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Replace the nth index of a character

How can I replace the nth index of a character using only Regex.
string input = "%fdfdfdfdfdfdfdfdfdfdfdffd";
string result = Regex.Replace(input, "^%", "");
The above code, replaces the first character with an empty string, But, I want to specify an index: like nth index, so that character gets replaced with an empty string.
Can someone help me out here.

It's possible to create a regex pattern that captures all characters before and after the replaced character and then replace the whole string with the two captures separated by the new character. For example:
Regex.Replace("abcdefgh", #"^(.{4}).(.*)$", #"$1E$2") // returns "abcdEfgh"
You could then create a method that replaces the character at a specific index:
string ReplaceCharacter(string text, int index, char value)
=> Regex.Replace(text, $#"^(.{{{index}}}).(.*)$", $#"${{1}}{value}${{2}}");
// Usage:
ReplaceCharacter("Foo-bar", 3, 'l') // returns "Foolbar"

As Johan Wentholt said in the comments, you can perfectly use Regex.Replace to match a number of characters from the start of the line and replace it with a capture group that's one character less than the full matched piece:
String result = Regex.Replace(input, "^(.{" + index + "}).", "$1");
This matches "index times any character, followed by another character, at the start of the string", but replaces it by only the "index times any character" without that last character, since that last dot is outside of the capture group.
If you want to replace by something else than an empty string, you just concatenate it to the end of the "$1" replacement string. Though to be safe then, you should replace it with "${1}" to avoid problems if the piece you add behind it starts with a number, since that would change the capture group number.

What you want to do may not be possible with Regex alone. This is sort of a cheat:
var input = "%fdfd678dfdfdfdfdfdfdfdffd";
var result = Regex.Replace(input, "^.{7}", input.Substring(0,6));
Console.WriteLine($"result = {result}");

Splitting of a string using Regex

I have string of the following format:
string test = "test.BO.ID";
My aim is string that part of the string whatever comes after first dot.
So ideally I am expecting output as "BO.ID".
Here is what I have tried:
// Checking for the first occurence and take whatever comes after dot
var output = Regex.Match(test, #"^(?=.).*?");
The output I am getting is empty.
What is the modification I need to make it for Regex?

You get an empty output because the pattern you have can match an empty string at the start of a string, and that is enough since .*? is a lazy subpattern and . matches any char.
Use (the value will be in Match.Groups[1].Value)
\.(.*)
or (with a lookahead, to get the string as a Match.Value)
(?<=\.).*
See the regex demo and a C# online demo.
A non-regex approach can be use String#Split with count argument (demo):
var s = "test.BO.ID";
var res = s.Split(new[] {"."}, 2, StringSplitOptions.None);
if (res.GetLength(0) > 1)
Console.WriteLine(res[1]);

If you only want the part after the first dot you don't need a regex at all:
x.Substring(x.IndexOf('.'))

Regular Expression without braces

i have the following sample cases :
1) "Sample"
2) "[10,25]"
I want to form a(only one) regular expression pattern, to which the above examples are passed returns me "Sample" and "10,25".
Note: Input strings do not include Quotes.
I came up with the following expression (?<=\[)(.*?)(?=\]), this satisfies the second case and retreives me only "10,25" but when the first case is matched it returns me blank. I want "Sample" to be returned? can anyone help me.
C#.

here you go, a small regex using a positive lookbehind, sometime these are very handy
Regex
(?<=^|\[)([\w,]+)
Test string
Sample
[10,25]
Result
MATCH 1
[0-6] Sample
MATCH 2
[8-13] 10,25
try at regex101.com
if " is included in your original string, use this regex, this will look for " mark as well, you may choose to remove ^| from lookup if " mark is always included or you may choose to leave it as it is if your text has combination of with and without " marks
Regex
(?<=^|\[|\")([\w,]+)
try at regex101.com

As far as I can tell, the below regex should help:
Regex regex = new Regex(#"^\w+|[[](\w)+\,(\w)+[]]$");
This will match multiple words, or 2 words (alphanumeric) separated by commas and inside square brackets.

One Java example:
// String input = "Sample";
String input = "[10,25]";
String text = "[^,\\[\\]]+";
Pattern pMod = Pattern.compile("(" + text + ")|(?>\\[(" + text + "," + text + ")\\])");
Matcher mMod = pMod.matcher(input);
while (mMod.find()) {
if(mMod.group(1) != null) {
System.out.println(mMod.group(1));
}
if(mMod.group(2)!=null) {
System.out.println(mMod.group(2));
}
}
if input is "[hello&bye,25|35]", then the output is hello&bye,25|35

Replace all characters and first 0's (zeroes)

I am trying to replace all characters inside a Regular Expression expect the number, but the number should not start with 0
How can I achieve this using Regular Expression?
I have tried multiple things like #"^([1-9]+)(0+)(\d*)"and "(?<=[1-9])0+", but those does not work
Some examples of the text could be hej:\\\\0.0.0.22, hej:22, hej:\\\\?022 and hej:\\\\?22, and the result should in all places be 22

Rather than replace, try and match against [1-9][0-9]*$ on your string. Grab the matched text.
Note that as .NET regexes match Unicode number characters if you use \d, here the regex restricts what is matched to a simple character class instead.
(note: regex assumes matches at end of line only)

According to one of your comments hej:\\\\0.011.0.022 should yield 110022. First select the relevant string part from the first non zero digit up to the last number not being zero:
([1-9].*[1-9]\d*)|[1-9]
[1-9] is the first non zero digit
.* are any number of any characters
[1-9]\d* are numbers, starting at the first non-zero digit
|[1-9] includes cases consisting of only one single non zero digit
Then remove all non digits (\D)
Match match = Regex.Match(input, #"([1-9].*[1-9]\d*)|[1-9]");
if (match.Success) {
result = Regex.Replace(match.Value, "\D", "");
} else {
result = "";
}

Use following
[1-9][0-9]*$
You don't need to do any recursion, just match that.

Here is something that you can try The87Boy you can play around with or add to the pattern as you like.
string strTargetString = #"hej:\\\\*?0222\";
string pattern = "[\\\\hej:0.?*]";
string replacement = " ";
Regex regEx = new Regex(pattern);
string newRegStr = Regex.Replace(regEx.Replace(strTargetString, replacement), #"\s+", " ");
Result from the about Example = 22

RegEx Problem using .NET

I have a little problem on RegEx pattern in c#. Here's the rule below:
input: 1234567
expected output: 123/1234567
Rules:
Get the first three digit in the input. //123
Add /
Append the the original input. //123/1234567
The expected output should looks like this: 123/1234567
here's my regex pattern:
regex rx = new regex(#"((\w{1,3})(\w{1,7}))");
but the output is incorrect. 123/4567

I think this is what you're looking for:
string s = #"1234567";
s = Regex.Replace(s, #"(\w{3})(\w+)", #"$1/$1$2");
Instead of trying to match part of the string, then match the whole string, just match the whole thing in two capture groups and reuse the first one.

It's not clear why you need a RegEx for this. Why not just do:
string x = "1234567";
string result = x.Substring(0, 3) + "/" + x;

Another option is:
string s = Regex.Replace("1234567", #"^\w{3}", "$&/$&"););
That would capture 123 and replace it to 123/123, leaving the tail of 4567.
^\w{3} - Matches the first 3 characters.
$& - replace with the whole match.
You could also do #"^(\w{3})", "$1/$1" if you are more comfortable with it; it is better known.

Use positive look-ahead assertions, as they don't 'consume' characters in the current input stream, while still capturing input into groups:
Regex rx = new Regex(#"(?'group1'?=\w{1,3})(?'group2'?=\w{1,7})");
group1 should be 123, group2 should be 1234567.