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How to replace all linebreaks in the start of string c#

I've seen a lot of questions on how to replace all linebreaks.
However I have a scenario where I would like to replace all linebreaks in the start of the string only, there can be none, one or multiple and there is no way to know beforehand how many linebreaks it will be.
Question:
How do I remove all linebreaks that may occour before the string?
any help or input appreciated, thanks!

If you want to remove all leading occurrences of a set of characters, use String.TrimStart. Its prototype is:
public string TrimStart (params char[] trimChars);
In trimChars you put \n, in your case.

How to write a C# Regex to ensure a string starts with one char followed by 6 digits

I need to write a Regex in C# to ensure that a string starts with one character of either S, R, or V, and then has six following digits, like this: "S123456". Here is the regular expression I'm trying to use:
#"(S|R|V)[0-9]{6}?"
But if I pass a string with one too many digits, like "S1234567", it fails. What am I doing wrong here?

var regex = new Regex(#"^[SRV]\d{6}$");

Three options, that depend on what you want to achieve:
Just for matching the string:
[SRV]\d{6}
For finding that string as a separated "word":
\b[SRV]\d{6}\b
For the regex to match the full string (I think this is what you need):
^[SRV]\d{6}$
EDIT:
Your regexp "fails" because it's just looking for your pattern in the string (as my first example). If the string is S1234567, the engine matches the bold part (from S to 6), so it reports a success. You have to use anchors (my third example) if you want the string just to contain your pattern and nothing else (i.e. the string matches the pattern from start to end).

^[sSvVnN]\d{6}$
Will match lower and upper case of your specified characters

C# regex not matching string

I have a string which is formatted like this: $20,$40,$AA,$FF. Basically, hex numbers and they can be of many bytes. I want to check if a string is in the above format, so I tried something like this:
string a = "$20,$30,$40";
Regex reg = new Regex(#"$[0-9a-fA-F],");
if (a.StartsWith(string.Format("{0}{1}", reg, reg)))
MessageBox.Show("A");
It doesn't seem to work though, is there anything I'm missing?

$ is a special character in regular expressions and means end of string. That regex won't match anything at all since you're specifying stuff after the string end. Escape the $ character like
"\$[0-9a-fA-F]{2},"
Anyway AFAIK this will not work with your string since it doesn't end with an ",". You might try:
"^(\$[0-9a-fA-F]{2},?)+$"
You can even simplify the regex by using case-insensitive regex matching:
Regex reg = new Regex(#"^(\$[0-9A-F]{2},?)+$", RegexOptions.IgnoreCase);
EDIT: corrected to match exactly 2 hexadecimal digits.
EDIT: maybe you should write your regex checking like:
if (Regex.IsMatch(a,#"^(\$[0-9A-F]{2},?)+$",RegexOptions.IgnoreCase))
{
// Do whatever
}

I think you are missing a quantifier:
"\$[0-9a-fA-F]+,"
For the problem with the comma at the end, I would simply append one at the end to keep the regex as simple as possible. But this is just the way I would do it.

There are 3 things that need to be changed:
Need to escape your $ symbol as it represents end of line.
\$
Need to tweak your regex pattern to match the entire string instead of parts.
^(\$[0-9a-fA-F]{2},+)+\$[0-9a-fA-F]{2}$
Need to change your code to use Regex.IsMatch.
string a = "$20,$30,$40";
if (Regex.IsMatch(a,#"^(\$[0-9a-fA-F]{2},+)+\$[0-9a-fA-F]{2}$",RegexOptions.IgnoreCase))
MessageBox.Show("A");
PS:
If the input string has white space like a tab or a space in between, then this regex will need to be modified. In such cases, you have to use "\s" at the right positions. For example, if you have white space around the commas like
string a = "$20 ,$30, $40";
then you need to tweak your RegEx this way:
^(\$[0-9a-fA-F]{2}\s*,+\s*)+\$[0-9a-fA-F]{2}\s*$
References:
C# Regex Testers
A Better .NET Regular Expression Tester
RegexHero tester
about Regex.IsMatch (instead of using Match)
MSDN Regex.isMatch
Usage example
C# Regular Expression Cheat Sheet
Old answer below (Ignore):
Try this:
"\$[0-9a-fA-F]{2}?[,]{0,1}"

You might also want to add a repeat modifier to your set such that it becomes;
"\$[0-9a-fA-F]+,"

Does this regex expression allow "*"?

I really know very little about regex's.
I'm trying to test a password validation.
Here's the regex that describes it (I didn't write it, and don't know what it means):
private static string passwordField = "[^A-Za-z0-9_.\\-!##$%^&*()=+;:'\"|~`<>?\\/{}]";
I've tried a password like "dfgbrk*", and my code, using the above regex, allowed it.
Is this consistent with what the regex defines as acceptable, or is it a problem with my code?
Can you give me an example of a string that validation using the above regex isn't suppose to allow?
Added: Here's how the original code uses this regex (and it works there):
public static bool ValidateTextExp(string regexp, string sText)
{
if ( sText == null)
{
Log.WriteWarning("ValidateTextExp got null text to validate against regExp {0} . returning false",regexp);
return false;
}
return (!Regex.IsMatch(sText, regexp));
}
It seems I'm doing something wrong..
Thanks.

Your regex matches a value that contains any single character which is not in that list.
Your test value matches because it has spaces in it, which do not appear to be in your expression.
The reason it's not is because your character class starts with ^. The reason it matches any value that contains any single character that is not that is because you did not specify the beginning or end of the string, or any quantifiers.
The above assumes I'm not missing the importance of any of the characters in the middle of the character soup :)
This answer is also dependent on how you actually use the Regex in code.
If your intention was for that Regex string to represent the only characters that are actually allowed in a password, you would change the regex like so:
string pattern = "^[A-Z0-9...etc...]+$";
The important parts there are:
The ^ has been removed from inside the bracket, to outside; where it signifies the start of the whole string.
The $ has been added to the end, where it signifies the end of the whole string.
Those are needed because otherwise, your pattern will match anything that contains the valid values anywhere inside - even if invalid values are also present.
finally, I've added the + quantifier, which means you want to find any one of those valid characters, one or more times. (this regex would not permit a 0-length password)
If you wanted to permit the ^ character also as part of the password, you would add it back in between the brackets, but just *not as the first thing right after the opening bracket [. So for example:
string pattern = "^[A-Z0-9^...etc...]+$";
The ^ has special meaning in different places at different times in Regexes.

[^A-Za-z0-9_.\-!##$%^&*()=+;:'\"|~`?\/{}]
----------------------^
Looks fine to me, at least in regards to your question title. I'm not clear yet on why the spaces in your sample don't trip it up.
Note that I'm assuming the purpose of this expression is to find invalid characters. Thus, if the expression is a positive match, you have a bad password that you must reject. Since there appears to be some confusion about this, perhaps I can clear it up with a little psuedo-code:
bool isGoodPassword = !Regex.IsMatch(#"[^A-Za-z0-9_.\-!...]", requestedPassword);
You could re-write this for a positive match (without the negation) like so:
bool isGoodPassword = Regex.IsMatch(#"^[A-Za-z0-9_.\-!...]+$", requestedPassword);
The new expression matches a string that from the beginning of the string is filled with one or more of any of the characters in the list all the way the way to end. Any character not in the list would cause the match to fail.

You regular expression is just an inverted character class and describes just one single character (but that can’t be *). So it depends on how you use that character class.

Depends on how you apply it. It describes exactly one character, however, the ^ in the beginning buggs me a little, as it prohibits every other character, so there is probably something terribly fishy there.
Edit: as pointed out in other answers, the reason for your string to match is the space, not the explanation that was replaced by this line.

Simple regex pattern

i'm using C# and i'm trying to allow only alphabetical letters and spaces. my expression at the moment is:
string regex = "^[A-Za-z\s]{1,40}$";
my IDE says that \s is an "Unrecognized escape sequence"
what am i missing?

"\" is a c# escape character as well as a regex escape character. Try:
string regex = #"^[A-Za-z\s]{1,40}$";

You need to put an # in front of your string to turn it into a verbatim string literal:
string regex = #"^[A-Za-z\s]{1,40}$";
Right now, the \ in your regex is being interpreted as trying to escape the following s, which the compiler doesn't understand.
Alternatively, you can just escape the backslash with another one:
string regex = "^[A-Za-z\\s]{1,40}$";
but in general, prefer the first approach to the second.

An additional note, your regex doesn't do what you describe. You say a max of 1 space in between words. In order to do that, you need to move the "\s" out of the character list. The pattern you're currently using allows "any alphanumeric or space from 1 to 40 times" which allows for multiple successive spaces. You'll need something more like the following:
string regex = #"^(?:[A-Za-z]+\s?)+$";
This means "any alphanumeric 1 or more times followed by an optional space, this whole thing one or more times". I don't know how to limit the whole string to 40 characters when you don't know the size of the first expression in advance. Maybe this can be achieved with a "look behind" expression, but I'm not sure. You might have to do it in two steps.