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Find regex pattern match string have multiple condition?

I have some strings formatted as follows:
1=case1,case2,..caseN;2=case1,..,caseN;3=case1, ..,caseN
Note: comma ";" is used to separate cases and case1, case2 are anything like strings, number doesn't matter their type.
I want to find regex pattern to match string
1=home,house;2=abc;3=2019,2021
however, it will not match the following:
1=home,;2=abc;3=2019,2021 (Excess comma mark at case 1)
1=;2=abc,2012;3= (must 1=..; not 1=;)
1=home,age;2 (must 2=.. not 2)
2=home;;3=sea (must ;3 not ;;3)
4=flower;k3=sea (must 3= , not k3)
I tried with the pattern: (\d+={1}[^;]+;). However, it will match if the backstring is not.
Please show me the way.
Many thanks!

Maybe this pattern helps you out:
^\b(?:(?:^|;)\d+=[^,;]+(?:,[^,;]+)*)+$
See the Online Demo
^ - Start string ancor.
\b - Word-boundary.
(?: - Opening 1st non-capture group.
(?:- Opening 2nd non-capture group.
^|; - Alternation between start string ancor or semi-colon.
) - Closing 2nd non-capture group.
\d+= - One or more digits followed by a =.
[^,;]+ - Negated character class, any character other than comma or semicolon one or more times.
(?: - Opening 3rd non-capture group.
, - A comma.
[^,;]+ - Negated character class, any character other than comma or semicolon one or more times.
)* - Close 3rd non-capture group and make it match zero or more times.
)+ - Close 1st non-capture group and make sure it's matches one or more times.
$ - End string ancor.
Note: I went with a negated character class since you mentioned "case1, case2 are anything like strings, number doesn't matter their type", therefor I read there can be spaces, special characters or any kind other than comma and semicolon.

This works on regex101
^(?:\d=(?:\w{1,},)*(?:\w{1,});)*(?:\d=(?:\w{1,},)*\w{1,})$

^(?:\d+=[a-z\d]+(?:,[a-z\d]+)*(?:;|$))+$
Demo
^ : match beginning of string
(?: : begin nc group
\d+=[a-z\d]+ : match 1+ digits, then '=' then 1+ lc letters or digits
(?:,[a-z\d]+) : match ',' then 1+ lc letters or digits in nc group
* : execute nc group 0+ times
(?:;|$) : match ';' or end of string
)+ : end nc group and execute 1+ times
$ : match end of string

I don't know if c# supports recursive pattern, but, if it does, use:
^(\d+=\w+(?:,\w+)*)(?:;(?1))*$
if it doesn't:
^\d+=\w+(?:,\w+)*(?:;\d+=\w+(?:,\w+)*)*$
Demo & explanation

Regex to get digits from a string when there is no separator between digits

I have a string like Acc:123-456-789 and another string like -1234567, I need your help to write an expression to match digits in case there is no separator between the digits.
-*(?!\d*(?:\d*-)$)\d*$
Input strings:
Acc:123-456-789 -12323232 7894596
Desired result:
group 1 12323232
group 2 7894596

I think this ought to work:
(?<=^|\s|\s-)(\d+)(?=\s|$)
Breaking it down:
(?<=^|\s|\s-) - A positive lookbehind that matches the start of the string, whitespace, or whitespace followed by a -.
(\d+) - Matches and captures number sequences.
(?=\s|$) - A positive lookahead that matches whitespace or the end of the string.
** Note: If you need to capture negative number sequences, replace (\d+) with (\-?\d+).
Try it online
Regex reference
Remember for use in C# that you need to escape backslashes or use the # prefix to a string literal (#" ").

Regex with Letters, Numbers, # and spaces

I would like to know how to write a regex in c# that allows only
numbers, letters, spaces and the # symbol.
Valid inputs are:
Abc
Abc def
#Abc
Abc#
Example I tried so far: #"[^\w-\s-#]"

You may use
#"^[A-Za-z0-9\s#]*$"
See the regex demo
The pattern matches:
^ - start of the string
[A-Za-z0-9\s#]* - zero or more (*, if you do not want to match an empty string, use +, 1 or more) occurrences of ASCII letters, digits, any whitespace and # chars
$ - end of string (replace with \z if you do not want to match if the matching string ends with \n char).

RegEx to match a number in the second line

I need a regex to match a number in the second line. Similar input is like this:
^C1.1
xC20
SS3
M 4
Decimal pattern (-?\d+(\.\d+)?) matches all numbers and second number can be get in a loop on the code behind but I need a regular expression to get directly the number in the second line.

/^[^\r\n]*\r?\n\D*?(-?\d+(\.\d+)?)/
This operates by capturing a single line at the beginning of the input:
^ Beginning of the string
[^\r\n]* Anything that isn't a line terminator
\r?\n A newline, optionally preceded by a carriage return
Then all the non digit characters, then your numbers.
Since you've now repeatedly changed your needs, try this on for size:
/(?<=\n\D*)-?\d+(\.\d+)?/

I was able to capture it with this regex.
.*\n\D*(\d*).*\n

Check out group 1 of anything that this matches:
^.*?\r\n.*?(\d+)
If that doesn't work, try this:
^.*?\r\n.*?(\d+)
Both are with multiline NOT set...

I would probably use the captured group in /^.*?\r?\n.*?(-?\d+(?:\.\d+)?)/ where…
^ # beginning of string
.*? # anything...
\r?\n # followed by a new line
.*? # anything...
( # followed by...
-? # an optional negative sign (minus)
\d+ # a number
(?: # -this part not captured explicitly-
\.\d+ # a dot and a number
)? # -and is optional-
)
If it is a flavor that supports lookbehind then there are other alternatives.

Regular expression to find separator dots in formula

The C# expression library I am using will not directly support my table/field parameter syntax:
The following are table/field parameter names that are not directly supported:
TableName1.FieldName1
[TableName1].[FieldName1]
[Table Name 1].[Field Name 1]
It accepts alphanumeric parameters without spaces, or most characters enclosed within square brackets. I would like to use C# regular expressions to replace the dot separators and neighboring brackets to a different delimiter, so the results would be as follows:
[TableName1|FieldName1]
[TableName1|FieldName1]
[Table Name 1|Field Name 1]
I also need to skip any string literals within single quotes, like:
'TableName1.FieldName1'
And, of course, ignore any numeric literals like:
12345.6789
EDIT: Thank you for your feedback on improving my question. Hopefully it is clearer now.

I've written a completely new answer, now that the problem is clarified:
You can do this in a single regex. It is quite bulletproof, I think, but as you can see, it's not exactly self-explanatory, which is why I've commented it liberally. Hope it makes sense.
You're lucky that .NET allows re-use of named capturing groups, otherwise you would have had to do this in several steps.
resultString = Regex.Replace(subjectString,
#"(?: # Either match...
(?<before> # (and capture into backref <before>)
(?=\w*\p{L}) # (as long as it contains at least one letter):
\w+ # one or more alphanumeric characters,
) # (End of capturing group <before>).
\. # then a literal dot,
(?<after> # (now capture again, into backref <after>)
(?=\w*\p{L}) # (as long as it contains at least one letter):
\w+ # one or more alphanumeric characters.
) # (End of capturing group <after>) and end of match.
| # Or:
\[ # Match a literal [
(?<before> # (now capture into backref <before>)
[^\]]+ # one or more characters except ]
) # (End of capturing group <before>).
\]\.\[ # Match literal ].[
(?<after> # (capture into backref <after>)
[^\]]+ # one or more characters except ]
) # (End of capturing group <after>).
\] # Match a literal ]
) # End of alternation. The match is now finished, but
(?= # only if the rest of the line matches either...
[^']*$ # only non-quote characters
| # or
[^']*'[^']*' # contains an even number of quote characters
[^']* # plus any number of non-quote characters
$ # until the end of the line.
) # End of the lookahead assertion.",
"[${before}|${after}]", RegexOptions.Multiline | RegexOptions.IgnorePatternWhitespace);