I am trying to write a pattern that matches {^xyz^} as bellow,
#"\b\{\^\S*\^\}\b
But I am not getting success and wondering what is problem with my pattern.
You can use:
#"\{\^([^}]*)\^\}"
and extract captured group #1 for your string.
Use a captured group to get the substring you want to extract from a larger match.
Word boundaries or \b won't work here because { and } are non-word characters.
Use of negated character class [^}]* is more efficient and accurate than greedy \S*.
I would simply use \{\^(\S*?)\^\}. This way you are capturing the contents between the carets and curly brackets. The ? is to make the * quantifier lazy, so it matches as little characters as possible (in order to prevent matching the beginning of one block until the end of another block in the same line).
With those \b you need a word-type character right before and after the curly braces for the regex to match. Is that really a requirement? Or can there be a space?
Related
I have a string and would like to match a part of it.
The string is Accept: multipart/mixedPrivacy: nonePAI: <sip:4168755400#1.1.1.238>From: <sip:4168755400#1.1.1.238>;tag=5430960946837208_c1b08.2.3.1602135087396.0_1237422_3895152To: <sip:4168755400#1.1.1.238>
I want to match PAI: <sip:4168755400#
the whitespace can be a word so i would like to use .* but if i used that it matches most of the string
The example on that link is showing what i'm matching if i use the whitespace instead of .*
(PAI: <sip:)((?:\([2-9]\d{2}\)\ ?|[2-9]\d{2}(?:\-?|\ ?))[2-9]\d{2}[- ]?\d{4})#
The example on that link is showing what i'm trying to achieve with .* but it should only match PAI: <sip:4168755400#
(PAI:.*<sip:)((?:\([2-9]\d{2}\)\ ?|[2-9]\d{2}(?:\-?|\ ?))[2-9]\d{2}[- ]?\d{4})#
I tried lookaround but failing.
Any idea?
thanks
Matching the single space can be updated by using a character class matching either a space or a word character and repeat that 1 or more times to match at least a single occurrence.
Note that you don't have to escape the spaces, and in both occasions you can use an optional character class matching either a space or hyphen [ -]?
If you want the match only, you can omit the 2 capturing groups if you want to.
(PAI:[ \w]+<sip:)((?:\([2-9]\d{2}\) ?|[2-9]\d{2}[ -]?)[2-9]\d{2}[- ]?\d{4})#
Regex demo
The regex should be like
PAI:.*?(<sip:.*?#)
Explanation:
PAI:.*? find the word PAI: and after the word it can be anything (.*) but ? is used to indicate that it should match as few as possible before it found the next expression.
(<sip:.*?#) capturing group that we want the result.
<sip:.*?# find <sip: and after the word it can be anything .*? before it found #.
Example
I am new to regex and playing around with writing regex to match markdown syntaxes, particularly italic text like:
this is markdown with some *italic text*
After writing some naive implementations I found this regex which seems to do the job quite nicely (dealing with edge-cases) and matches the entire string:
(?<!\*)\*([^ ][^*\n]*?)\*(?!\*)
However, I don't want to match the entire string - I only want to match the beginning and end * characters (so that I can do some special formatting to those characters). How might I go about doing that?
The tricky thing is that I only want to the match the * characters when the rest of the string matches the correct format of a string in italics (i.e. meets the requirements of that regex above). So a simple regex like (\*|\*) isn't going to cut it.
Except from using a capturing group for the asterix at the start and at the end, you can add an asterix to the first negated character class to prevent matching a double **.
Note that as pointed out by #toto you don't really need the capturing groups around the asterix (\*). You can also match them and add the replacement characters before and after the single capturing group for the content in the middle.
It also means that it should match at least a single character other then an asterix.
You don't have to make the first character class non greedy *? as it can not cross the * boundary that follows.
(?<!\*)(\*)([^*\s][^*\r\n]*)(\*)(?!\*)
Regex demo
If there can also not be a space before the ending asterix, you can repeat matching a space followed by matching any non whitespace char except an asterix (?: [^*\s]+)*
The \r\n in the negated character class is to prevent newline boundaries which are also matched by \s. If that should not be the case, you can replace that by a space or tab and space.
(?<!\*)(\*)([^*\s]+(?: [^*\s]+)*)(\*)(?!\*)
Regex demo
Just change the first and second \* to capturing groups and you can change at will:
(?<!\*)(\*)([^ ][^*\n]*?)(\*)(?!\*)
Demo
I have the following regex:
\b[_\.][0-9]{1,}[a-zA-Z]{0,}[_]{0,}\b
My input string is:
_49791626567342fYbYzeRESzHsQUgwjimkIfW
.49791626567342fYbYzeRESzHsQUgwjimkIfW
I would assume that it matches 1. and 2., but it is only matching in the first scenario. Can you help me find the mistake in the regex?
A word boundary is a border between a word character (letters, digits, underscore) and either a non-word-character or the start or end of the string. So there simply is no word boundary between dot (non-word-character) and the start of the string.
You can use an anchor in this case, to signal the start of the string, like
^[_\.][0-9]{1,}[a-zA-Z]{0,}[_]{0,}$
You can also shorten your regex a bit by using * and + quantifiers and avoiding unnecessary escape sequences, as suggested by Toto
^[_.][0-9]+[a-zA-Z]*_*$
You can also use lookahead and lookbehind (if available) to build yourself a custom boundary.
I have <autorpodpis>([^;,<\n\r]*?)\s*[;,<\n\r] to catch everything inside
<autorpodpis>_this_is_an_example_of_what_I'd_like_to_match< If there is a space, a colon (;) or a semicolon (;) or a space before a colon or a semicolon, my RegEx catches everything but including these characters – see my link. It works as it is expected to.
Overall, the RegEx works fine with substitution \1 (or in AutoHotKey I use – $1). But I'd like match without using substitution.
You seem to mix the terms substitution (regex based replacement operation) and capturing (storing a part of the matched value captured with a part of a pattern enclosed with a pair of unescaped parentheses inside a numbered or named stack).
If you want to just match a substring in specific context without capturing any subvalues, you might consider using lookarounds (lookbehind or lookahead).
In your case, since you need to match a string after some known string, you need a lookbehind. A lookbehind tells the regex engine to temporarily step backwards in the string, to check if the text inside the lookbehind can be matched there.
So, you could use
pos := RegExMatch(input, "(?<=<autorpodpis>)\p{L}+(?:\s+\p{L}+)*", Res)
So, the Res should have WOJCIECH ZAŁUSKA if you supply <autorpodpis>WOJCIECH ZAŁUSKA</autorpodpis> as input.
Explanation:
(?<=<autorpodpis>) - check if there is <autorpodpis> right before the currently tested location. If there is none, fail this match, go on to the next location in string
\p{L}+ - 1+ Unicode letters
(?:\s+\p{L}+)* - 0+ sequences of 1+ whitespaces followed with 1+ Unicode letters.
However, in most cases, and always in cases like this when the pattern in the lookbehind is known, the lookbehind is unanchored (say, when it is the first subpattern in the pattern) and you do not need overlapping matches, use capturing.
The version with capturing in place:
pos := RegExMatch(input, "<autorpodpis>(\p{L}+(?:\s+\p{L}+)*)", Res)
And then Res[1] will hold the WOJCIECH ZAŁUSKA value. Capturing is in most cases (96%) faster.
Now, your regex - <autorpodpis>([^;,<\n\r]*?)\s*[;,<\n\r] - is not efficient as the [^;,<\n\r] also matches \s and \s matches [;,<\n\r]. My regex is linear, each subsequent subpattern does not match the previous one.
We have tab spaced list of "key=value" pairs.
How we can split it, using regexp?
Case key=value must be transformed into value. Case key=value=value2 must be transformed into value=value2.
https://regex101.com/r/dR5dT0/1 - I've started solution like this, but can't find beautiful way to remove only "key=" part from text.
UPD BTW, do you know cool crash courses on regular expressions?
You can just use
=(\S*)
See regex demo
Since the list is already formatted, the = in the pattern will always be the name/value delimiter.
The \S matches any non-whitespace character.
The * is a quantifier meaning that the \S should occur zero or more times (\S* matches zero or more non-whitespace characters).
You can use this regex for matching:
/\w+=(\S+)/
and grab captured group #1
RegEx Demo