I want to regex match the last word in a string where the string ends in ... The match should be the word preceding the ...
Example: "Do not match this. This sentence ends in the last word..."
The match would be word. This gets close: \b\s+([^.]*). However, I don't know how to make it work with only matching ... at the end.
This should NOT match: "Do not match this. This sentence ends in the last word."
If you use \s+ it means there must be at least a single whitespace char preceding so in that case it will not match word... only.
If you want to use the negated character class, you could also use
([^\s.]+)\.{3}$
( Capture group 1
[^\s.]+ Match 1+ times any char except a whitespace char or dot
) Close group
\.{3} Match 3 dots
$ End of string
Regex demo
You can anchor your regex to the end with $. To match a literal period you will need to escape it as it otherwise is a meta-character:
(\S+)\.\.\.$
\S matches everything everything but space-like characters, it depends on your regex flavor what it exactly matches, but usually it excludes spaces, tabs, newlines and a set of unicode spaces.
You can play around with it here:
https://regex101.com/r/xKOYa4/1
Related
I want to make a regex that recognize some patterns and some not.
_*[a-zA-Z][a-zA-Z0-9_][^-]*.*(?<!_)
The sample of patterns that i want to recognize:
a100__version_2
_a100__version2
And the sample of patterns that i dont want to recognize:
100__version_2
a100__version2_
_100__version_2
a100--version-2
The regex works for all of them except this one:
a100--version-2
So I don't want to match the dashes.
I tried _*[a-zA-Z][a-zA-Z0-9_][^-]*.*(?<!_)
so the problem is at [^-]
You could write the pattern like this, but [^-]* can also match newlines and spaces.
To not match newlines and spaces, and matching at least 2 characters:
^_*[a-zA-Z][a-zA-Z0-9_][^-\s]*$(?<!_)
Regex demo
Or matching only word characters, matching at least a single character repeating \w* zero or more times:
^_*[a-zA-Z]\w*$(?<!_)
^ Start of string
_* Match optional underscores
[a-zA-Z] Match a single char a-zA-Z
\w* Match optional word chars (Or [a-zA-Z0-9_]*)
$ End of string
(?<!_) Assert not _ to the left at the end of the string
Regex demo
Given a string such as below:
word.hi. bla. word.
I want to construct a regex which will match all "."s preceded by "word" and any other non space character
So, in the above example I would want the the first, second and last dots to be matched.
While matching the first and last dots would be easy with global flag (/(?:word.*)\K./gU), I'm not sure how to construct a regex that would also match the second dot.
Appreciate any pointers.
You might match word and then get all consecutive matches using the \G anchor excluding matching whitespace chars or a dot.
(?:\bword|\G(?!\A))[^.\s]*\K\.
In parts
(?: Non capture group
\bword Match word preceded by a word boundary
| Or
\G(?!\A) Assert the position at the end of the previous match, not at the start
) Close non capture group
[^.\s]* Match 0+ occurrences of any char except . or a whitespace char
\K Clear the match buffer (forget what is matched until now)
\. Match a dot
Regex demo
I'm attempting to match the last character in a WORD.
A WORD is a sequence of non-whitespace characters
'[^\n\r\t\f ]', or an empty line matching ^$.
The expression I made to do this is:
"[^ \n\t\r\f]\(?:[ \$\n\t\r\f]\)"
The regex matches a non-whitespace character that follows a whitespace character or the end of the line.
But I don't know how to stop it from excluding the following whitespace character from the result and why it doesn't seem to capture a character preceding the end of the line.
Using the string "Hi World!", I would expect: the "i" and "!" to be captured.
Instead I get: "i ".
What steps can I take to solve this problem?
"Word" that is a sequence of non-whitespace characters scenario
Note that a non-capturing group (?:...) in [^ \n\t\r\f](?:[ \$\n\t\r\f]) still matches (consumes) the whitespace char (thus, it becomes a part of the match) and it does not match at the end of the string as the $ symbol is not a string end anchor inside a character class, it is parsed as a literal $ symbol.
You may use
\S(?!\S)
See the regex demo
The \S matches a non-whitespace char that is not followed with a non-whitespace char (due to the (?!\S) negative lookahead).
General "word" case
If a word consists of just letters, digits and underscores, that is, if it is matched with \w+, you may simply use
\w\b
Here, \w matches a "word" char, and the word boundary asserts there is no word char right after.
See another regex demo.
In Word text, if I want to highlight the last a in para. I search for all the words that have [space][para][space] to make sure I only have the word I want, then when it is found it should be highlighted.
Next, I search for the last [a ] space added, in the selection and I will get only the last [a] and I will highlight it or color it differently.
I am using the regex
(.*)\d.txt
on the expression
MyFile23.txt
Now the online tester says that using the above regex the mentioned string would be allowed (selected). My understanding is that it should not be allowed because there are two numeric digits 2 and 3 while the above regex expression has only one numeric digit in it i.e \d.It should have been \d+. My current expression reads. Zero of more of any character followed by one numeric digit followed by .txt. My question is why is the above string passing the regex expression ?
This regex (.*)\d.txt will still match MyFile23.txt because of .* which will match 0 or more of any character (including a digit).
So for the given input: MyFile23.txt here is the breakup:
.* # matches MyFile2
\d # matched 3
. # matches a dot (though it can match anything here due to unescaped dot)
txt # will match literal txt
To make sure it only matches MyFile2.txt you can use:
^\D*\d\.txt$
Where ^ and $ are anchors to match start and end. \D* will match 0 or more non-digit.
The pattern you have has one group (.*) which would match using your example:MyFile2
because the . allows any character.
Furthermore the . in the pattern after this group is not escaped which will result in allowing another character of any kind.
To avoid this use:
(\D*)\d+\.txt
the group (\D*) would now match all non digit characters.
Here is the explanation, your "MyFile23.txt" matches the regex pattern:
A literal period . should always be escaped as \. else it will match "any character".
And finally, (.*) matches all the string from the beginning to the last digit (MyFile2). Have a look at the "MATCH INFORMATION" area on the right at this page.
So, I'd suggest the following fix:
^\D*\d\.txt$ = beginning of a line/string, non-digit character, any number of repetitions, a digit, a literal period, a literal txt, and the end of the string/line (depending on the m switch, which depends on the input string, whether you have a list of words on separate lines, or just a separate file name).
Here is a working example.
How to regex match words that have digits or any non-characters inside words, excluding when digits and non-characters (\/°†#*()'\s+&;±|-\^) are at the end of word? I need to match dAS2a but not dASI6. Could not adapt the Regex to match string not ending with pattern solution.
dA/Sa
dAS2a
dASI/
dASI6
http://regex101.com/r/qM4dV7/1 failed.
This should work just fine (if you use the gmi modifiers):
^.*[a-z]$
Demo
You said each word is on a new line. Using the m modifier we can anchor each expression to the beginning/end of a line with ^ and $ anchors (without the modifier, this means beginning/end of the string). Then you said a word can essentially be anything (.*) as long as it ends in a non-digit or non-special character (I took that to mean a "letter", [a-z] with the i modifier).