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Using RegEx, what's the best way to capture groups of digits, ignoring any whitespace in them

Given the following string...
ABC DEF GHI: 319 022 6543 QRS : 531 450
I'm trying to extract all ranges that start/end with a digit, and which may contain whitespace, but I want that whitespace itself removed.
For instance, the above should yield two results (since there are two 'ranges' that match what I aim looking for)...
3190226543
531450
My first thought was this, but this matches the spaces between the letters...
([\d\s])
Then I tried this, but it didn't seem to have any effect...
([\d+\s*])
This one comes close, but its grabbing the trailing spaces too. Also, this grabs the whitespace, but doesn't remove it.
(\d[\d\s]+)
If it's impossible to remove the spaces in a single statement, I can always post-process the groups if I can properly extract them. That most recent statement comes close, but how do I say it doesn't end with whitespace, but only a digit?
So what's the missing expression? Also, since sometimes people just post an answer, it would be helpful to explain out the RegEx too to help others figure out how to do this. I for one would love not just the solution, but an explanation. :)
Note: I know there can be some variations between RegEx on different platforms so that's fine if those differences are left up to the reader. I'm more interested in understanding the basic mechanics of the regex itself more so than the syntax. That said, if it helps, I'm using both Swift and C#.

You cannot get rid of whitespace from inside the match value within a single match operation. You will need to remove spaces as a post-processing step.
To match a string that starts with a digit and then optionally contains any amount of digits or whitespaces and then a digit you can use
\d(?:[\d\s]*\d)?
Details:
\d - a digit
(?:[\d\s]*\d)? - an optional non-capturing group matching
[\d\s]* - zero or more whitespaces / digits
\d - a digit.
See the regex demo.

Update regular expression to only allow single spaces [duplicate]

Right now I have a regex that prevents the user from typing any special characters. The only allowed characters are A through Z, 0 through 9 or spaces.
I want to improve this regex to prevent the following:
No leading/training spaces - If the user types one or more spaces before or after the entry, do not allow.
No double-spaces - If the user types the space key more than once, do not allow.
The Regex I have right now to prevent special characters is as follows and appears to work just fine, which is:
^[a-zA-Z0-9 ]+$
Following some other ideas, I tried all these options but they did not work:
^\A\s+[a-zA-Z0-9 ]+$\A\s+
/s*^[a-zA-Z0-9 ]+$/s*
Could I get a helping hand with this code? Again, I just want letters A-Z, numbers 0-9, and no leading or trailing spaces.
Thanks.

You can use the following regex:
^[a-zA-Z0-9]+(?: [a-zA-Z0-9]+)*$
See regex demo.
The regex will match alphanumerics at the start (1 or more) and then zero or more chunks of a single space followed with one or more alphanumerics.
As an alternative, here is a regex based on lookaheads (but is thus less efficient):
^(?!.* {2})(?=\S)(?=.*\S$)[a-zA-Z0-9 ]+$
See the regex demo
The (?!.* {2}) disallows consecutive spaces and (?=.*\S$) requires a non-whitespace to be at the end of the string and (?=\S) requires it at the start.

Conditional match without false force a match?

I'm using the following regex in c# to match some input cases:
^
(?<entry>[#])?
(?(entry)(?<id>\w+))
(?<value>.*)
$
The options are ignoring pattern whitespaces.
My input looks as follows:
hello
#world
[xxx]
This all can be tested here: DEMO
My problem is that this regex will not match the last line. Why?
What I'm trying to do is to check for an entry character. If it's there I force an identifier by \w+. The rest of the input should be captured in the last group.
This is a simplyfied regex and simplyfied input.
The problem can be fixed if I change the id regex to something like (?(entry)(?<id>\w+)|), (?(entry)(?<id>\w+))? or (?(entry)(?<id>\w+)?).
I try to understand why the conditional group doesn't match as stated in original regex.
I'm firm in regex and know that the regex can be simplyfied to ^(\#(?<id>\w+))?(?<value>.*)$ to match my needs. But the real regex contains two more optional groups:
^
(?<entry>[#])?
(\?\:)?
(\(\?(?:\w+(?:-\w+)?|-\w+)\))?
(?(entry)(?<id>\w+))
(?<value>.*)
$
That's the reason why I'm trying to use a conditional match.
UPDATE 10/12/2018
I tested a little arround it. I found the following regex that should match on every input, even an empty one - but it doesn't:
(?(a)a).*
DEMO
I'm of the opinion that this is a bug in .net regex and reported it to microsoft: See here for more information

There is no error in the regex parser, but in one's usage of the . wildcard specifier. The . specifier will consume all characters, wait for it, except the linefeed character \n. (See Character Classes in Regular Expressions "the any character" .])
If you want your regex to work you need to consume all characters including the linefeed and that can be done by specify the option SingleLine. Which to paraphrase what is said
Singline tells the parser to handle the . to match all characters including the \n.
Why does it still fail when not in singleline mode for the other lines are consumed? That is because the final match actually places the current position at the \n and the only option (as specified is use) is the [.*]; which as we mentioned cannot consume it, hence stops the parser. Also the $ will lock in the operations at this point.
Let me demonstrate what is happening by a tool I have created which illustrates the issue. In the tool the upper left corner is what we see of the example text. Below that is what the parser sees with \r\n characters represented by ↵¶ respectively. Included in that pane is what happens to be matched at the time in yellow boxes enclosing the match. The middle box is the actual pattern and the final right side box shows the match results in detail by listening out the return structures and also showing the white space as mentioned.
Notice the second match (as index 1) has world in group capture id and value as ↵.
I surmise your token processor isn't getting what you want in the proper groups and because one doesn't actually see the successful match of value as the \r, it is overlooked.
Let us turn on Singline and see what happens.
Now everything is consumed, but there is a different problem. :-)

Regex pattern in C# with empty space

I am having issue with a reg ex expression and can't find the answer to my question.
I am trying to build a reg ex pattern that will pull in any matches that have # around them. for example #match# or #mt# would both come back.
This works fine for that. #.*?#
However I don't want matches on ## to show up. Basically if there is nothing between the pound signs don't match.
Hope this makes sense.
Thanks.

Please use + to match 1 or more symbols:
#+.+#+
UPDATE:
If you want to only match substrings that are enclosed with single hash symbols, use:
(?<!#)#(?!#)[^#]+#(?!#)
See regex demo
Explanation:
(?<!#)#(?!#) - a # symbol that is not preceded with a # (due to the negative lookbehind (?<!#)) and not followed by a # (due to the negative lookahead (?!#))
[^#]+ - one or more symbols other than # (due to the negated character class [^#])
#(?!#) - a # symbol not followed with another # symbol.

Instead of using * to match between zero and unlimited characters, replace it with +, which will only match if there is at least one character between the #'s. The edited regex should look like this: #.+?#. Hope this helps!
Edit
Sorry for the incorrect regex, I had not expected multiple hash signs. This should work for your sentence: #+.+?#+
Edit 2
I am pretty sure I got it. Try this: (?<!#)#[^#].*?#. It might not work as expected with triple hashes though.

Try:
[^#]?#.+#[^#]?
The [^ character_group] construction matches any single character not included in the character group. Using the ? after it will let you match at the beginning/end of a string (since it matches the preceeding character zero or more times. Check out the documentation here

Does regex + symbol apply to previous element only?

In order to match all strings beginning with 04 and only containing digits, will the following work?
Regex.IsMatch(str, "^04[0-9]+$")
Or is another set of brackets necessary?
Regex.IsMatch(str, "^04([0-9])+$")

In Regex:
[character_group]
Matches any single character in character_group.
\d
Matches any decimal digit.
+
Matches the previous element one or more times.
(subexpression)
Captures the matched subexpression and assigns it a ordinal number.
^
The match must start at the beginning of the string or line.
$
The match must occur at the end of the string or before \n at the end of the line or string.
so that this code could be helpful:
Regex.IsMatch(str, "^04\d+$")
and all of your code works correctly.

Your first regex is correct, but the second one isn't. It matches the same things as the first regex, but it does a lot of unnecessary work in the process. Check it out:
Regex.IsMatch("04123", #"^04([0-9])+$")
In this example, the 1 is captured in group #1, only to be overwritten by 2 and again by 3. It's almost never a good idea to add a quantifier to a capturing group. For a detailed explanation, read this.
But maybe it's precedence rules you're asking about. Quantifiers have higher precedence than concatenation, so there's no need to isolate the character class with parentheses (if that's what you're doing).