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I'm searching for a pattern for matching numbers with hyphen at the end like this :
125,000-
1.234,567-
60,000-
Just try with following regex:
/\d[.,\d]*-/
Or even:
/\d([.,]?\d+)*-/
NOTE Aleš Krajník's answer is basically the same as the answer I finally came to, except that his uses non-capturing grouping (as captures are not required)... he should get the votes IMHO as he was first
Note that in the following answer I'm assuming that , comma is the decimal separator, and that the . point is the thousands separator (eg for European numbering).
I believe the following is "correct":
^\d{1,3}(.\d{3})*(,\d+)?-$
This matches eg:
1-
12-
123-
123.456-
123.456.789-
1,0-
1,01-
1,001-
1,0001-
123.456,01-
123.456.789,0001-
etc
But will not match eg
1234-
123,-
123.4-
123.1,001-
123.45-
1..1..1-
1.1.1-
1,1,1-
.,-
etc.
The exact regex should read: \d{1,3}(?:\.\d{3})*(?:,\d+)?-
Try something like this:
[0-9.,]+-
\d{1,3}(?:[,]\d{3})*- takes internationalisation into account. The one below allows strings like 1..9 to match, which really should not.
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The problem in Akeneo seems to be that simple regex combinations not working. I think the functionality (a single or group regex combination) is not integrated/implemented proper in Akeneo. If there is anybody out there who knows a trick to do a regex combination please let me know.
Tried to figure out how to make regex with | OR working in Akeneo "attributes".
the simple Example not working either a syntax error or no matching in Akeneo:
find this "323"
or find "123456"
\d{3}|\d{6}
Can anybody help?
According to the documentation, you need to use regex literal notation, and anchor the match both at the start and end of the string (so, add a grouping):
/^(\d{6}|\d{3})$/
Here, / are regex delimiters, ^ matches the start of string, (...) is a capturing group that contains two alternatives, six digits or three digits, and then end of string anchor, $, follows.
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I'm trying to work with regex but I'm still not capable of. Asking for your help!
I have links like these:
https://open.spotify.com/track/1Q07lxRM6aQJYtRFzQUtwu?si=LrEcPs3pSxaznY2GLH4V8Q
https://open.spotify.com/album/7lyxArCeA4kkHRiYpnh8eA
open.spotify.com/artist/1mBlZPMpRL8wT9aHBnBBph
I'd like to match the "artist" part in the last link. How can I do it? I thought about using slashes as "separator" than get the string from there but I have no idea.
Assuming you want to extract the top directory name following the
domain name, how about:
import re
url = 'open.spotify.com/artist/1mBlZPMpRL8wT9aHBnBBph'
m = re.search(r'(?:https?://)?[^/]+/([^/]+)', url, re.IGNORECASE)
if m:
print(m.group(1))
Output:
artist
Below, I'm using Python.
It's just build a case insensitive regex that advance any chars (.) with +(1 or more chars) and than use a prefix that occurs always before artist link artist/, finally use ()s to group. Inside ()s uses a class for letters and digits with + suffix (1 or more chars)
.+artist/([a-z1-9]+)
The replaced string it just
/1
The details vary a little bit according to the programming language adopted
See here:
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I have this regex string:
("prodcssstart.*prodcssend","xx")
and my file like this:
prodcssstart
<!--<link href="content/bundles/css.min.css" rel="stylesheet" />-->
prodcssend
But when I run it then it fails to find the expression.
Can someone suggest what I might be doing wrong, I've simplified it so much but I am thinking maybe there is a problem with the .* that I use to match everything. Any help would be much appreciated.
First of all, you can't parse HTML with regex.
But in your case the problem is most likely caused by newlines, because by default the dot doesn't match the newline. You need to pass the proper switch to disable this (e.g. re.DOTALL in Python and s in Perl).
You could come up with a tempered greedy token solution:
prodcssstart
(?:(?!prodcssend).)*
prodcssend
Breakdown:
look for prodcssstart
match any character as long as it is not followed by prodcssend
match prodcssend
See a demo on regex101.com (mind the different modifiers!).
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How to process search results using regex?
E.g., I have a file with many strings like AB.
I want to get: 'AB'.
The letters always differ.
So I would search for the regex pattern ^\w+\n and want to use the search result, let me use '$#' to depict it, to get '$#'.
Regex:
(?<=:)([^,]+)
REplacement string:
'$1'
DEMO
Just to clarify: You want 71:A,72:BC,73:ABD to become 71:'A',72:'BC',73:'ABD' right?
Do a find/replace in whatever language or program you are using:
Find: (\w+)
Replace: '$1'
This finds ANY multi-letter string in your file and puts ' around it. if you only want to do the ones with [number:string] you will need to use a look-ahead like (?=\d+:) in front of the (\w+). So the whole Find would then look like (?=\d+:)(\w+), similar to what Avinash Raj has posted.
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I would use regex to select one changing number
For example in the following example I would select 12090343 that I could change
I use:
preg_match (/(?<=Dossier.N..)(.*)(?=-)/)
It works but it's not so clean I think because the number of spaces could change and so it will not detect the number any more
Dossier N° 11110144-001 Pvt du : 03/09/2013 à 7:16
It looks fine, but you could slightly clean it up by .not grouping the number and making it match digits:
(?<=Dossier.{0,3}N.{0,3})\d+(?=-)
Most regex engines can't handle look behinds of arbitrary length, so rather than using the simpler (but unbounded) expression \s*, you must use a limited length expression like \s{0,3} to allow "some" whitespace.
You can try the following expression:
/Dossier[^0-9]*\K([0-9]*)(?=-)/
Regex101 Demo