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
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 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
i Have a string of words. I want get a word which begins and ends with 3 back ticks ```. how to I use regular expressions to accomplish this in flutter. I have tried this(^```.*\.```$)\w+but its not working on a sentence like Hello there, ```friend```, how are you doing?
The pattern you tried (^```.*\.```$)\w+ uses anchors to assert the start ^ and the end $ of the string and in between match any char except a newline followed by a literal dot around triple backticks.
After that it tries to match 1+ word characters which will not match.
You could use a capturing group and match 1+ word characters in between
```(\w+)```
Regex demo
I need to match only those words which doesn't have special characters like # and :.
For example:
git#github.com shouldn't match
list should return a valid match
show should also return a valid match
I tried it using a negative lookahead \w+(?![#:])
But it matches gi out of git#github.com but it shouldn't match that too.
You may add \w to the lookahead:
\w+(?![\w#:])
The equivalent is using a word boundary:
\w+\b(?![#:])
Besides, you may consider adding a left-hand boundary to avoid matching words inside non-word non-whitespace chunks of text:
^\w+(?![\w#:])
Or
(?<!\S)\w+(?![\w#:])
The ^ will match the word at the start of the string and (?<!S) will match only if the word is preceded with whitespace or start of string.
See the regex demo.
Why not (?<!\S)\w+(?!\S), the whitespace boundaries? Because since you are building a lexer, you most probably have to deal with natural language sentences where words are likely to be followed with punctuation, and the (?!\S) negative lookahead would make the \w+ match only when it is followed with whitespace or at the end of the string.
You can use negative lookbehind and negative lookahead patterns around a word pattern to make sure that the word is not preceded or followed by a non-space character, or in other words, to make sure that it is surrounded by either a space or a string boundary:
(?<!\S)\w+(?!\S)
Demo: https://regex101.com/r/cjhUUM/2
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).