Repeated variable length regexp matching - regex

I have an expression
AA-BB/CC/DD
I want to convert this to
<AA-BB> <AA-CC> <AA-DD>
All I can do is configure this as a regexp substitution. I can't figure it out.
AA should match at the beginning of a line. - and / are literal characters, BB,CC and DD are numbers, i.e \d+
So a first draft is ...
^(\w+)([\-/]\d+)+
but I want all matches, not just the greedy one.
(actually this one matches AA-BB-CC-DD too, but that's ok although it's not according to spec)

No, you can't do that with regex. Probably with .net, because there you can access all intermediate results of repeated capturing groups ...
Repeating a Capturing Group vs. Capturing a Repeated Group
That is the problem, if you do something like ^(\w+)([\-/]\w+)+ the value stored in group2 is always only the last pattern it matched. Your task is not possible with regex/replace.
I would do something like:
^(\w+)-([\w+\/]+)
Then split the content of group 2 by "/" and combine group1 with each element of the array resulting from the split.

Related

RegEx to compare two substrings within a line

I have a consistently formatted lines that are essentially delimited field/value pairs and I need to use a single regex expression to compare the values of two fields and match if they are the same. I don't need the values, just the match/no match comparison. Here are two examples, the first should match the second should not.
##field1##someValue##field2##someValue##otherField##otherValue
##field1##someValue##field2##someDIFFERENTValue##otherField##otherValue
I can match any field or value on a known pattern, and I can use lookarounds to do AND operations.
But I don't know how to extract and save the value of some pattern match, like field1##(.*?)##
, and then use that "someValue" in an expression comparing it to the "contents" of field2##(.*?)##
Thanks in advance for any assistance.
You may use
##field1##([^#]*)##field2##(?=\1##)
##field1##((?:(?!##).)*?)##field2##(?=\1##)
Use the former if there can be no # in value contents. Else, use the latter regex.
See the regex demo
Details
##field1## - a literal string
([^#]*) - Capturing group 1 (later in the pattern, it can be referred to with \1 backreference): any 0 or more chars other than #
##field2## - a literal string
(?=\1##) - immediately to the right of the current location, there must be the same value as captured in Group 1 followed with ##. If end of string can occur, replace with (?=\1(?:##|$)).
The (?:(?!##).)*? pattern "roughly" means any text but ##, it actually matches any single char (other than a line break char), as few occurrences as possible, that does not start a ## char sequence.

Regular Expression Replace Time Value between Date-Time Format

I have an XML file with date-time formats looking like this:
<published>2019-01-03T23:54:00.000+10:00</published>
and this
<published>2019-01-07T14:22:00.001+10:00</published>
and so on, where the time value is 23:54:00.000 and 14:22:00.001.
How do I replace just the time value between the <published></published> tags with regular expressions? For example, I want to replace both time values with 03:00:00.000 so the first example becomes
<published>2019-01-03T03:00:00.000+10:00</published>
My aim is to use any existing tools/apps Notepad++ or websites since it is much faster, not any specific programming languages.
First, the obligatory warning to not try to parse xml/html with regex. It's fine if this is a once-off reformatting task and you have control over the data. A regex solution will not be very robust...
That out of the way, you will need a tool that can handle capture groups with regex, so you can match on the whole published tag and avoid false positives. A regex like this might do the trick (adjust the capture grouping as appropriate for your tool):
(\<published\>\d\d\d\d-\d\d-\d\dT)\d\d:\d\d:\d\d\.\d\d\d(\+\d\d:\d\d\<\/published\>)
Note that the above is a regex in PCRE format - demo on regex101. You may need to adjust to suit the format your tool uses.
In this regex, there are two capture groups, one before and one after the time you want to replace. An example string that you could use in the replace field of your chosen tool would be: \103:00:00.000\2 (using \1 syntax for backreferences).
Try this regex:
(<published>\d{4}(?:-\d{2}){2}T)\d{2}(?::\d{2}){2}\.\d{3}([^<]*<\/published>)
Click for Demo
Replace each match with \103:00:00.000\2 i.e. Group 1 contents followed by 03:00:00.000 followed by Group 2 contents.
Explanation:
(<published>\d{4}(?:-\d{2}){2}T) - matches <published> followed by 4 digits followed by - followed by 2 digits followed by - followed by 2 digits followed by the letter T. This sub-match is captured in Group 1
\d{2}(?::\d{2}){2}\.\d{3} - matches time of the format XX:XX:XX.XXX where X is a digit.
([^<]*<\/published>) - matches 0+ occurrences of any character that is not a < followed by </published>. This sub-match is captured in Group 2.
Before Replace:
After Replace:

Detect multiple periods in Regex and kill entire match

I'm trying to detect a price in regex with this:
^\-?[0-9]+(,[0-9]+)?(\.[0-9]+)?
This covers:
12
12.5
12.50
12,500
12,500.00
But if I pass it
12..50 or 12.5.0 or 12.0.
it still returns a match on the 12 . I want it to negate the entire string and return no match at all if there is more than one period in the entire string.
I've been trying to get my head around negative lookaheads for an hour and have searched on Stack Overflow but can't seem to find the right answer. How do I do this?
What you are looking for, is this:
^\d+(,\d{3})*(\.\d{1,2})?$
What it does:
^ Start of Line
\d+ one or more Digits followed by
(,\d{3})* zero, one or more times a , followed by three Digits followed by
(\.\d{1,2})? one or zero . followed by one or two Digits followed by
$ End of Line
This will only match valid Prices. The Comma (,) is not obligatory in this Regex, but it will be matched.
Look here: http://www.regextester.com/?fam=98001
If you work with Prices and want to store them in a Database I recommend saving them as INT. So 1,234,56 becomes 123456 or 1,234 becomes 123400. After you matched the valid price, all you have to do is to remove the ,s, split the Value by the Dot, and fill the Value of [1] with str_pad() (STR_PAD_RIGHT) with Zeros. This makes Calculations easier, in special when you work with Javascript or other different Languages.
Your regex:
^\-?[0-9]+(,[0-9]+)?(\.[0-9]+)?
Note: The regex you provided does not seem to work for 12 (without "."). Since you didn't add a quantifier after \., it tries to match that pattern literally (.).
While there are multiple ways to solve this and the most "correct" answer will depend on your specific requirements, here's a regex that will not match 12..1, but will match 12.1:
(^\-?[0-9]+(?:,[0-9]+)?(?:\.[0-9]+))+
I surrounded the entire regex you provided in a capturing group (...), and added a one or more quantifier + at the end, so that the entire regex will fail if it does not satisfy that pattern.
Also (this may or may not be what you want), I modified the inner groups into non-capturing groups (?: ... ) so that it does not return unnecessary groups.
This site offers a deconstruction of regexes and explains them:
For the regex provided: https://regex101.com/r/EDimzu/2
Unit tests: https://regex101.com/r/EDimzu/2/tests (Note the 12 one's failure for multiple languages).
You can limit it by requiring there is only 0 or 1 periods like this:
^[0-9,]+[\.]{0,1}?[0-9,]+$

Regex to match number specific number in a string

I'm trying to fix a regex I create.
I have an url like this:
http://www.demo.it/prodotti/822/Panasonic-TXP46G20E.html
and I have to match the product ID (822).
I write this regex
(?<=prodotti\/).*(?<=\/)
and the result is "822/"
My match is always a group of numbers between two / /
You're almost there!
Simply use:
(?<=prodotti\/).*?(?=\/)
instead of:
(?<=prodotti\/).*(?<=\/)
And you're good ;)
See it working here on regex101.
I've actually just changed two things:
replaced that lookbehind of yours ((?<=\/)) by its matching lookahead... so it asserts that we can match a / AFTER the last character consumed by .*.
changed the greediness of your matching pattern, by using .*? instead of .*. Without that change, in case of an url that has several / following prodotti/, you wouldn't have stopped to the first one.
i.e., given the input string: http://www.demo.it/prodotti/822/Panasonic/TXP46G20E.html, it would have matched 822/Panasonic.

Fetch one out of two Numbers out of String

I hav a list of strings, such as: Ø20X400
I need to extract the first of the numbers - between Ø and X
I've come so far to match the numbers in general with \d+ - as simple as it is...
But I need an expression to get the first value separated, not both of them...
You can use lookarounds (?<=..) and (?=..):
(?<=Ø)\d+(?=X)
or in Java style:
(?<=Ø)\\d+(?=X)
A second way is to use a capture group:
Ø(\d+)X
or
Ø(\\d+)X
Then you can extract the content of the group.
The regex engines I know parse \n as a newline. \d is used for numbers.
The following regex gives you the first number between a Ø and a X in a capture group:
^.*?Ø(\d+)X.*
Edit live on Debuggex
This Regex will do it for you, (\d+?)X, and here is a Rubular to prove it. See, you want to group digits together, but make it non-greedy, ending the evaluation on X.
Try this one:
\d+(?=\D)
Should find first number wich has some not a number ahead
With normal regular expressions, I would say:
Ø(\d+)X
This finds the Ø character, followed by one or more numbers, followed by an X. Also, the numbers will be stored in the first capture group. Capture groups differ from one regex implementation to another, but this would typically be denoted by \1. Capture group zero, \0, is usually the matched string itself. In this version, \d denotes digits 0-9, but if your regex engine uses \n for that purpose, use:
Ø(\n+)X