I have a string looks like this
#123##1234###2356####69
It starts with # and followed by any digits, every time the # appears, the number of # increases, first time 1, second time 2, etc.
It's similar to this regex, but since I don't know how long this pattern goes, so it's not very useful.
^#\d+##\d+###\d+$
I'm using PCRE regex engine, it allows recursion (?R) and conditions (?(1)...) etc.
Is there a regex to validate this pattern?
Valid
#123
#12##235
#1234##12###368
#1234##12###368####22235#####723356
Invalid
##123
#123###456
#123##456##789
I tried ^(?(1)(?|(#\1)|(#))\d+)+$ but it doesn't seem to work at all
You can do this using PCRE conditional sub-pattern matching:
^(?:((?(1)\1)#)\d+)++$
RegEx Demo
RegEx Details:
^: Start
(?:: Start non-capture group
(: Start capture group #1
(?(1)\1): if/then/else directive that means match back-reference \1 only if 1st capture group is available otherwise match null
#: Match an additional #
): End capture group #1
\d+: Match 1+ digits
)++: End non-capture group. Match 1+ of this non-capture group.
$: End
One option could be optionally matching a backreference to group 1 inside group 1 using a possessive quantifier \1?+# adding # on every iteration.
^(?:(\1?+#)\d+)++$
^ Start of string
(?: Non capture group
(\1?+#)\d+ Capture group 1, match an optional possessive backreference to what is already captured in group 1 and add matching a # followed by 1+ digits
)++ Close the non capture group and repeat 1+ times possessively
$ End of string
Regex demo
I think you can use forward-referencing here:
^(?:((?:\1(?!^)|^)#)\d+)+$
See the regex demo.
Details:
^ - start of string
(?:((?:\1(?!^)|^)#)\d+)+ - one or more occurrences of
((?:\1(?!^)|^)#) - Group 1 (the \1 value): start of string or an occurrence of the Group 1 value if it is not at the string start position
\d+ - one or more digits
$ - end of string.
NOTE: This technique does not work in regex flavors that do not support forward referencing, like ECMAScript based flavors (e.g. JavaScript, VBA, C++ std::regex)
Despite there are already working answers, and inspired by Wiktor's answer, I came up this idea:
(?:(^#|#\1)\d+)+$
Which is also quite short and effective(also works for non pcre environment).
See the test cases
Related
I have the following example of numbers, and I need to add a zero after the second period (.).
1.01.1
1.01.2
1.01.3
1.02.1
I would like them to be:
1.01.01
1.01.02
1.01.03
1.02.01
I have the following so far:
Search:
^([^.])(?:[^.]*\.){2}([^.].*)
Substitution:
0\1
but this returns:
01 only.
I need the 1.01. to be captured in a group as well, but now I'm getting confuddled.
Does anyone know what I am missing?
Thanks!!
You may try this regex replacement with 2 capture groups:
Search:
^(\d+\.\d+)\.([1-9])
Replacement:
\1.0\2
RegEx Demo
RegEx Details:
^: Start
(\d+\.\d+): Match 1+ digits + dot followed by 1+ digits in capture group #1
\.: Match a dot
([1-9]): Match digits 1-9 in capture group #2 (this is to avoid putting 0 before already existing 0)
Replacement: \1.0\2 inserts 0 just before capture group #2
You could try:
^([^.]*\.){2}\K
Replace with 0. See an online demo
^ - Start line anchor.
([^.]*\.){2} - Negated character 0+ times (greedy) followed by a literal dot, matched twice.
\K - Reset starting point of reported match.
EDIT:
Or/And if \K meta escape isn't supported, than see if the following does work:
^((?:[^.]*\.){2})
Replace with ${1}0. See the online demo
^ - Start line anchor.
( - Open 1st capture group;
(?: - Open non-capture group;
`Negated character 0+ times (greedy) followed by a literal dot.
){2} - Close non-capture group and match twice.
) - Close capture group.
Using your pattern, you can use 2 capture groups and prepend the second group with a dot in the replacement like for example \g<1>0\g<2> or ${1}0${2} or $10$2 depending on the language.
^((?:[^.]*\.){2})([^.])
^ Start of string
((?:[^.]*\.){2}) Capture group 1, match 2 times any char except a dot, then match the dot
([^.].*) Capture group 2, match any char except a dot
Regex demo
A more specific pattern could be matching the digits
^(\d+\.\d+\.)(\d)
^ Start of string
(\d+\.\d+\.) Capture group 1, match 2 times 1+ digits and a dot
(\d) Capture group 2, match a digit
Regex demo
For example in JavaScript
const regex = /^(\d+\.\d+\.)(\d)/;
[
"1.01.1",
"1.01.2",
"1.01.3",
"1.02.1",
].forEach(s => console.log(s.replace(regex, "$10$2")));
Obviously, there will be tons of solutions for this, but if this pattern holds (i.e. always the trailing group that is a single digit)... \.(\d)$ => \.0\1 would suffice - to merely insert a 0, you don't need to match the whole thing, only just enough context to uniquely identify the places targeted. In this case, finding all lines ending in a . followed by a single digit is enough.
Hello I am trying to capture a repetitive pattern x.x.x.x where x is a number (one digit or more) but to capture only if the whole line or string starts or contains a "#" before the group:
This is a test # 1.2.3.4 which 11.2.38.49 should pass 3.2.4.5 and capture the 3 groups
This # is a test 1.2.3.4 which 1.2.3.4 should pass 3.2.4.5 too
#This is a test 1.2.3.4 which 1.2.3.4 should pass 3.2.4.5 too
This is a test 1.2.3.4 which # 1.2.3.4 pass 3.2.4.5 but don't capture the first group
This test 1.2.3.4 1.2.3.4 should # pass but capture nothing
This test 1.2.3.4 1.2.3.4 should test fail
so far I have (\d*\.\d+)+ is it even possible to find a regex for it?
https://regex101.com/r/G81BBj/1
There are different regex engines where some support different features than others.
If you want to match all occurrences of a dot-separated chunk, you could make use of a quantifier in a lookbehind assertion.
Match 1+digits, and repeat matching a dot and 1+ digits at least
one or more times to prevent matching only digits.
(?<=#.*)\d+(?:\.\d+)+
(?<=#.*) Positive lookbehind, assert that there is an # on the left
\d+ Match 1+ digits
(?: Non capture group
\.\d+ Match a dot and 1+ digits
)+ Close group and repeat 1+ times to make sure there is at least 1 dot present
.NET regex demo
Another option could be when the engine supports the \G anchor which will assert either at the start of the string, or asserts the position at the end of the previous match.
(?:^[^\r\n#]*#|\G(?!^)).*?(\d+(?:\.\d+)+)
(?: Non capture group
^[^\r\n#]*# Match from the start of the string until the first #
| Or
\G(?!^) Assert the position at the end of the previous match, not at the start
) Close group
.*? Match as least char as possible
( Capture group 1
\d+(?:\.\d+)+ Match a dot-separated chunk
) Close group 1
Regex demo
If the engine does not support \G and the engine will recognize \G as a G char, it will first try to match until the first occurrence of an # followed by matching until the dot-separated chunk.
After matching the first dot-separated chunk, it tries for all the following positions to match the first part of the alternation, which can not match due to the ^ anchor. It tries the second part, but that will not match because there is no G char in the example data so eventually there will be only a single match.
If \K to is also supported to clear the starting point of the reported match, for example in pcre, you could omit the capturing group and get the match only:
(?:^[^\r\n#]*#|\G(?!^)).*?\K\d+(?:\.\d+)+
Regex demo
With any regex engine that supports unknown width patterns in lookbehind (.NET, ECMAScript 2018+, PyPi regex module in Python, JGSoft Software), you may use
(?<=#.*?)\d+(?:\.\d+)*
See the regex demo.
Details
(?<=#.*?) - a location that is immediately preceded with # char that may be followed with any 0 or more chars other than line break chars, as few as possible
\d+(?:\.\d+)* - one or more digits followed with 0 or more occurrences of a dot and then 1+ digits.
I'm trying the create a regex to catch my url and his, optionnals, groups. The regex works fine if the url is complete. The optionnals groups are not optionnals at all.
Regex :
\/(.+)(?:\/(.+))(?:(?:\?(.+)))
Urls to catch :
/taxi
/taxi/lyon
/taxi/lyon?coordinates=7542
https://regex101.com/r/NKFkwq/4/
As you can see, the third line is catched. But i'd like the first and second too.
I thought the ?: will be enought to do that, but i missed something...
Thanks a lot for your help !
Cheers
EDIT and answer
Thanks in the comments for helping me. Here the great regex (the one i expected) : https://regex101.com/r/NKFkwq/8
Indeed ?: is about ignoring a match, not made him optionnal.
Your pattern consists of capturing and non capturing groups. The (?: denotes a non capturing group.
If you want to match all 3 lines, you could use match the part starting from the first forward slash and make the part starting from the second forward slash optional.
^/[^\s/]+(?:/[^\s/]+)?$
^ Start of string
/[^\s/]+ Match / and match 1+ times any char except a whitespace or /
(?: Non capturing group
/[^\s/]+ Match / and match 1+ times any char except a whitespace or /
)? Close non capturing group and make it optional
$ End of string
Regex demo
If you want to have capturing groups, but don't want to match /taxi?coordinates=7542 you could nest the groups and make them optional as well.
^/\w+(/\w+(\?\S*)?)?$
^ Start of string
/\w+ Match / and 1+ word chars
( Capture group 1
/\w+ Match / and 1+ word chars
( Capture group 2
\?\S* Match ? and 0+ times a non whitespace char
)? Close group 2
)? Close group 1
$ End of string
Regex demo
I need to extract 1234567 from below URLs
http://www.test.in/some--wonders-1234567---2
http://www.test.in/some--wonders-1234567
I tried with .*\-([0-9]+)(?:-{2,}2)?.
but for the first URL it returned 2, but this is in non-capturing group.
Please give me a solution. I am digging it for so long. not getting any idea.
Try this one:
.*?\-([0-9]+)(?:-{2,}2|$)
It sets lazy mode for first .* pattern, you can also remove it at all with same effect:
\-([0-9]+)(?:-{2,}2|$)
If your regex engine supports negative look behinds (some do not), you can do it this way:
(?<!\d+-+)\d+
It gives you any non-empty digit string, which is not preceded by (minuses followed by digits).
Big advantage is that you don't have to use groups here - regex itself returns what you want.
You could match a - followed by one or more digits which you could capture in a group ([0-9]+). This group will contain the value you want to extract.
Then an optional part (?:-{2,}[0-9]+)? that would match ---2 followed by asserting the end of the line $.
-(\d+)(?:-{2,}\d+)?$
Explanation
- Match literally
(\d+) Capture one or more digits in a group
(?: Non capturing group
-{2,} Match 2 or more times -
\d+ Match one or more digits
)? close non capturing group and make it optional
$ Assert position at the end of the line
Hello i want to match with regex this word
(Parc Installé)
from this text:
31/1/2017 17:19:23,4245986,ct0001#Intotel.int,Parc Installé,100.100.30.100
I did this regex ',[A-Za-zA-zÀ-ú+ \/\w+0-9._%+-]+,'
But the result is : 4245986 ans Parc Installé.
How can i match only Parc Installé
You may try a regex based on a lookahead that will require a comma and digits/commas after it up to the end of string:
[^,]+(?=\s*,[\d.]+$)
See this regex demo
Details:
[^,]+ - 1 or more chars other than ,
(?=\s*,[\d.]+$) - a lookahead requiring
\s* - zero or more whitespaces
, - a comma
[\d.]+ - 1+ digits or dots up to...
$ - ... the end of string
To make it a bit more restrictive, you may replace the lookahead with (?=\s*,\d+(?:\.\d+){3}$) to require 4 sequences of dot-separated 1+ digits. See this regex demo.
If a lookahead is not supported (case with a RE2 engine), you might want to use a capturing group based solution:
([^,]+)\s*,[\d.]+$
Here, the part within (...) will be captured into Group 1 and will be accessible via a backreference or a function like =REGEXEXTRACT in Google Spreasheets that only retrieves the contents of a capturing group if the latter is present in the pattern.