I'm trying to match all #mentions and #hashtags on a String using this RegEx expression:
(^|\s)([##][a-z\d-]+)
According to regex101.com, since the + is there it should match all occurances
"+" Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed
But when I run it through a String with more than one occurrance, it only matches the first.
What's going on?
Thanks for your attention.
Add the g (global) flag at the end for multiple matches.
/(^|\s)([##][a-z\d-]+)/g
^ this symbol defines beginning of string. That is why it only match with first string.
Use /[##]\w+/ regex.
Related
I want to match exactly one occurrence of # in a string. I found that /^[^#]+#[^#]+$/ working but not /[^#]+#[^#]+/
Why should I include the search from beginning to end? Wont the pattern anyway check throughout the string? Can someone explain it for me?
Without providing ^ and $, your RegEx will match parts of your string.
Let's demonstrate with some examples :
/^[^#]+#[^#]+$/
matches test#String.
doesn't match test#Str#ing
/[^#]+#[^#]+/
matches test#String
matches the part test#Str of test#Str#ing
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_]|$)
The problem is that the 'not' matches at the beginning and end are not really working like I hoped they would. If I use them on the example
a_d Dd_da 4_d d_4
they would block matching the second Dd_da because the space was used in the first match.Sadly I can't use lookarounds because I am using JS.
So the input:
a_d Dd_da 4_d d_4
should match: a_d, Dd_da and d_4
but matches: a_d (there is a space at the end)
Is there another way to match the needed sections, or to not consume the 'anchor' matches?
I really appreciate your help.
You can make use of \b:
\b[a-zA-Z]+(_[a-zA-Z0-9]+)?\b
\b matches the (zero-width) point where either the preceding character or following character is a letter, digit or underscore, but not both. It also matches with the start/end of the string if the first/last character is a letter, digit or underscore.
I am trying to use regexp to match some specific key words.
For those codes as below, I'd like to only match those IFs at first and second line, which have no prefix and postfix. The regexp I am using now is \b(IF|ELSE)\b, and it will give me all the IFs back.
IF A > B THEN STOP
IF B < C THEN STOP
LOL.IF
IF.LOL
IF.ELSE
Thanks for any help in advance.
And I am using http://regexr.com/ for test.
Need to work with JS.
I'm guessing this is what you're looking for, assuming you've added the m flag for multiline:
(?:^|\s)(IF|ELSE)(?:$|\s)
It's comprised of three groups:
(?:^|\s) - Matches either the beginning of the line, or a single space character
(IF|ELSE) - Matches one of your keywords
(?:$|\s) - Matches either the end of the line, or a single space character.
Regexr
you can do it with lookaround (lookahead + lookbehind). this is what you really want as it explicitly matches what you are searching. you don't want to check for other characters like string start or whitespaces around the match but exactly match "IF or ELSE not surrounded by dots"
/(?<!\.)(IF|ELSE)(?!\.)/g
explanation:
use the g-flag to find all occurrences
(?<!X)Y is a negative lookbehind which matches a Y not preceeded by an X
Y(?!X) is a negative lookahead which matches a Y not followed by an X
working example: https://regex101.com/r/oS2dZ6/1
PS: if you don't have to write regex for JS better use a tool which supports the posix standard like regex101.com
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 want to write regexp which allows some special characters like #-. and it should contain at least one letter. I want to understand below things also:
/(?=^[A-Z0-9. '-]{1,45}$)/i
In this regexp what is the meaning of ?=^ ? What is a subexpression in regexp?
(?=) is a lookahead, it's looking ahead in the string to see if it matches without actually capturing it
^ means it matches at the BEGINNING of the input (for example with the string a test, ^test would not match as it doesn't start with "test" even though it contains it)
Overall, your expression is saying it has to ^ start and $ end with 1-45 {1,45} items that exist in your character group [A-Z0-9. '-] (case insensitive /i). The fact it is within a lookahead in this case just means it's not going to capture anything (zero-length match).
?= is a positive lookahead
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