I have a string text which is like:
"ruf": "the text I want",
"puf":
I want extract the text inside the quotes.
tried this :
string cg="?<=\"ruf\":\")(.*?)(?=\",puf";
Regex g = new Regex(cg);
It didnt work.
Try with below regex:
(?<="ruf":\s\")[^"]*
Online demo
String literals for use in programs:
C#
#"(?<=""ruf"":\s\"")[^""]*"
output:
the text I want
Pattern description:
(?<= look behind to see if there is:
"ruf": '"ruf":'
\s whitespace (\n, \r, \t, \f, and " ")
\" '"'
) end of look-behind
[^"]* any character except: '"' (0 or more times
(matching the most amount possible))
Debuggex Demo
EDIT
Can you add puf. Because it is a long text which has multiple quotes in it
If you are looking till "puf" is found then try below regex:
(?<="ruf":\s\")[\s\S]*(?=",\s*"puf")
Online demo
String literals for use in programs:
C#
#"(?<=""ruf"":\s\"")[\s\S]*(?="",\s*""puf"")"
You could try the below regex with s modifier,
/(?<=\"ruf\": \")[^\"]*(?=\",.*?\"puf\":)/s
DEMO
With the s modifier, dot matches even newline character also.
Do it like this:
var myRegex = new Regex(#"(?s)(?<=""ruf"": "")[^""]*(?=\s*""puf"")");
string resultString = myRegex.Match(yourString).Value;
Console.WriteLine(resultString);
Related
So I have a regex here:
var text = new Regex(#"(?<=Paybacks).*", RegexOptions.IgnoreCase);
This looks for the line where it starts with Paybacks. Now it currently prints ": blah".
The context sometimes can be "Paybacks" or "Paybacks:" or "Paybacks " or I don't know "Paybacks (with thousands of whitespaces). How can I modify this regex to be like.. after "Paybacks" ignore a colon and a whitespace (or whitespaces) that may or may not exist.
I've been playing with it in regex101 and this seems to be working, but is there a better way?
(?<=Volatility(:\s)).*
In these situations, you'd better use a regex with a capturing group:
var pattern = new Regex(#"Paybacks[\s:]*(.*)", RegexOptions.IgnoreCase);
Then, you can use
var output = Regex.Match(text, pattern)?.Groups[1].Value;
See the .NET regex demo:
See the C# demo:
var texts = new List<string> { "Paybacks: blah","Paybacks:blah","Paybacks blah"};
var pattern = new Regex(#"Paybacks[\s:]*(.*)", RegexOptions.IgnoreCase);
texts.ForEach(text => Console.WriteLine(pattern.Match(text)?.Groups[1].Value));
printing 3 blahs.
You might also match optional colons and whitspace chars in the lookbehind, and start matching the first chars being any non whitspace char other than :
(?<=Paybacks[:\s]*)[^\s:].*
The pattern matches:
(?<= Positive lookbehind, assert what is on the left is
Paybacks Match literally
[:\s]* Optionally match either : or a whitespace char using a character class
) Close lookbehind
[^\s:].* Match a single non whitespace char other than : and the rest of the line
Regex demo | C# demo
var regex = new Regex(#"(?<=Paybacks[:\s]*)[^\s:].*", RegexOptions.IgnoreCase);
string[] strings = {"Paybacks: blah", "Paybacks blah", "Paybacks blah"};
foreach (String s in strings)
{
Console.WriteLine(regex.Match(s)?.Value);
}
Output
blah
blah
blah
If the order should be a single optional colon and optional whitespace chars, you can make the colon optional and the quantifier for the whitespace chars 0 or more using :?\s*
(?<=Paybacks:?\s*)[^\s:].*
Regex demo
I have the following input string:
string val = "[01/02/70]\nhello world ";
I want to get the all words after the last ] character.
Example output for a sample string above:
\nhello world
In C#, use Substring() with IndexOf:
string val = val.Substring(val.IndexOf(']') + 1);
If you have multiple ] symbols, and you want to get all the string after the last one, use LastIndexOf:
string val = "[01/02/70]\nhello [01/02/80] world ";
string val = val.Substring(val.LastIndexOf(']') + 1); // => " world "
If you are a fan of Regex, you might want to use a Regex.Replace like
string val = "[01/02/70]\nhello [01/02/80] world ";
val = Regex.Replace(val, #"^.*\]", string.Empty, RegexOptions.Singleline); // => " world "
See demo
Notes on REGEX:
RegexOptions.Singleline makes . match a linebreak
^ - matches beginning of string
.* - matches 0 or more characters but as many as possible (greedy matching)
\] - matches literal ] (as it is a special regex metacharacter, it must be escaped).
You need to use lookbehind assertion. And not only that, you have to enable DOTALL modifier also, so that it would also match the newline character present inbetween.
"(?s)(?<=\\]).*"
(?s) - DOTALL modifier.
(?<=\\]) - lookbehind which asserts that the match must be preceeded by a close bracket
.* - Matches any chracater zero or more times.
or
"(?s)(?<=\\])[\\s\\S]*"
Try this if you don't want to match the following newline character.
#"(?<=\][\n\r]*).*"
I'm checking to see if my regular expression matches my string.
I have a filename that looks like somename_somthing.txt and I want to match it to somename_*.txt, but my code is failing when I try to pass something that should match. Here is my code.
string pattern = "somename_*.txt";
Regex r = new Regex(pattern, RegexOptions.IgnoreCase);
using (ZipFile zipFile = ZipFile.Read(fullPath))
{
foreach (ZipEntry e in zipFile)
{
Match m = r.Match("somename_something.txt");
if (!m.Success)
{
throw new FileNotFoundException("A filename with format: " + pattern + " not found.");
}
}
}
The asterisk is matching the underscore and throwing it off.
Try:
somename_(\w+).txt
The (\w+) here will match the group at this location.
You can see it match here: https://regex101.com/r/qS8wA5/1
In General
Regex give in this code matches the _ with an * meaning zero or more underscores instead of what you intended. The * is used to denote zero or more of the previous item. Instead try
^somename_(.*)\.txt$
This matches exactly the first part "somename_".
Then anything (.*)
And finally the end ".txt". The backslash escapes the 'dot'.
More Specific
You can also say if you only want letters and not numbers or symbols in the middle part of the match with:
^somename_[a-z]*\.txt$
As written, your regular expression
somename_*.txt
matches (in a case-insensitive manner):
the literal text somename, followed by
zero or more underscore characters (_), followed
any character (other than newline), followed
the literal text txt
And it will match that anywhere in the source text. You probably want to write something like
Regex myPattern = new Regex( #"
^ # anchor the match to start-of-text, followed by
somename # the literal 'somename', followed by
_ # a literal underscore character, followed by
.* # zero or of any character (except newline), followed by
\. # a literal period/fullstop, followed by
txt # the literal text 'txt'
$ # with the match anchored at end-of-text
" , RegexOptions.IgnoreCase|RegexOptions.IgnorePatternWhitespace
) ;
Hi I think the pattern should be
string pattern = "somename_.*\\.txt";
Regards
I just wanna know whats the Regex for singlequote and doublequote specifically something like this 1:
openquote(startswith) + word + closequote(endswith)
(singlequote)word(/singlequote) sample-> 'asdasdasdass'
(doublequote)word(/doublequote) sample-> "asdasdasdass"
in c#winforms /thanks .
--- updated:
replacing regex within this line:
string hoveredWord = r.GetFragment("[a-zA-Z]").Text;
thanks!
The following RegEx is for Single Quotation sign + Some Texts + Single Quotation sign ('asdasdasdass'):
Regex regexString = new Regex(#"'[^']*'");
The following RegEx is for Double Quotation sign + Some Texts + Double Quotation sign ("asdasdasdass"):
Regex regexString = new Regex(#"""[^""]*""");
Regex pattern :
Former example : ^\'\w+\'
Later example : ^\"\w+\"
^ : match the beginning of the string
\' or \" : match the single quote or double quote
\w+ : match the alpha-numeric characters and underscore
I have a string:
Hello "quoted string" and 'tricky"stuff' world
and want to get the string minus the quoted parts back. E.g.,
Hello and world
Any suggestions?
resultString = Regex.Replace(subjectString,
#"([""'])# Match a quote, remember which one
(?: # Then...
(?!\1) # (as long as the next character is not the same quote as before)
. # match any character
)* # any number of times
\1 # until the corresponding closing quote
\s* # plus optional whitespace
",
"", RegexOptions.IgnorePatternWhitespace);
will work on your example.
resultString = Regex.Replace(subjectString,
#"([""'])# Match a quote, remember which one
(?: # Then...
(?!\1) # (as long as the next character is not the same quote as before)
\\?. # match any escaped or unescaped character
)* # any number of times
\1 # until the corresponding closing quote
\s* # plus optional whitespace
",
"", RegexOptions.IgnorePatternWhitespace);
will also handle escaped quotes.
So it will correctly transform
Hello "quoted \"string\\" and 'tricky"stuff' world
into
Hello and world
Use a regular expression to match any quoted strings with the string and replace them with the empty string. Use the Regex.Replace() method to do the pattern matching and replacement.
In case, like me, you're afraid of regex, I've put together a functional way to do it, based on your example string. There's probably a way to make the code shorter, but I haven't found it yet.
private static string RemoveQuotes(IEnumerable<char> input)
{
string part = new string(input.TakeWhile(c => c != '"' && c != '\'').ToArray());
var rest = input.SkipWhile(c => c != '"' && c != '\'');
if(string.IsNullOrEmpty(new string(rest.ToArray())))
return part;
char delim = rest.First();
var afterIgnore = rest.Skip(1).SkipWhile(c => c != delim).Skip(1);
StringBuilder full = new StringBuilder(part);
return full.Append(RemoveQuotes(afterIgnore)).ToString();
}