Develop Reference

.NET Regex To Remove Line Breaks Within Quotes

I am trying to clean up a text file so that it can be imported into Excel but the text file contains line breaks within several of the double quoted fields. The file is tab delimited.
Example would be:
"12313"\t"1234"\t"123
5679"
"test"\t"test"\t"test"
"test"\t"test"\t"test"
"12313"\t"1234"\t"123
5679"
I need to remove the line breaks so that it will ultimately display like:
"12313"\t"1234"\t"1235679"
"test"\t"test"\t"test"
"test"\t"test"\t"test"
"12313"\t"1234"\t"1235679"
The "\t" is the tab delimiter.
I've looked at several other solutions on SO but they don't seem to deal with multiple lines. We've tried using several CSV parser solutions but can't seem to get them to work for this scenario. The goal is to pass the entire string into a REGEX expression and have it return with all line breaks between quotes removed while the line breaks outside of the quotes remain.

You can use this regex:
(?!(([^"]*"){2})*[^"]*$)\n+
Working Demo
This one matches one or more newline character that are not followed by even number of quotes (It assumes there is no escaping exceptions in the data).

This worked for me:
var fixedCsvFileContent = Regex.Replace(csvFileContent, #"(?!(([^""]*""){2})*[^""]*$)\n+", string.Empty);
This didnt work:
var fixedCsvFileContent = Regex.Replace(csvFileContent, #"(?!(([^""]*""){2})*[^""]*$)\n+", string.Empty, RegexOptions.Multiline);
Thus one must not add RegexOptions.Multiline when doing the check on the input string.

If just removing blank lines works:
string text = Regex.Replace( inputString, #"\n\n", "" , RegexOptions.None | RegexOptions.Multiline );

I have been running into a similar problem, but also some of the files might be really large. So using a RegEx on everything would be a heavy solution, and instead I wanted to try to make something a bit like ReadLine except that it would ignore breaklines within quotes. This is the solution I am using.
It is an extension to the StreamReader class, used to reading the CSV files and like some of the RegEx solutions here, it ensures there is an even number of quotes. So it uses ReadLine, checks if there is an odd number of quotes and if there is it does another ReadLine until the number of quotes is even:
public static class Extensions
{
public static string ReadEntry(this StreamReader sr)
{
string strReturn = "";
//get first bit
strReturn += sr.ReadLine();
//And get more lines until the number of quotes is even
while (strReturn.GetNumberOf("\"").IsOdd())
{
string strNow = sr.ReadLine();
strReturn += strNow;
}
//Then return what we've gotten
if (strReturn == "")
{
return null;
}
else
{
return strReturn;
}
}
public static int GetNumberOf(this string s, string strSearchString)
{
return s.Length - s.Replace(strSearchString, "").Length;
}
public static Boolean IsOdd(this int i)
{
return i % 2 != 0;
}
}

string output = Regex.Replace(input, #"(?<=[^""])\r\n", string.Empty);
Demo with the input provided

Check if a string ends with another string or a part of another string

I wanted to know if there is a solution to the problem mentioned in the topic.
Example:
In my project I have to parse a lot of messages. These messages contain formatting characters like "\n" or "\r".
The end of this message is always signed with the name of the author.
Now I want to remove the signatures from each message. The problem is that the end of the message could look like
\r\n\rDaniel Walters\n\r\n
\n\r\n\r\n\rDaniel
or something else
The problem is that I don't know how to identifiy these varying endings.
I tried to only remove the last "\n\r\n"'s by calling string.EndsWith() in a loop but this solution only removes everything except "\r\n\rDaniel Walter".
Then I tried to remove the author (I parsed it prior to this step) but this does not work either. Sometimes the parsed author is "Daniel Walters" and the signature is only "Daniel".
Any ideas how to solve this?
Are there maybe some easier and smarter solutions than looping through the string?

You can make a regular expression to replace the name with an optional last name, and any number of whitespace characters before and after.
Example:
string message = "So long and thanks for all the fish \t\t\r Arthur \t Dent \r\r\n ";
string firstName = "Arthur";
string lastName = "Dent";
string pattern = "\\s+" + Regex.Escape(firstName) + "(\\s+" + Regex.Escape(lastName) + ")?\\s*$";
message = Regex.Replace(message, pattern, String.Empty);
(Yes, I know it was really the dolphins saying that.)

you could try something like the following (untested) :-
string str="\r\n\rDaniel Walters\n\r\n";
while(str.EndsWith("\r") || str.EndsWith("\n"))
{
// \r and \n have the same length. So, we can use either \r or \n in the end
str=str.SubString(0,str.Length - ("\r".Length));
}
while(str.StartsWith("\r") || str.StartsWith("\n"))
{
// \r and \n have the same length
str=str.SubString("\r".Length,str.length);
}

You'll have to determine what "looks like" a signature. Are there specific criteria that always apply?
Always followed by at least 3 newlines (\r or \n)
Starts with a capital letter
Has no following text
A regex like this might work for those criteria:
/[\r\n]{3,}[A-Z][\w ]+[\r\n]*(?!\w)/
Adjust according to your needs.
Edited to add: This should match the last "paragraph" of a document.
/([\r\n]+[\w ]+[\r\n]*)(?!.)/

you can do this as well but I am not sure if your pattern changes but this will return Daniel Walter
string replaceStr = "\r\n\rDaniel Walters\n\r\n";
replaceStr = replaceStr.TrimStart(new char[] { '\r', '\n' });
replaceStr = replaceStr.TrimEnd(new char[] { '\r', '\n' });
or if you want to use the trim method you can do the following
string replaceStr = "\r\n\rDaniel Walters\n\r\n";
replaceStr = replaceStr.Trim();

A different approach could be to split your message at the newline chars removing the empty newline entries. Then reassembling the expected string excluding the last line where I assume there is always the signature.
string removeLastLine = "Text on the firstline\r\ntest on second line\rtexton third line\r\n\rDaniel Walters\n\r\n";
string[] lines = removeLastLine.Split(new char[] {'\r', '\n'}, StringSplitOptions.RemoveEmptyEntries);
lines = lines.Take(lines.Length - 1).ToArray();
string result = string.Join(Environment.NewLine, lines);

Removing all whitespace lines from a multi-line string efficiently

In C# what's the best way to remove blank lines i.e., lines that contain only whitespace from a string? I'm happy to use a Regex if that's the best solution.
EDIT: I should add I'm using .NET 2.0.
Bounty update: I'll roll this back after the bounty is awarded, but I wanted to clarify a few things.
First, any Perl 5 compat regex will work. This is not limited to .NET developers. The title and tags have been edited to reflect this.
Second, while I gave a quick example in the bounty details, it isn't the only test you must satisfy. Your solution must remove all lines which consist of nothing but whitespace, as well as the last newline. If there is a string which, after running through your regex, ends with "/r/n" or any whitespace characters, it fails.

If you want to remove lines containing any whitespace (tabs, spaces), try:
string fix = Regex.Replace(original, #"^\s*$\n", string.Empty, RegexOptions.Multiline);
Edit (for #Will): The simplest solution to trim trailing newlines would be to use TrimEnd on the resulting string, e.g.:
string fix =
Regex.Replace(original, #"^\s*$\n", string.Empty, RegexOptions.Multiline)
.TrimEnd();

string outputString;
using (StringReader reader = new StringReader(originalString)
using (StringWriter writer = new StringWriter())
{
string line;
while((line = reader.ReadLine()) != null)
{
if (line.Trim().Length > 0)
writer.WriteLine(line);
}
outputString = writer.ToString();
}

off the top of my head...
string fixed = Regex.Replace(input, "\s*(\n)","$1");
turns this:
fdasdf
asdf
[tabs]
[spaces]
asdf
into this:
fdasdf
asdf
asdf

Using LINQ:
var result = string.Join("\r\n",
multilineString.Split(new string[] { "\r\n" }, ...None)
.Where(s => !string.IsNullOrWhitespace(s)));
If you're dealing with large inputs and/or inconsistent line endings you should use a StringReader and do the above old-school with a foreach loop instead.

Alright this answer is in accordance to the clarified requirements specified in the bounty:
I also need to remove any trailing newlines, and my Regex-fu is
failing. My bounty goes to anyone who can give me a regex which passes
this test: StripWhitespace("test\r\n \r\nthis\r\n\r\n") ==
"test\r\nthis"
So Here's the answer:
(?<=\r?\n)(\s*$\r?\n)+|(?<=\r?\n)(\r?\n)+|(\r?\n)+\z
Or in the C# code provided by #Chris Schmich:
string fix = Regex.Replace("test\r\n \r\nthis\r\n\r\n", #"(?<=\r?\n)(\s*$\r?\n)+|(?<=\r?\n)(\r?\n)+|(\r?\n)+\z", string.Empty, RegexOptions.Multiline);
Now let's try to understand it. There are three optional patterns in here which I am willing to replace with string.empty.
(?<=\r?\n)(\s*$\r?\n)+ - matches one to unlimited lines containing only white space and preceeded by a line break (but does not match the first preceeding line breaks).
(?<=\r?\n)(\r?\n)+ - matches one to unlimited empty lines with no content that are preceeded by a line break (but does not match the first preceeding line breaks).
(\r?\n)+\z - matches one to unlimited line breaks at the end of the tested string (trailing line breaks as you called them)
That satisfies your test perfectly! But also satisfies both \r\n and \n line break styles! Test it out! I believe this will be the most correct answer, although simpler expression would pass your specified bounty test, this regex passes more complex conditions.
EDIT: #Will pointed out a potential flaw in the last pattern match of the above regex in that it won't match multiple line breaks containing white space at the end of the test string. So let's change that last pattern to this:
\b\s+\z The \b is a word boundry (beginning or END of a word), the \s+ is one or more white space characters, the \z is the end of the test string (end of "file"). So now it will match any assortment of whitespace at the end of the file including tabs and spaces in addition to carriage returns and line breaks. I tested both of #Will's provided test cases.
So all together now, it should be:
(?<=\r?\n)(\s*$\r?\n)+|(?<=\r?\n)(\r?\n)+|\b\s+\z
EDIT #2: Alright there is one more possible case #Wil found that the last regex doesn't cover. That case is inputs that have line breaks at the beginning of the file before any content. So lets add one more pattern to match the beginning of the file.
\A\s+ - The \A match the beginning of the file, the \s+ match one or more white space characters.
So now we've got:
\A\s+|(?<=\r?\n)(\s*$\r?\n)+|(?<=\r?\n)(\r?\n)+|\b\s+\z
So now we have four patterns for matching:
whitespace at the beginning of the file,
redundant line breaks containing white space, (ex: \r\n \r\n\t\r\n)
redundant line breaks with no content, (ex: \r\n\r\n)
whitespace at the end of the file

not good. I would use this one using JSON.net:
var o = JsonConvert.DeserializeObject(prettyJson);
new minifiedJson = JsonConvert.SerializeObject(o, Formatting.None);

In response to Will's bounty, which expects a solution that takes "test\r\n \r\nthis\r\n\r\n" and outputs "test\r\nthis", I've come up with a solution that makes use of atomic grouping (aka Nonbacktracking Subexpressions on MSDN). I recommend reading those articles for a better understanding of what's happening. Ultimately the atomic group helped match the trailing newline characters that were otherwise left behind.
Use RegexOptions.Multiline with this pattern:
^\s+(?!\B)|\s*(?>[\r\n]+)$
Here is an example with some test cases, including some I gathered from Will's comments on other posts, as well as my own.
string[] inputs =
{
"one\r\n \r\ntwo\r\n\t\r\n \r\n",
"test\r\n \r\nthis\r\n\r\n",
"\r\n\r\ntest!",
"\r\ntest\r\n ! test",
"\r\ntest \r\n ! "
};
string[] outputs =
{
"one\r\ntwo",
"test\r\nthis",
"test!",
"test\r\n ! test",
"test \r\n ! "
};
string pattern = #"^\s+(?!\B)|\s*(?>[\r\n]+)$";
for (int i = 0; i < inputs.Length; i++)
{
string result = Regex.Replace(inputs[i], pattern, "",
RegexOptions.Multiline);
Console.WriteLine(result == outputs[i]);
}
EDIT: To address the issue of the pattern failing to clean up text with a mix of whitespace and newlines, I added \s* to the last alternation portion of the regex. My previous pattern was redundant and I realized \s* would handle both cases.

string corrected =
System.Text.RegularExpressions.Regex.Replace(input, #"\n+", "\n");

I'll go with:
public static string RemoveEmptyLines(string value) {
using (StringReader reader = new StringReader(yourstring)) {
StringBuilder builder = new StringBuilder();
string line;
while ((line = reader.ReadLine()) != null) {
if (line.Trim().Length > 0)
builder.AppendLine(line);
}
return builder.ToString();
}
}

Here's another option: use the StringReader class. Advantages: one pass over the string, creates no intermediate arrays.
public static string RemoveEmptyLines(this string text) {
var builder = new StringBuilder();
using (var reader = new StringReader(text)) {
while (reader.Peek() != -1) {
string line = reader.ReadLine();
if (!string.IsNullOrWhiteSpace(line))
builder.AppendLine(line);
}
}
return builder.ToString();
}
Note: the IsNullOrWhiteSpace method is new in .NET 4.0. If you don't have that, it's trivial to write on your own:
public static bool IsNullOrWhiteSpace(string text) {
return string.IsNullOrEmpty(text) || text.Trim().Length < 1;
}

In response to Will's bounty here is a Perl sub that gives correct response to the test case:
sub StripWhitespace {
my $str = shift;
print "'",$str,"'\n";
$str =~ s/(?:\R+\s+(\R)+)|(?:()\R+)$/$1/g;
print "'",$str,"'\n";
return $str;
}
StripWhitespace("test\r\n \r\nthis\r\n\r\n");
output:
'test
this
'
'test
this'
In order to not use \R, replace it with [\r\n] and inverse the alternative. This one produces the same result:
$str =~ s/(?:(\S)[\r\n]+)|(?:[\r\n]+\s+([\r\n])+)/$1/g;
There're no needs for special configuration neither multi line support. Nevertheless you can add s flag if it's mandatory.
$str =~ s/(?:(\S)[\r\n]+)|(?:[\r\n]+\s+([\r\n])+)/$1/sg;

if its only White spaces why don't you use the C# string method
string yourstring = "A O P V 1.5";
yourstring.Replace(" ", string.empty);
result will be "AOPV1.5"

char[] delimiters = new char[] { '\r', '\n' };
string[] lines = value.Split(delimiters, StringSplitOptions.RemoveEmptyEntries);
string result = string.Join(Environment.NewLine, lines)

Here is something simple if working against each individual line...
(^\s+|\s+|^)$

Eh. Well, after all that, I couldn't find one that would hit all the corner cases I could figure out. The following is my latest incantation of a regex that strips
All empty lines from the start of a string
Not including any spaces at the beginning of the first non-whitespace line
All empty lines after the first non-whitespace line and before the last non-whitespace line
Again, preserving all whitespace at the beginning of any non-whitespace line
All empty lines after the last non-whitespace line, including the last newline
(?<=(\r\n)|^)\s*\r\n|\r\n\s*$
which essentially says:
Immediately after
The beginning of the string OR
The end of the last line
Match as much contiguous whitespace as possible that ends in a newline*
OR
Match a newline and as much contiguous whitespace as possible that ends at the end of the string
The first half catches all whitespace at the start of the string until the first non-whitespace line, or all whitespace between non-whitespace lines. The second half snags the remaining whitespace in the string, including the last non-whitespace line's newline.
Thanks to all who tried to help out; your answers helped me think through everything I needed to consider when matching.
*(This regex considers a newline to be \r\n, and so will have to be adjusted depending on the source of the string. No options need to be set in order to run the match.)

String Extension
public static string UnPrettyJson(this string s)
{
try
{
// var jsonObj = Json.Decode(s);
// var sObject = Json.Encode(value); dont work well with array of strings c:['a','b','c']
object jsonObj = JsonConvert.DeserializeObject(s);
return JsonConvert.SerializeObject(jsonObj, Formatting.None);
}
catch (Exception e)
{
throw new Exception(
s + " Is Not a valid JSON ! (please validate it in http://www.jsoneditoronline.org )", e);
}
}

Im not sure is it efficient but =)
List<string> strList = myString.Split(new string[] { "\n" }, StringSplitOptions.None).ToList<string>();
myString = string.Join("\n", strList.Where(s => !string.IsNullOrWhiteSpace(s)).Distinct().ToList());

Try this.
string s = "Test1" + Environment.NewLine + Environment.NewLine + "Test 2";
Console.WriteLine(s);
string result = s.Replace(Environment.NewLine, String.Empty);
Console.WriteLine(result);

s = Regex.Replace(s, #"^[^\n\S]*\n", "");
[^\n\S] matches any character that's not a linefeed or a non-whitespace character--so, any whitespace character except \n. But most likely the only characters you have to worry about are space, tab and carriage return, so this should work too:
s = Regex.Replace(s, #"^[ \t\r]*\n", "");
And if you want it to catch the last line, without a final linefeed:
s = Regex.Replace(s, #"^[ \t\r]*\n?", "");

What does char 160 mean in my source code?

I am formatting numbers to string using the following format string "# #.##", at some point I need to turn back these number strings like (1 234 567) into something like 1234567. I am trying to strip out the empty chars but found that
value = value.Replace(" ", "");
for some reason and the string remain 1 234 567. After looking at the string I found that
value[1] is 160.
I was wondering what the value 160 means?

The answer is to look in Unicode Code Charts - where you'll find the Latin-1 supplement chart; this shows that U+00A0 (160 as per your title, not 167 as per the body) is a non-breaking space.

char code 160 would be

Maybe you could to use a regex to replace those empty chars:
Regex.Replace(input, #"\p{Z}", "");
This will remove "any kind of whitespace or invisible separator".

value.Replace(Convert.ToChar(160).ToString(),"")

This is a fast (and fairly readable) way of removing any characters classified as white space using Char.IsWhiteSpace:
StringBuilder sb = new StringBuilder (value.Length);
foreach (char c in value)
{
if (!char.IsWhiteSpace (c))
sb.Append (c);
}
string value= sb.ToString();
As dbemerlin points out, if you know you will only need numbers from your data, you would be better use Char.IsNumber or the even more restrictive Char.IsDigit:
StringBuilder sb = new StringBuilder (value.Length);
foreach (char c in value)
{
if (char.IsNumber(c))
sb.Append (c);
}
string value= sb.ToString();
If you need numbers and decimal seperators, something like this should suffice:
StringBuilder sb = new StringBuilder (value.Length);
foreach (char c in value)
{
if (char.IsNumber(c)|c == System.Globalization.NumberFormatInfo.CurrentInfo.NumberDecimalSeparator )
sb.Append (c);
}
string value= sb.ToString();

I would suggest using the char overload version:
value = value.Replace(Convert.ToChar(160), ' ')

Solution with extended methods:
public static class ExtendedMethods
{
public static string NbspToSpaces(this string text)
{
return text.Replace(Convert.ToChar(160), ' ');
}
}
And it can be used with this code:
value = value.NbspToSpaces();

Wouldn't be the preferred method to replace all empty characters (and this is what the questioner wanted to do) with the Regex Method which Rubens already posted?
Regex.Replace(input, #"\p{Z}", "");
or what Expresso suggests:
Regex.Replace(input, #"\p{Zs}", "");
The difference here is that \p{Z} replaces any kind of whitespace or invisible separator whereas the \p{Zs} replaces a whitespace character that is invisible, but does take up space.
You can read it here (Section Unicode Categories):
http://www.regular-expressions.info/unicode.html
Using RegEx has the advantage that only one command is needed to replace also the normal whitespaces and not only the non-breaking space like explained in some answers above.
If performance is the way to go then of course other methods should be considered but this is out of scope here.

Replace Line Breaks in a String C#

How can I replace Line Breaks within a string in C#?

Use replace with Environment.NewLine
myString = myString.Replace(System.Environment.NewLine, "replacement text"); //add a line terminating ;
As mentioned in other posts, if the string comes from another environment (OS) then you'd need to replace that particular environments implementation of new line control characters.

The solutions posted so far either only replace Environment.NewLine or they fail if the replacement string contains line breaks because they call string.Replace multiple times.
Here's a solution that uses a regular expression to make all three replacements in just one pass over the string. This means that the replacement string can safely contain line breaks.
string result = Regex.Replace(input, #"\r\n?|\n", replacementString);

To extend The.Anyi.9's answer, you should also be aware of the different types of line break in general use. Dependent on where your file originated, you may want to look at making sure you catch all the alternatives...
string replaceWith = "";
string removedBreaks = Line.Replace("\r\n", replaceWith).Replace("\n", replaceWith).Replace("\r", replaceWith);
should get you going...

I would use Environment.Newline when I wanted to insert a newline for a string, but not to remove all newlines from a string.
Depending on your platform you can have different types of newlines, but even inside the same platform often different types of newlines are used. In particular when dealing with file formats and protocols.
string ReplaceNewlines(string blockOfText, string replaceWith)
{
return blockOfText.Replace("\r\n", replaceWith).Replace("\n", replaceWith).Replace("\r", replaceWith);
}

If your code is supposed to run in different environments, I would consider using the Environment.NewLine constant, since it is specifically the newline used in the specific environment.
line = line.Replace(Environment.NewLine, "newLineReplacement");
However, if you get the text from a file originating on another system, this might not be the correct answer, and you should replace with whatever newline constant is used on the other system. It will typically be \n or \r\n.

if you want to "clean" the new lines, flamebaud comment using regex #"[\r\n]+" is the best choice.
using System;
using System.Text.RegularExpressions;
class MainClass {
public static void Main (string[] args) {
string str = "AAA\r\nBBB\r\n\r\n\r\nCCC\r\r\rDDD\n\n\nEEE";
Console.WriteLine (str.Replace(System.Environment.NewLine, "-"));
/* Result:
AAA
-BBB
-
-
-CCC
DDD---EEE
*/
Console.WriteLine (Regex.Replace(str, #"\r\n?|\n", "-"));
// Result:
// AAA-BBB---CCC---DDD---EEE
Console.WriteLine (Regex.Replace(str, #"[\r\n]+", "-"));
// Result:
// AAA-BBB-CCC-DDD-EEE
}
}

Use new in .NET 6 method
myString = myString.ReplaceLineEndings();
Replaces ALL newline sequences in the current string.
Documentation:
ReplaceLineEndings

Don't forget that replace doesn't do the replacement in the string, but returns a new string with the characters replaced. The following will remove line breaks (not replace them). I'd use #Brian R. Bondy's method if replacing them with something else, perhaps wrapped as an extension method. Remember to check for null values first before calling Replace or the extension methods provided.
string line = ...
line = line.Replace( "\r", "").Replace( "\n", "" );
As extension methods:
public static class StringExtensions
{
public static string RemoveLineBreaks( this string lines )
{
return lines.Replace( "\r", "").Replace( "\n", "" );
}
public static string ReplaceLineBreaks( this string lines, string replacement )
{
return lines.Replace( "\r\n", replacement )
.Replace( "\r", replacement )
.Replace( "\n", replacement );
}
}

To make sure all possible ways of line breaks (Windows, Mac and Unix) are replaced you should use:
string.Replace("\r\n", "\n").Replace('\r', '\n').Replace('\n', 'replacement');
and in this order, to not to make extra line breaks, when you find some combination of line ending chars.

Why not both?
string ReplacementString = "";
Regex.Replace(strin.Replace(System.Environment.NewLine, ReplacementString), #"(\r\n?|\n)", ReplacementString);
Note: Replace strin with the name of your input string.

I needed to replace the \r\n with an actual carriage return and line feed and replace \t with an actual tab. So I came up with the following:
public string Transform(string data)
{
string result = data;
char cr = (char)13;
char lf = (char)10;
char tab = (char)9;
result = result.Replace("\\r", cr.ToString());
result = result.Replace("\\n", lf.ToString());
result = result.Replace("\\t", tab.ToString());
return result;
}

var answer = Regex.Replace(value, "(\n|\r)+", replacementString);

As new line can be delimited by \n, \r and \r\n, first we’ll replace \r and \r\n with \n, and only then split data string.
The following lines should go to the parseCSV method:
function parseCSV(data) {
//alert(data);
//replace UNIX new lines
data = data.replace(/\r\n/g, "\n");
//replace MAC new lines
data = data.replace(/\r/g, "\n");
//split into rows
var rows = data.split("\n");
}

Use the .Replace() method
Line.Replace("\n", "whatever you want to replace with");

Best way to replace linebreaks safely is
yourString.Replace("\r\n","\n") //handling windows linebreaks
.Replace("\r","\n") //handling mac linebreaks
that should produce a string with only \n (eg linefeed) as linebreaks.
this code is usefull to fix mixed linebreaks too.

Another option is to create a StringReader over the string in question. On the reader, do .ReadLine() in a loop. Then you have the lines separated, no matter what (consistent or inconsistent) separators they had. With that, you can proceed as you wish; one possibility is to use a StringBuilder and call .AppendLine on it.
The advantage is, you let the framework decide what constitutes a "line break".

string s = Regex.Replace(source_string, "\n", "\r\n");
or
string s = Regex.Replace(source_string, "\r\n", "\n");
depending on which way you want to go.
Hopes it helps.

If you want to replace only the newlines:
var input = #"sdfhlu \r\n sdkuidfs\r\ndfgdgfd";
var match = #"[\\ ]+";
var replaceWith = " ";
Console.WriteLine("input: " + input);
var x = Regex.Replace(input.Replace(#"\n", replaceWith).Replace(#"\r", replaceWith), match, replaceWith);
Console.WriteLine("output: " + x);
If you want to replace newlines, tabs and white spaces:
var input = #"sdfhlusdkuidfs\r\ndfgdgfd";
var match = #"[\\s]+";
var replaceWith = "";
Console.WriteLine("input: " + input);
var x = Regex.Replace(input, match, replaceWith);
Console.WriteLine("output: " + x);

This is a very long winded one-liner solution but it is the only one that I had found to work if you cannot use the the special character escapes like "\r" and "\n" and \x0d and \u000D as well as System.Environment.NewLine as parameters to thereplace() method
MyStr.replace( System.String.Concat( System.Char.ConvertFromUtf32(13).ToString(), System.Char.ConvertFromUtf32(10).ToString() ), ReplacementString );
This is somewhat offtopic but to get it to work inside Visual Studio's XML .props files, which invoke .NET via the XML properties, I had to dress it up like it is shown below.
The Visual Studio XML --> .NET environment just would not accept the special character escapes like "\r" and "\n" and \x0d and \u000D as well as System.Environment.NewLine as parameters to thereplace() method.
$([System.IO.File]::ReadAllText('MyFile.txt').replace( $([System.String]::Concat($([System.Char]::ConvertFromUtf32(13).ToString()),$([System.Char]::ConvertFromUtf32(10).ToString()))),$([System.String]::Concat('^',$([System.Char]::ConvertFromUtf32(13).ToString()),$([System.Char]::ConvertFromUtf32(10).ToString())))))

Based on #mark-bayers answer and for cleaner output:
string result = Regex.Replace(ex.Message, #"(\r\n?|\r?\n)+", "replacement text");
It removes \r\n , \n and \r while perefer longer one and simplify multiple occurances to one.