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.
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.
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
I am trying to match a string the 2nd word after "Vores ref.:" using positive lookbehind. It works in online testers like https://regexr.com/, but my tool Alteryx dont allow quantifiers like + in a lookbehind.
"ABC This is an example Vores ref.: 23244-2234 LW782837673 Test 2324324"
(?<=Vores\sref.:\s\d+-\d+\s+)\w+ is correctly matching the LW78283767, on regexr.com but not in Alteryx.
How can I rewrite the lookahead expression by using quantifiers but still get what I want?
You can use a replacement approach here using
.*?\bVores\s+ref\.:\s+\d+-\d+\s+(\w+).*
Replace with $1.
See the regex demo.
Details:
.*? - any 0+ chars other than line break chars, as few as possible
\bVores - whole word Vores
\s+ - one or more whitespaces
ref\.: - ref.: substring
\s+ - one or more whitespaces
\d+-\d+ - one or more digits, - and one or more digits
\s+ - one or more whitespaces
(\w+) - Capturing group 1: one or more word chars.
.* - any 0+ chars other than line break chars, as many as possible.
You can use a capture group instead.
Note to escape the dot \. to match it literally.
\bVores\sref\.:\s\d+-\d+\s+(\w+)
The pattern matches:
\bVores\sref\.:\s\d+-\d+\s+ Your pattern turned into a match
(\w+) Capture group 1, match 1+ word characters
Regex demo
I am trying to find a regex expression to match strings with 4 repeating digits and optional hyphens in between, such as:
-3-3-3-3-
-1111-
2222-
0-0-00
Currently I have:
\-?(\d(\-*))\1{3}\-?
which matches the first two but not the last two. Any suggestions?
You may use
^-?(\d)(?:-?\1){3}-?$
See the regex demo. To find the pattern in a larger string, remove the ^ and $ anchors:
-?(\d)(?:-?\1){3}-?
If the pattern is a part of a longer pattern, you might have to adjust the backreference number (if there are other capturing groups in the pattern).
Details
^ - start of string
-? - an optional -
(\d) - Group 1: any digit
(?:-?\1){3} - three occurrences of an optional - and then the same value as captured in Group 1
-? - an optional hyphen
$ - end of the string.
Here is my samples string
Google Chrome-Helper -type=renderer -field-trial-handle=1
But I want just Google Chrome-Helper
Ive tried: ^.*[ ][-] but captures up to the last parameter.
Example Here
You need to use lazy dot matching and either use capturing or a lookahead:
^(.*?)\s+-
(your value will be in Group 1) or
^.*?(?=\s+-)
See the regex demo with capturing and with a lookahead.
Details:
^ - start of string anchor
.*? - any 0+ chars other than a newline, as few as possible (i.e. the subsequent subpatterns are tried first, this one is skipped, the regex engine only comes back here if they fail to find a match)
(?=\s+-) - a positive lookahead that requires 1+ whitespace and then a hyphen.