensitive option is set and it matches the last example:
(\w+): \1
You might be able to use case-sensitivity switch and lookahead. eg.
\b(?=[A-Z]+:\s*[a-z]+)(?i)(\w+):\s*\1\b
or
\b(?=\p{Lu}+:\s*\p{Ll}+)(?i)(\p{L}+):\s*\1\b
Essentially you use 2 regexes at once.
The first (i.e. everything within (?=...)) asserts that the first word is all uppercase ([A-Z]+ or \p{Lu}+) and that the second word is all lowercase ([a-z]+ or \p{Ll}+).
Then you turn on case-insensitivity with (?i).
Then the second regex looks for 2 words that are equal (ignoring case).
The \b prevent matches on input like: xxAAA: aaayy
Note: As the question mentioned VSCode, this answer uses .NET-style regex and assumes that the i modifier is initially turned off but can be toggled. However, I don't think this is possible in ECMAScript ("flags are an integral part of a regular expression. They cannot be added or removed later").
Related
There is a transform of multiline string.
!a! b!
should become
.a. b.
And
!a! b!
c!
!d!
should become
.a. b.
c!
!d!
I approached it with a lookbehind:
str(/(?<!\n)([^\n!]*)!+/g, '$1.')
It didn't work as intended:
.a. b.
c.
!d.
Splitting a string and transforming the first line seems straightforward. But is there a reliable way to do replacements only in the first line of multiline string with a regex only?
Also would appreciate an explanation what exactly goes wrong with my approach so it fails.
The question is not limited to JS regex flavour but I'm interested in this one in the first place.
About the pattern you tried:
(?<!\n) Negative lookbehind, assert what is directly to the left is not a newline or !
([^\n!]*) Capture group 1, match 0+ times any char except a newline or !
!+ Match 1+ times ! (What you want to remove)
The pattern will match too much, as it will match all the individual parts. There is for example no rule that says match this pattern 2 times, so you will replace with group 1 for every time that pattern has a match.
Note that the quantifier in this part is 0+ times ([^\n!]*) it will also match a single ! except when preceded by a newline.
If you can make use of SKIP FAIL, you can first match what you want to avoid, which in this case is a line that optionally starts with an exclamation mark and ends with an exclamation mark with none in between.
After that match all the other exclamation marks and replace them with a dot.
^!?[^\r\n!]*!$(*SKIP)(*FAIL)|!
See a regex demo
Another option could be using 2 capturing groups.
The first group will match between the first set of exclamation marks, and the second group will match the whitespaces after followed by a char other than !.
Then match the ! at the end so it is not in the replacement
!([^\s!]+)!([^\S\r\n]+[^\s!])!
See another regex demo
In the replacement use the 2 capturing groups with the dots
.$1.$2.
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've got these strings I want to parse:
?forum=Jiné akce a jiné#comments
?trening=140#$|Pralinka
?novinka=87#comments
?forum=Mimo mísu#comments
?forum=Členské forum#comments
?trening=139#comments
and I want to output array like
1. forum
2. Jiné akce a jiné
3. comments
or
1. trening
2. 140
3. Pralinka
So I wrote following regexp:
\?([a-z]{4,})\=(.+)\#(\$\|)?([a-z]+)
Regex101
But It's not working in second case (optional string part).
Remember that by default, regex are case sensitive... So [a-z] can't match Pralinka. You can fix that by using the i (case insensitive) flag, or with:
\?([a-z]{4,})=(.+)#(?:\$\|)?([A-Za-z]+)
Notice that there is no need to escape the = or the # (we're not in free spacing mode), and I added a non capturing group (?:...) so that Pralinka will be in the same capturing group as comment.
The demo is here
[a-z]+ does not match Pralinka because P is an uppercase letter.
Fixed regex
You need to add a global flag: /g.
http://regex101.com/r/vR0oM4