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Building a regular expression in C#

How to check the following text in C# with Regex:
key_in-get { 43243225543543543 };
or
key_in_set { password123 : 34980430943834 };
I tried to build a regular expression, but I failed after few hours.
Here is my code:
string text1 = "key_in-get { 322389238237 };";
string text2 = "key_in-set { password123 : 322389238237 };";
string pattern = "key_in-(get|set) { .* };";
var result1 = Regex.IsMatch(text, pattern);
Console.Write("Is valid: {0} ", result1);
var result2 = Regex.IsMatch(text, pattern);
Console.Write("Is valid: {0} ", result2);
I have to check if there is "set" or "get".
If the pattern finds "set" then it can only accept following pattern "text123 : 123456789", and if it finds "get" then should accept only "123456789".

You can use
key_in-(?:get|(set)) {(?(1) \w+ :) \w+ };
key_in-(?:get|(set))\s*{(?(1)\s*\w+\s*:)\s*\w+\s*};
key_in-(?:get|(set))\s*{(?(1)\s*\w+\s*:)\s*\d+\s*};
See the regex demo. The second one allows any amount of any whitespace between the elements and the third one allows only digits after : or as part of the get expression.
If the whole string must match, add ^ at the start and $ at the end of the pattern.
Details:
key_in- - a substring
(?:get|(set)) - get or set (the latter is captured into Group 1)
\s* - zero or more whitespaces
{ - a { char
(?(1)\s*\w+\s*:) - a conditional construct: if Group 1 matched, match one or more word chars enclosed with zero or more whitespaces and then a colon
\s*\w+\s* - one or more word chars enclosed with zero or more whitespaces
}; - a literal substring.

In the pattern that you tried key_in-(get|set) { .* }; you are matching either get or set followed by { until the last occurrence of } which could possibly also match key_in-get { }; };
As an alternative solution, you could use an alternation | specifying each of the accepted parts for the get and the set.
key_in-(?:get\s*{\s*\w+|set\s*{\s*\w+\s*:\s*\w+)\s*};
The pattern matches
key_in- Match literally
(?: Non capture group
get\s*{\s*\w+ Match get, { between optional whitespace chars and 1+ word chars
| Or
set\s*{\s*\w+\s*:\s*\w+ Match set, { between optional whitespace chars and word chars on either side with : in between.
) Close non capture group
\s*}; Match optional whitespace chars and };
Regex demo

Match only the nth occurrence using a regular expression

I have a string with 3 dates in it like this:
XXXXX_20160207_20180208_XXXXXXX_20190408T160742_xxxxx
I want to select the 2nd date in the string, the 20180208 one.
Is there away to do this purely in the regex, with have to resort to pulling out the 2 match in code. I'm using C# if that matters.
Thanks for any help.

You could use
^(?:[^_]+_){2}(\d+)
And take the first group, see a demo on regex101.com.
Broken down, this says
^ # start of the string
(?:[^_]+_){2} # not _ + _, twice
(\d+) # capture digits
C# demo:
var pattern = #"^(?:[^_]+_){2}(\d+)";
var text = "XXXXX_20160207_20180208_XXXXXXX_20190408T160742_xxxxx";
var result = Regex.Match(text, pattern)?.Groups[1].Value;
Console.WriteLine(result); // => 20180208

Try this one
MatchCollection matches = Regex.Matches(sInputLine, #"\d{8}");
string sSecond = matches[1].ToString();

You could use the regular expression
^(?:.*?\d{8}_){1}.*?(\d{8})
to save the 2nd date to capture group 1.
Demo
Naturally, for n > 2, replace {1} with {n-1} to obtain the nth date. To obtain the 1st date use
^(?:.*?\d{8}_){0}.*?(\d{8})
Demo
The C#'s regex engine performs the following operations.
^ # match the beginning of a line
(?: # begin a non-capture group
.*? # match 0+ chars lazily
\d{8} # match 8 digits
_ # match '_'
) # end non-capture group
{n} # execute non-capture group n (n >= 0) times
.*? # match 0+ chars lazily
(\d{8}) # match 8 digits in capture group 1
The important thing to note is that the first instance of .*?, followed by \d{8}, because it is lazy, will gobble up as many characters as it can until the next 8 characters are digits (and are not preceded or followed by a digit. For example, in the string
_1234abcd_efghi_123456789_12345678_ABC
capture group 1 in (.*?)_\d{8}_ will contain "_1234abcd_efghi_123456789".

You can use System.Text.RegularExpressions.Regex
See the following example
Regex regex = new Regex(#"^(?:[^_]+_){2}(\d+)"); //Expression from Jan's answer just showing how to use C# to achieve your goal
GroupCollection groups = regex.Match("XXXXX_20160207_20180208_XXXXXXX_20190408T160742_xxxxx").Groups;
if (groups.Count > 1)
{
Console.WriteLine(groups[1].Value);
}

Regular expression matching a given structure

I need to generate a regex to match any string with this structure:
{"anyWord"}{"aSpace"}{"-"}{"anyLetter"}
How can I do it?
Thanks
EDIT
I have tried:
string txt="print -c";
string re1="((?:[a-z][a-z]+))"; // Word 1
Regex r = new Regex(re1,RegexOptions.IgnoreCase|RegexOptions.Singleline);
Match m = r.Match(txt);
if (m.Success)
{
String word1=m.Groups[1].ToString();
Console.Write("("+word1.ToString()+")"+"\n");
}
Console.ReadLine();
but this only matches the word "print"

This would be pretty straight-forward :
[a-zA-Z]+\s\-[a-zA-Z]
explained as follows :
[a-zA-Z]+ # Matches 1 or more letters
\s # Matches a single space
\- # Matches a single hyphen / dash
[a-zA-Z] # Matches a single letter
If you needed to implement this in C#, you could just use the Regex class and specifically the Regex.Matches() method:
var matches = Regex.Matches(yourString,#"[a-zA-Z]+\s\-[a-zA-Z]");
Some example matching might look like this :

Regex to find special pattern

I have a string to parse. First I have to check if string contains special pattern:
I wanted to know if there is substrings which starts with "$(",
and end with ")",
and between those start and end special strings,there should not be
any white-empty space,
it should not include "$" character inside it.
I have a little regex for it in C#
string input = "$(abc)";
string pattern = #"\$\(([^$][^\s]*)\)";
Regex rgx = new Regex(pattern, RegexOptions.IgnoreCase);
MatchCollection matches = rgx.Matches(input);
foreach (var match in matches)
{
Console.WriteLine("value = " + match);
}
It works for many cases but failed at input= $(a$() , which inside the expression is empty. I wanted NOT to match when input is $().[ there is nothing between start and end identifiers].
What is wrong with my regex?

Note: [^$] matches a single character but not of $
Use the below regex if you want to match $()
\$\(([^\s$]*)\)
Use the below regex if you don't want to match $(),
\$\(([^\s$]+)\)
* repeats the preceding token zero or more times.
+ Repeats the preceding token one or more times.
Your regex \(([^$][^\s]*)\) is wrong. It won't allow $ as a first character inside () but it allows it as second or third ,, etc. See the demo here. You need to combine the negated classes in your regex inorder to match any character not of a space or $.

Your current regex does not match $() because the [^$] matches at least 1 character. The only way I can think of where you would have this match would be when you have an input containing more than one parens, like:
$()(something)
In those cases, you will also need to exclude at least the closing paren:
string pattern = #"\$\(([^$\s)]+)\)";
The above matches for example:
abc in $(abc) and
abc and def in $(def)$()$(abc)(something).

Simply replace the * with a + and merge the options.
string pattern = #"\$\(([^$\s]+)\)";
+ means 1 or more
* means 0 or more

How can I use lookbehind in a C# Regex in order to skip matches of repeated prefix patterns?

How can I use lookbehind in a C# Regex in order to skip matches of repeated prefix patterns?
Example - I'm trying to have the expression match all the b characters following any number of a characters:
Regex expression = new Regex("(?<=a).*");
foreach (Match result in expression.Matches("aaabbbb"))
MessageBox.Show(result.Value);
returns aabbbb, the lookbehind matching only an a. How can I make it so that it would match all the as in the beginning?
I've tried
Regex expression = new Regex("(?<=a+).*");
and
Regex expression = new Regex("(?<=a)+.*");
with no results...
What I'm expecting is bbbb.

Are you looking for a repeated capturing group?
(.)\1*
This will return two matches.
Given:
aaabbbb
This will result in:
aaa
bbbb
This:
(?<=(.))(?!\1).*
Uses the above principal, first checking that the finding the previous character, capturing it into a back reference, and then asserting that that character is not the next character.
That matches:
bbbb

I figured it out eventually:
Regex expression = new Regex("(?<=a+)[^a]+");
foreach (Match result in expression.Matches(#"aaabbbb"))
MessageBox.Show(result.Value);
I must not allow the as to me matched by the non-lookbehind group. This way, the expression will only match those b repetitions that follow a repetitions.
Matching aaabbbb yields bbbb and matching aaabbbbcccbbbbaaaaaabbzzabbb results in bbbbcccbbbb, bbzz and bbb.

The reason the look-behind is skipping the "a" is because it is consuming the first "a" (but no capturing it), then it captures the rest.
Would this pattern work for you instead? New pattern: \ba+(.+)\b
It uses a word boundary \b to anchor either ends of the word. It matches at least one "a" followed by the rest of the characters till the word boundary ends. The remaining characters are captured in a group so you can reference them easily.
string pattern = #"\ba+(.+)\b";
foreach (Match m in Regex.Matches("aaabbbb", pattern))
{
Console.WriteLine("Match: " + m.Value);
Console.WriteLine("Group capture: " + m.Groups[1].Value);
}
UPDATE: If you want to skip the first occurrence of any duplicated letters, then match the rest of the string, you could do this:
string pattern = #"\b(.)(\1)*(?<Content>.+)\b";
foreach (Match m in Regex.Matches("aaabbbb", pattern))
{
Console.WriteLine("Match: " + m.Value);
Console.WriteLine("Group capture: " + m.Groups["Content"].Value);
}