Regex that matches a newline (\n) in C# - c#

OK, this one is driving me nuts....
I have a string that is formed thus:
var newContent = string.Format("({0})\n{1}", stripped_content, reply)
newContent will display like:
(old text)
new text
I need a regular expression that strips away the text between parentheses with the parenthesis included AND the newline character.
The best I can come up with is:
const string regex = #"^(\(.*\)\s)?(?<capture>.*)";
var match= Regex.Match(original_content, regex);
var stripped_content = match.Groups["capture"].Value;
This works, but I want specifically to match the newline (\n), not any whitespace (\s)
Replacing \s with \n \\n or \\\n does NOT work.
Please help me hold on to my sanity!
EDIT: an example:
public string Reply(string old,string neww)
{
const string regex = #"^(\(.*\)\s)?(?<capture>.*)";
var match= Regex.Match(old, regex);
var stripped_content = match.Groups["capture"].Value;
var result= string.Format("({0})\n{1}", stripped_content, neww);
return result;
}
Reply("(messageOne)\nmessageTwo","messageThree") returns :
(messageTwo)
messageThree

If you specify RegexOptions.Multiline then you can use ^ and $ to match the start and end of a line, respectively.
If you don't wish to use this option, remember that a new line may be any one of the following: \n, \r, \r\n, so instead of looking only for \n, you should perhaps use something like: [\n\r]+, or more exactly: (\n|\r|\r\n).

Actually it works but with opposite option i.e.
RegexOptions.Singleline

You are probably going to have a \r before your \n. Try replacing the \s with (\r\n).

Think I may be a bit late to the party, but still hope this helps.
I needed to get multiple tokens between two hash signs.
Example i/p:
## token1 ##
## token2 ##
## token3_a
token3_b
token3_c ##
This seemed to work in my case:
var matches = Regex.Matches (mytext, "##(.*?)##", RegexOptions.Singleline);
Of course, you may want to replace the double hash signs at both ends with your own chars.
HTH.

Counter-intuitive as it is, you can use both Multiline and Singleline option.
Regex.Match(input, #"(.+)^(.*)", RegexOptions.Multiline | RegexOptions.Singleline)
First capturing group will contain first line (including \r and \n) and second group will have second line.
Why:
First of all RegexOptions enum is flag so it can be combined with bitwise operators, then
Multiline:
^ and $ match the beginning and end of each line (instead of the beginning and end of the input string).
Singleline:
The period (.) matches every character (instead of every character except \n)
see docs

Related

C# Regex for validating words with no space, no special character

I have written the following Regex for matching only those words with no space and no special character. But it is matching with words containing space too. What is wrong in it?
Regex rgx = new Regex("[a-zA-Z0-9]+");
if (!rgx.IsMatch(TextBox_EntityType.Text))
{
}
You can change the logic of your check so it does the opposite, and you take the appropriate action:
Regex rgx = new Regex("[^a-zA-Z0-9]");
# Match if there is something that is not alphanumeric
if (rgx.IsMatch(TextBox_EntityType.Text))
{
# Do what should be done if the text contains non-alphanumeric
}
This one works just as well because .IsMatch() looks for a match anywhere in a string (it tries its best to find a match), so either you make it match the whole string with anchors like Nikhil suggested, or invert the logic like I did (and which I believe should be slightly more efficient, but not benchmarked).
It should be ^[a-zA-Z0-9]+$
Added ^ and $.
The ^ matches the start of the string and $ matches the end.

Replace a string in multiline regex with end of line token

I got the following regex
var fixedString = Regex.Replace(subject, #"(:[\w]+ [\d]+)$", "",
RegexOptions.Multiline);
which doesn't work. It works if I use \r\n, but I would like to support all types of line breaks. As another answer states I have to use RegexOptions.Multiline to be able to use $ as end of line token (instead of end of string). But it doesn't seem to help.
What am I doing wrong?
I am not sure what you want to achieve, I think I understood, you want to replace also the newline character at the end of the row.
The problem is the $ is a zero width assertion. It does not match the newline character, it matches the position before \n.
You could do different other things:
If it is OK to match all following newlines, means also all following empty rows, you could do this:
var fixedString = Regex.Replace(subject, #"(:[\w]+ [\d]+)[\r\n]+", "");
If you only want to match the newline after the row and keep following empty rows, you have to make a pattern for all possible combinations, e.g.:
var fixedString = Regex.Replace(subject, #"(:[\w]+ [\d]+)\r?\n", "");
This would match the combination \n and \r\n

Matching Negative Sequence

I want to match text between word1 and first occurrence of word2. What's the best way to do that, considering that text may include newline characters? Is there a pattern like this: (word1)(not word2)*(word2)?
You could use a lazy quantifier to match as few characters as possible between word1 and word2.
(word1).*?(word2)
See quantifiers topic on MSDN.
You can match them using the SingleLine option:
//use '*' or '*?' depending on what you want for "word1 aaa word2 bbb word2"
string pattern = "word1(.*)word2";
var m = Regex.Match(text1, pattern, RegexOptions.Singleline);
Console.WriteLine(m.Groups[1]); // the result
MSDN about SingleLine :
... causes the regular expression engine to treat the input string as if
it consists of a single line. It does this by changing the behavior of
the period (.) language element so that it matches every character,
instead of matching every character except for the newline character
\n or \u000A.

.net regex match line

Why does ^.*$ does not match a line in:
This is some sample text
this is another line
this is the third line
how can I create a regular expression that will match an entire line so that when finding the next match it will return me the next line.
In other words I will like to have a regex so that the first match = This is some sample text , next match = this is another line etc...
^ and $ match on the entire input sequence. You need to use the Multiline Regex option to match individual lines within the text.
Regex rgMatchLines = new Regex ( #"^.*$", RegexOptions.Multiline);
See here for an explanation of the regex options. Here's what it says about the Multiline option:
Multiline mode. Changes the meaning of ^ and $ so they match at the
beginning and end, respectively, of any line, and not just the
beginning and end of the entire string.
use regex options
Regex regex = new Regex("^.*$", RegexOptions.Multiline);
You have to enable RegexOptions.Multiline to make ^ and $ matches the start and end of line. Otherwise, ^ and $ will match the start and end of the whole input string.

Regex to match full lines of text excluding crlf

How would a regex pattern to match each line of a given text be?
I'm trying ^(.+)$ but it includes crlf...
Just use RegexOptions.Multiline.
Multiline mode. Changes the meaning of
^ and $ so they match at the beginning
and end, respectively, of any line,
and not just the beginning and end of
the entire string.
Example:
var lineMatches = Regex.Matches("Multi\r\nlines", "^(.+)$", RegexOptions.Multiline);
I'm not sure what you mean by "match each line of a given text" means, but you can use a character class to exclude the CR and LF characters:
[^\r\n]+
The wording of your question seems a little unclear, but it sounds like you want RegexOptions.Multiline (in the System.Text.RegularExpressions namespace). It's an option you have to set on your RegEx object. That should make ^ and $ match the beginning and end of a line rather than the entire string.
For example:
Regex re = new Regex("^(.+)$", RegexOptions.Compiled | RegexOptions.Multiline);
Have you tried:
^(.+)\r?\n$
That way the match group includes everything except the CRLF, and requires that a new line be present (Unix default), but accepts the carriage return in front (Windows default).
I assume you're using the Multiline option? In that case you'll want to match the newline explicitly with "\n". (substitute "\r\n" as appropriate.)

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