I am looking for a regular expression to catch a whole word or expression within a sentence that contains dots:
this is an example test.abc.123 for what I am looking for
In this case i want to catch "test.abc.123"
I tried with this regex:
(.*)(\b.+\..++\b)(.*)
(.*) some signs or not
(\b.+\..++\b) a word containing some signs followed by at least on dot that is followed by some signs and this at least once
(.*) some more signs nor not#
but it gets me: "abc.123 for what I am looking for"
I see that I got something completely wrong, can anyone enlighten me?
If you need to match part of a string you don't need to match entire string (unless you are restricted by a functionality).
Your regex is so greedy. It also has dots every where (.+ is not a good choice most of the time). It doesn't have a precise point to start and finish either. You only need:
\w+(?:\.+\w+)+
It looks for strings that begin and end with word characters and contain at least a period. See live demo here
This regex pattern matches strings with two or more dots:
.*\..*\..*
"." matches any character except line-breaks
"*" repeats previous tokens 0 or more times
"." matches a single dot, slash is used for escape
.* Match any character and continue matching until next token
test.abc.123
(.) Match a single dot
test. abc.123
.* Again, any character and continue matching until next token
test.example.com
. Matches a single dot
test.example. com
.* Matches any character and continue matching until next token
test.example.com
Try this pattern: (?=\w+\.{1,})[^ ]+.
Details: (?=\w+\.{1,}) - positive lookahead to locate starting of a word with at least one dot (.). Then, start matching from that position, until space with this pattern [^ ]+.
Demo
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 need to create regex to find last underscore in string like 012344_2.0224.71_3 or 012354_5.00123.AR_3.335_8
I have wanted find last part with expression [^.]+$ and then find underscore at found element but I can not handle it.
I hope you can help me :)
Just use a negative character class [^_] that will match everything except an underscore (this helps to ensure no other underscores are found afterwards) and end of string $
Pattern would look as such:
(_)[^_]*$
The final underscore _ is in a capturing group, so you are wanting to return the submatch. You would replace the group 1 (your underscore).
See it live: Regex101
Notice the green highlighted portion on Regex101, this is your submatch and is what would be replaced.
The simplest solution I can imagine is using .*\K_, however not all regex flavours support \K.
If not, another idea would be to use _(?=[^_]*$)
You have a demo of the first and second option.
Explanation:
.*\K_: Fetches any character until an underscore. Since the * quantifier is greedy, It will match until the last underscore. Then \K discards the previous match and then we match the underscore.
_(?=[^_]*$): Fetch an underscore preceeded by non-underscore characters until the end of the line
If you want nothing but the "net" (i.e., nothing matched except the last underscore), use positive lookahead to check that no more underscores are in the string:
/_(?=[^_]*$)/gm
Demo
The pattern [^.]+$ matches not a dot 1+ times and then asserts the end of the string. The will give you the matches 71_3 and 335_8
What you want to match is an underscore when there are no more underscores following.
One way to do that is using a negative lookahead (?!.*_) if that is supported which asserts what is at the right does not match any character followed by an underscore
_(?!.*_)
Pattern demo
I'm stuck on a regex. I'm trying to match words in any language to the right of a colon without matching the colon itself.
The basic rule:
For a line to be valid, it must not begin with or contain any characters outside of [a-z0-9_] until after :.
Any characters to the right of : should match as long as the line begins with the set of characters defined above.
For instance, given a string such as these:
this string should not match
bob_1:Hi. I'm Bob. I speak русский and this string should match
alice:Hi Bob. I speak 한국어 and this string should also match
http://example.com - would prefer to not match URLs
This string:should not match because no spaces or capital letters are allowed left of the colon
Only 2 of the 5 strings above need to match. And only to the right of the colon.
Hi. I'm Bob. I speak русский and this string should match
Hi Bob. I speak 한국어 and this string should also match
I'm currently using (^[a-z0-9_]+(?=:)) to match characters to the left of :. I just can't seem to reverse the logic.
The closest I have at the moment is (?!(?!:)).+. This seems to match everything to right of the colon as well as the colon itself. I just can't figure out how to not include : in the match.
Can one of you regex wizards help me out? If anything is unclear please let me know.
Short regex pattern (case insensitive):
^\w+:(\w.*)
\w - matches any word character (equal to [a-zA-Z0-9_])
https://regex101.com/r/MZhqSL/6
As you marked pcre, here's the pattern you need (only to the right of the colon):
^\w+:\K\w.*
\K - resets the starting point of the reported match. Any previously consumed characters are no longer included in the final match
https://regex101.com/r/E1yHVY/1
You can use this regex:
^[a-z0-9_]+:\K(?!//).*
RegEx Demo
RegEx Breakup:
^: Start
[a-z0-9_]+: Match 1+ of [a-z0-9_] characters
:: Match a colon
\K: Reset matched info so far
(?!//): Negative lookahead to disallow // right after colon to avoid matching potential URLs
.*: Match anything until end
You can use the regex: ^.*?:(.*)$
^.*?: - from the beginning of the line, any character until the colon (non-greedy) included
(.*)$ - use a matching group to anything that follows it till the end of the line
Link to DEMO
I have
/\d+/
Using the string >"tom666tom"
It matches the 666. Shouldnt it fail when it hits the first t in tom?
How exactly is the regex engine working here. I know the plus sign means one or more.
it will fail if you tell the regex is should start and end with a number like so
/^\d+$/
the ^ defines the start of the string and $ the end.
Pattern search one or more digits (+) in the input string
You are not telling your expression to match the entire string. If any part of the string contains one or more digits, it will match. Use the ^ (zero-length start of line marker) and $ (zero-length end of line marker) to delimit your regex and indicate that the only thing on the line should be digits: /^\d+$/.
It shouldn't fall when it encounters first t in "tom" because a +
matches 1 or more of the preceeding token. This is a greedy match, and
will match as many characters as possible before satisfying the next
token.
In your regex /\d+/, the + is placed after \d which matches any digit.
As said in the definition, the regex engine is working perfectly, because it is matching the previous token (\d) as many times it could.
So it will match the digits till it encounters a mismatch.
So the preceeding token here is \d and hence, regex engine is working fine.
I have this regex:
(?:\S)\++(?:\S)
Which is supposed to catch all the pluses in a query string like this:
?busca=tenis+nike+categoria:"Tenis+e+Squash"&pagina=4&operador=or
It should have been 4 matches, but there are only 3:
s+n
e+c
s+e
It is missing the last one:
e+S
And it seems to happen because the "e" character has participated in a previous match (s+e), because the "e" character is right in the middle of two pluses (Teni s+e+S quash).
If you test the regex with the following input, it matches the last "+":
?busca=tenis+nike+categoria:"Tenis_e+Squash"&pagina=4&operador=or
(changed "s+e" for "s_e" in order not to cause the "e" character to participate in the match).
Would someone please shed a light on that?
Thanks in advance!
In a consecutive match the search for the next match starts at the position of the end of the previous match. And since the the non-whitespace character after the + is matched too, the search for the next match will start after that non-whitespace character. So a sequence like s+e+S you will only find one match:
s+e+S
\_/
You can fix that by using look-around assertions that don’t match the characters of the assumption like:
\S\++(?=\S)
This will match any non-whitespace character followed by one or more + only if it is followed by another non-whitespace character.
But tince whitespace is not allowed in a URI query, you don’t need the surrounding \S at all as every character is non-whitespace. So the following will already match every sequence of one or more + characters:
\++
You are correct: The fourth match doesn't happen because the surrounding character has already participated in the previous match. The solution is to use lookaround (if your regex implementation supports it - JavaScript doesn't support lookbehind, for example).
Try
(?<!\s)\++(?!\s)
This matches one or more + unless they are surrounded by whitespace. This also works if the plus is at the start or the end of the string.
Explanation:
(?<!\s) # assert that there is no space before the current position
# (but don't make that character a part of the match itself)
\++ # match one or more pluses
(?!\s) # assert that there is no space after the current position
If your regex implementation doesn't support lookbehind, you could also use
\S\++(?!\s)
That way, your match would contain the character before the plus, but not after it, and therefore there will be no overlapping matches (Thanks Gumbo!). This will fail to match a plus at the start of the string, though (because the \S does need to match a character). But this is probably not a problem.
You can use the regex:
(?<=\S)\++(?=\S)
To match only the +'s that are surrounded by non-whitespace.