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.
Related
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);
How do i search a string for a newline character? Both of the below seem to be returning -1....!
theJ = line.IndexOf(Environment.NewLine);
OR
theJ = line.IndexOf('\n');
The string it's searching is "yo\n"
the string i'm parsing contains this "printf("yo\n");"
the string i see contained during the comparison is this: "\tprintf(\"yo\n\");"
"yo\n" // output as "yo" + newline
"yo\n".IndexOf('\n') // returns 2
"yo\\n" // output as "yo\n"
"yo\\n".IndexOf('\n') // returns -1
Are you sure you're searching yo\n and not yo\\n?
Edit
Based on your update, I can see that I guessed correctly. If your string says:
printf("yo\n");
... then this does not contain a newline character. If it did, it would look like this:
printf("yo
");
What it actually has is an escaped newline character, or in other words, a backslash character followed by an 'n'. That's why the string you're seeing when you debug is "\tprintf(\"yo\\n\");". If you want to find this character combination, you can use:
line.IndexOf("\\n")
For example:
"\tprintf(\"yo\\n\");" // output as " printf("yo\n");"
"\tprintf(\"yo\\n\");".IndexOf("\\n") // returns 11
Looks like your line does not contain a newline.
If you are using File.ReadAllLines or string.Split on newline, then each line in the returned array will not contain the newline. If you are using StreamReader or one of the classes inheriting from it, the ReadLine method will return the string without the newline.
string lotsOfLines = #"one
two
three";
string[] lines = lotsOfLines.Split('\n');
foreach(string line in lines)
{
Console.WriteLine(line.IndexOf('\n'); // prints -1 three times
}
That should work although in Windows you'll have to search for '\r\n'.
-1 simply means that no enter was found.
It depends what you are trying to do. Both may no be identical on some platforms.
Environment.NewLine returns:
A string containing "\r\n" for non-Unix platforms, or a string
containing "\n" for Unix platforms.
Also:
If you want to search for the \n char (new line on Unix), use \n
If you want to search for the \r\n chars (new line on Windows), use \r\n
If your search depend on the current platform, use Environment.NewLine
If it returns -1 in both cases you mentioned, then you don't have a new line in your string.
When I was in college and I did a WebForms aplication to order referencies.
And the line break/carriage return it was what I used to break a referense.
//Text from TextBox
var text = textBox1.Text;
//Create an array with the text between the carriage returns
var references = text.Split(new string[] { "\r\n", "\r" }, StringSplitOptions.RemoveEmptyEntries);
//Simple OrderBy(Alphabetical)
var ordered = references.ToList<string>().OrderBy(ff => ff);
//Return the entry text ordered alphabetical em with a carriage return between every result
var valueToReturn = String.Join(Environment.NewLine, ordered);
textBox1.Text = valueToReturn;
The Environment.NewLine is not the same as \n. It is a CRLF (\r\n). However, I did try with the \n using IndexOf and my test did find the value. Are you sure what you're searching for is a \n rather than a \r? View your text in hexadecimal format and see what the hex value is.
I have a need to get rid of all line breaks that appear in my strings (coming from db).
I do it using code below:
value.Replace("\r\n", "").Replace("\n", "").Replace("\r", "")
I can see that there's at least one character acting like line ending that survived it. The char code is 8232.
It's very lame of me, but I must say this is the first time I have a pleasure of seeing this char. It's obvious that I can just replace this char directly, but I was thinking about extending my current approach (based on replacing combinations of "\r" and "\n") to something much more solid, so it would not only include the '8232' char but also all others not-found-by-me yet.
Do you have a bullet-proof approach for such a problem?
EDIT#1:
It seems to me that there are several possible solutions:
use Regex.Replace
remove all chars if it's IsSeparator or IsControl
replace with " " if it's IsWhiteSpace
create a list of all possible line endings ( "\r\n", "\r", "\n",LF ,VT, FF, CR, CR+LF, NEL, LS, PS) and just replace them with empty string. It's a lot of replaces.
I would say that the best results will be after applying 1st and 4th approaches but I cannot decide which will be faster. Which one do you think is the most complete one?
EDIT#2
I posted anwer below.
Below is the extension method solving my problem. LineSeparator and ParagraphEnding can be of course defined somewhere else, as static values etc.
public static string RemoveLineEndings(this string value)
{
if(String.IsNullOrEmpty(value))
{
return value;
}
string lineSeparator = ((char) 0x2028).ToString();
string paragraphSeparator = ((char)0x2029).ToString();
return value.Replace("\r\n", string.Empty)
.Replace("\n", string.Empty)
.Replace("\r", string.Empty)
.Replace(lineSeparator, string.Empty)
.Replace(paragraphSeparator, string.Empty);
}
According to wikipedia, there are numerous line terminators you may need to handle (including this one you mention).
LF: Line Feed, U+000A
VT: Vertical Tab, U+000B
FF: Form Feed, U+000C
CR: Carriage Return, U+000D
CR+LF: CR (U+000D) followed by LF (U+000A)
NEL: Next Line, U+0085
LS: Line Separator, U+2028
PS: Paragraph Separator, U+2029
8232 (0x2028) and 8233 (0x2029) are the only other ones you might want to eliminate. See the documentation for char.IsSeparator.
Props to Yossarian on this one, I think he's right. Replace all whitespace with a single space:
data = Regex.Replace(data, #"\s+", " ");
I'd recommend removing ALL the whitespace (char.IsWhitespace), and replacing it with single space.. IsWhiteSpace takes care of all weird unicode whitespaces.
This is my first attempt at this, but I think this will do what you want....
var controlChars = from c in value.ToCharArray() where Char.IsControl(c) select c;
foreach (char c in controlChars)
value = value.Replace(c.ToString(), "");
Also, see this link for details on other methods you can use: Char Methods
Have you tried string.Replace(Environment.NewLine, "") ? That usually gets a lot of them for me.
Check out this link: http://msdn.microsoft.com/en-us/library/844skk0h.aspx
You wil lhave to play around and build a REGEX expression that works for you. But here's the skeleton...
static void Main(string[] args)
{
StringBuilder txt = new StringBuilder();
txt.Append("Hello \n\n\r\t\t");
txt.Append( Convert.ToChar(8232));
System.Console.WriteLine("Original: <" + txt.ToString() + ">");
System.Console.WriteLine("Cleaned: <" + CleanInput(txt.ToString()) + ">");
System.Console.Read();
}
static string CleanInput(string strIn)
{
// Replace invalid characters with empty strings.
return Regex.Replace(strIn, #"[^\w\.#-]", "");
}
Assuming that 8232 is unicode, you can do this:
value.Replace("\u2028", string.Empty);
personally i'd go with
public static String RemoveLineEndings(this String text)
{
StringBuilder newText = new StringBuilder();
for (int i = 0; i < text.Length; i++)
{
if (!char.IsControl(text, i))
newText.Append(text[i]);
}
return newText.ToString();
}
If you've a string say "theString" then
use the method Replace and give it the arguments shown below:
theString = theString.Replace(System.Environment.NewLine, "");
Here are some quick solutions with .NET regex:
To remove any whitespace from a string: s = Regex.Replace(s, #"\s+", ""); (\s matches any Unicode whitespace chars)
To remove all whitespace BUT CR and LF: s = Regex.Replace(s, #"[\s-[\r\n]]+", ""); ([\s-[\r\n]] is a character class containing a subtraction construct, it matches any whitespace but CR and LF)
To remove any vertical whitespace, subtract \p{Zs} (any horizontal whitespace but tab) and \t (tab) from \s: s = Regex.Replace(s, #"[\s-[\p{Zs}\t]]+", "");.
Wrapping the last one into an extension method:
public static string RemoveLineEndings(this string value)
{
return Regex.Replace(value, #"[\s-[\p{Zs}\t]]+", "");
}
See the regex demo.
buildLetter.Append("</head>").AppendLine();
buildLetter.Append("").AppendLine();
buildLetter.Append("<style type="text/css">").AppendLine();
Assume the above contents resides in a file. I want to write a snippet that
removes any line which has empty string "" and put escape character before
the middle quotations. The final output would be:
buildLetter.Append("</head>").AppendLine();
buildLetter.Append("<style type=\"text/css\">").AppendLine();
The outer " .... " is not considered special chars. The special chars may be single
quotation or double quotation.
I could run it via find and replace feature of Visual Studio. However, in my case i
want it to be written in c# or VB.NET
Any help will be appreciated.
Perhaps this does what you want:
string s = File.ReadAllText("input.txt");
string empty = "buildLetter.Append(\"\").AppendLine();" + Environment.NewLine;
s = s.Replace(empty, "");
s = Regex.Replace(s, #"(?<="").*(?="")",
match => { return match.Value.Replace("\"", "\\\""); }
);
Result:
buildLetter.Append("</head>").AppendLine();
buildLetter.Append("<style type=\"text/css\">").AppendLine();
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?", "");