Would an anchor like "^" or "\A" at the beginning of this regex make any sense - any difference?
$string =~/(.*)([a-z])$/
Yes, either ^ or \A will cause the regex to not match if there is a newline anywhere before the letter, because .* (zero or more of any characters except newline) will no longer match up to the letter before the end.
Without the beginning anchor, the regex will match from after the last newline through the end of the string (or through the letter before the newline at the end, if there is a newline).
No, because of the greedy nature of regular expression matching that regex will pull everything before the final letter of the string, provided the last character is a letter.
It would make sense, just not any difference.
Related
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'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.
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).
/\ATo\:\s+(.*)/
Also, how do you work it out, what's the approach?
In multi-line regular expressions, \A matches the start of the string (and \Z is end of string, while ^/$ matches the start/end of the string or the start/end of a line). In single line variants, you just use ^ and $ for start and end of string/line since there is no distinction.
To is literal, \: is an escaped :.
\s means whitespace and the + means one or more of the preceding "characters" (white space in this case).
() is a capturing group, meaning everything in here will be stored in a "register" that you can use. Hence, this is the meat that will be extracted.
.* simply means any non newline character ., zero or more times *.
So, what this regex will do is process a string like:
To: paxdiablo
Re: you are so cool!
and return the text paxdiablo.
As to how to learn how to work this out yourself, the Perl regex tutorial(a) is a good start, and then practise, practise, practise :-)
(a) You haven't actually stated which regex implementation you're using but most modern ones are very similar to Perl. If you can find a specific tutorial for your particular flavour, that would obviously be better.
\A is a zero-width assertion and means "Match only at beginning of string".
The regex reads: On a line beginning with "To:" followed by one or more whitespaces (\s), capture the remainder of the line ((.*)).
First, you need to know what the different character classes and quantifiers are. Character classes are the backslash-prefixed characters, \A from your regex, for instance. Quantifiers are for instance the +. There are several references on the internet, for instance this one.
Using that, we can see what happens by going left to right:
\A matches a beginning of the string.
To matches the text "To" literally
\: escapes the ":", so it loses it's special meaning and becomes "just a colon"
\s matches whitespace (space, tab, etc)
+ means to match the previous class one or more times, so \s+ means one or more spaces
() is a capture group, anything matched within the parens is saved for later use
. means "any character"
* is like the +, but zero or more times, so .* means any number of any characters
Taking that together, the regex will match a string beginning with "To:", then at least one space, and the anything, which it will save. So, with the string "To: JaneKealum", you'll be able to extract "JaneKealum".
You start from left and look for any escaped (ie \A) characters. The rest are normal characters. \A means the start of the input. So To: must be matched at the very beginning of the input. I think the : is escaped for nothing. \s is a character group for all spaces (tabs, spaces, possibly newlines) and the + that follows it means you must have one or more space characters. After that you capture all the rest of the line in a group (marked with ( )).
If the input was
To: progo#home
the capture group would contain "progo#home"
It matches To: at the beginning of the input, followed by at least one whitespace, followed by any number of characters as a group.
The initial and trailing / characters delimit the regular expression.
A \ inside the expression means to treat the following character specially or treat it as a literal if it normally has a special meaning.
The \A means match only at the beginning of a string.
To means match the literal "To"
\: means match a literal ':'. A colon is normally a literal and has no special meaning it can be given.
\s means match a whitespace character.
+ means match as many as possible but at least one of whatever it follows, so \s+ means match one or more whitespace characters.
The ( and ) define a group of characters that will be captured and returned by the expression evaluator.
And finally the . matches any character and the * means match as many as possible but can be zero. Therefore the (.*) will capture all characters to the end of the input string.
So therefore the pattern will match a string that starts "To:" and capture all characters that occur after the first succeeding non-whitespace character.
The only way to really understand these things is to go through them one bit at a time and check the meaning of each component.
How can I use regex for all words beginning with : punctuation?
This gets all words beginning with a:
\ba\w*\b
The minute I change the letter a to :, the whole thing fails. Am I supposed to escape the colon, and if so, how?
\b matches between a non-alphanumeric and an alphanumeric character, so if you place it before :, it only matches if there is a letter/digit right before the colon.
So you either need to drop the \b here or specify what exactly constitutes a boundary in this situation, for example:
(?<!\w):\w*\b
That would ensure that there is no letter/digit/underscore right before the :. Of course this presumes a regex flavor that supports lookbehind assertions.
The problem is that \b won't match the start of a word when the word starts with a colon :, because colon is not a word character. Try this:
(?<=:)\w*\b
This uses a (non-capturing) look-behind to assert that the previous character is a colon.