I need to parse a text and check if between all squared brackets is a - and before and after the - must be at least one character.
I tried the following code, but it doesn't work. The matchcount is to large.
Regex regex = new Regex(#"[\.*-.*]");
MatchCollection matches = regex.Matches(textBox.Text);
SampleText:
Node
(Entity [1-5])
Figured I might as well provide an answer... To reiterate my points (with modifications):
* matches 0 or more occurences. You want + probably.
square brackets are special characters and will need to be escaped. They are used to define sets of characters.
You will probably want to exclude [ and ] from your "any character" matching
Put this all together and the following should do you better:
Regex regex = new Regex(#"\[[^-[\]]+-[^[\]]+\]");
Although its a little messy the key thing is that [^[\]] means any character except a square bracket. [^-[\]] means that but also disallows -. This is an optimisation and not required but it just reduces the work the regular expression engine has to do when working out the match. Thanks to ridgerunner for pointing out this optimisation.
Square brackets mean something special in Regexes, you'll need to escape them. Additionally, if you want at least one character then you need to use + rather than *.
Regex regex = new Regex(#"\[.+-.+\]");
MatchCollection matches = regex.Matches(textBox.Text);
string txt = "(Entity [1-5])";
Regex reg = new Regex(#"\[.+\-.+\]");
if it is for #:
string txt = "(Entity [1-5])";
Regex reg = new Regex(#"\[\d+\-\d+\]");
Related
I have a regex working to find anything between the square brackets in a text file, which is this:
Regex squareBrackets = new Regex(#"\[(.*?)\]");
And I want to create a regex that is basically the opposite way round to select whatever is after what's in the square brackets. So I thought just swap them round?
Regex textAfterTitles = new Regex(#"\](.*?)\[");
But this does not work and Regex's confuse me - can anyone help?
Cheers
You can use a lookbehind:
var textAfterTiles = new Regex(#"(?<=\[(.*?)\]).*");
You can combine it with a lookahead if you have multiple such bracketed groups, such as:
var textAfterTiles = "before [one] inside [two] after"
And you wanted to match " inside " and " after", you could do this:
new Regex(#"(?<=\[(.*?)\])[^\[]*");
The same \[(.*?)] regex (I'd just remove the redundant escaping backslash before ]), or even better regex is \[([^]]*)], can be used to split the text and get the text outside [...] (if used with RegexOptions.ExplicitCapture modifier):
var data = "A bracket is a tall punctuation mark[1] typically used in matched pairs within text,[2] to set apart or interject other text.";
Console.WriteLine(String.Join("\n", Regex.Split(data,#"\[([^]]*)]",RegexOptions.ExplicitCapture)));
Output of the C# demo:
A bracket is a tall punctuation mark
typically used in matched pairs within text,
to set apart or interject other text.
The RegexOptions.ExplicitCapture flag makes the capturing group inside the pattern non-capturing, and thus, the captured text is not output into the resulting split array.
If you do not have to keep the same regex, just remove the capture group, use \[[^]]*] for splitting.
You can try this one
\]([^\]]*)\[
I'm having a hard time with this one. First off, here is the difficult part of the string I'm matching against:
"a \"b\" c"
What I want to extract from this is the following:
a \"b\" c
Of course, this is just a substring from a larger string, but everything else works as expected. The problem is making the regex ignore the quotes that are escaped with a backslash.
I've looked into various ways of doing it, but nothing has gotten me the correct results. My most recent attempt looks like this:
"((\"|[^"])+?)"
In various test online, this works the way it should - but when I build my ASP.NET page, it cuts off at the first ", leaving me with just the a-letter, white space and a backslash.
The logic behind the pattern above is to capture all instances of \" or something that is not ". I was hoping this would search for \", making sure to find those first - but I got the feeling that this is overridden by the second part of the expression, which is only 1 single character. A single backslash does not match 2 characters (\"), but it will match as a non-". And from there, the next character will be a single ", and the matching is completed. (This is just my hypothesis on why my pattern is failing.)
Any pointers on this one? I have tried various combinations with "look"-methods in regex, but I didn't really get anywhere. I also get the feeling that is what I need.
ORIGINAL ANSWER
To match a string like a \"b\" c, you need to use following regex declaration:
(?:\\"|[^"])+
var rx = Regex(#"(?:\\""|[^""])+");
See RegexStorm demo
Here is an IDEONE demo:
var str = "a \\\"b\\\" c";
Console.WriteLine(str);
var rx = new Regex(#"(?:\\""|[^""])+");
Console.WriteLine(rx.Match(str).Value);
Please note the # in front of the string literal that lets us use verbatim string literals where we have to double quotes to match literal quotes and use single escape slashes instead of double. This makes regexps easier to read and maintain.
If you want to match any escaped entities in your input string, you can use:
var rx = new Regex(#"[^""\\]*(?:\\.[^""\\]*)*");
See demo on RegexStorm
UPDATE
To match the quoted strings, just add quotes around the pattern:
var rx = new Regex(#"""(?<res>[^""\\]*(?:\\.[^""\\]*)*)""");
This pattern yields much better performance than Tim Long's suggested regex, see RegexHero test resuls:
The following expression worked for me:
"(?<Result>(\\"|.)*)"
The expression matches as follows:
An opening quote (literal ")
A named capture (?<name>pattern) consisting of:
Zero or more occurences * of literal \" or (|) any single character (.)
A final closing quote (literal ")
Note that the * (zero or more) quantifier is non-greedy so the final quote is matched by the literal " and not the "any single character" . part.
I used ReSharper 9's built-in Regular Expression validator to develop the expression and verify the results:
I have used the "Explicit Capture" option to reduce cruft in the output (RegexOptions.ExplicitCapture).
One thing to note is that I am matching the whole string, but I am only capturing the substring, using a named capture. Using named captures is a really useful way to get at the results you want. In code, it might look something like this:
static string MatchQuotedString(string input)
{
const string pattern = #"""(?<Result>(\\""|.)*)""";
const RegexOptions options = RegexOptions.ExplicitCapture;
Regex regex = new Regex(pattern, options);
var matches = regex.Match(input);
var substring = matches.Groups["Result"].Value;
return substring;
}
Optimization: If you are planning on using the regex a lot, you could factor it out into a field and use the RegexOptions.Compiled option, this pre-compiles the expression and gives you faster throughput at the expense of longer initialization.
Greetings, I have file with the following strings:
string.Format("{0},{1}", "Having \"Two\" On The Same Line".Localize(), "Is Tricky For regex".Localize());
my goal is to get a match set with the two strings:
Having \"Two\" On The Same Line
Is Tricky For regex
My current regex looks like this:
private Regex CSharpShortRegex = new Regex("\"(?<constant>[^\"]+?)\".Localize\\(\\)");
My problem is with the escaped quotes in the first line I end up stopping at the quote and I get:
On The Same Line
Is Tricky For This Style Too
however attempting to ignore the escaped quotes is not working out because it makes the Regex greedy and I get
Having \"Two\" On The Same Line".Localize(), "Is Tricky For regex"
We seem to be caught between maximum and minimum munge. Is there any hope? I have some backup plans. Can you Regex backwards? that would make it easier because I can start with the "()ezilacoL."
EDIT:
To clarify. This is my lone edge case. Most of the time the string sits alone like:
var myString = "Hot Patootie".Localize()
This one works for me:
\"((?:[^\\"]|(?:\\\"))*)\"\.Localize\(\)
Tested on http://www.regexplanet.com/simple/index.html against a number of strings with various escaped quotes.
Looks like most of us who answered this one had the same rough idea, so let me explain the approach (comments after #s):
\" # We're looking for a string delimited by quotation marks
( # Capture the contents of the quotation marks
(?: # Start a non-capturing group
[^\\"] # Either read a character that isn't a quote or a slash
|(?:\\\") # Or read in a slash followed by a quote.
)* # Keep reading
) # End the capturing group
\" # The string literal ends in a quotation mark
\.Localize\(\) # and ends with the literal '.Localize()', escaping ., ( and )
For C# you'll need to escape the slashes twice (messy):
\"((?:[^\\\\\"]|(?:\\\\\"))*)\"\\.Localize\\(\\)
Mark correctly points out that this one doesn't match escaped characters other than quotation marks. So here's a better version:
\"((?:[^\\"]|(?:\\")|(?:\\.))*)\"\.Localize\(\)
And its slashed-up equivalent:
\"((?:[^\\\\\"]|(?:\\\\\")|(?:\\\\.))*)\"\\.Localize\\(\\)
Works the same way, except it has a special case that if encounters a slash but it can't match \", it just consumes the slash and the following character and moves on.
Thinking about it, it's better to just consume two characters at every slash, which is effectively Mark's answer so I won't repeat it.
Here's the regular expression you need:
#"""(?<constant>(\\.|[^""])*)""\.Localize\(\)"
A test program:
using System;
using System.Text.RegularExpressions;
using System.IO;
class Program
{
static void Main()
{
Regex CSharpShortRegex =
new Regex(#"""(?<constant>(\\.|[^""])*)""\.Localize\(\)");
foreach (string line in File.ReadAllLines("input.txt"))
foreach (Match match in CSharpShortRegex.Matches(line))
Console.WriteLine(match.Groups["constant"].Value);
}
}
Output:
Having \"Two\" On The Same Line
Is Tricky For regex
Hot Patootie
Notice that I have used #"..." to avoid having to escape backslashes inside the regular expression. I think this makes it easier to read.
Update:
My original answer (below the horizontal rule) has a bug: regular-expression matchers attempt alternatives in left-to-right order. Having [^"] as the first alternative allows it to consume the backslash, but then the next character to be matched is a quote, which prevents the match from proceeding.
Incompatibility note: Given the pattern below, perl backtracks to the other alternative (the escaped quote) and successfully finds a match for the Having \"Two\" On The Same Line case.
The fix is to try an escaped quote first and then a non-quote:
var CSharpShortRegex =
new Regex("\"(?<constant>(\\\\\"|[^\"])*)\"\\.Localize\\(\\)");
or if you prefer the at-string form:
var CSharpShortRegex =
new Regex(#"""(?<constant>(\\""|[^""])*)""\.Localize\(\)");
Allow for escapes:
private Regex CSharpShortRegex =
new Regex("\"(?<constant>([^\"]|\\\\\")*)\"\\.Localize\\(\\)");
Applying one level of escaping to make the pattern easier to read, we get
"(?<constant>([^"]|\\")*)"\.Localize\(\)
That is, a string starts and ends with " characters, and everything between is either a non-quote or an escaped quote.
Looks like you're trying to parse code so one approach might be to evaluate the code on the fly:
var cr = new CSharpCodeProvider().CompileAssemblyFromSource(
new CompilerParameters { GenerateInMemory = true },
"class x { public static string e() { return " + input + "}}");
var result = cr.CompiledAssembly.GetType("x")
.GetMethod("e").Invoke(null, null) as string;
This way you could handle all kinds of other special cases (e.g. concatenated or verbatim strings) that would be extremely difficult to handle with regex.
new Regex(#"((([^#]|^|\n)""(?<constant>((\\.)|[^""])*)"")|(#""(?<constant>(""""|[^""])*)""))\s*\.\s*Localize\s*\(\s*\)", RegexOptions.Compiled);
takes care of both simple and #"" strings. It also takes into account escape sequences.
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?)+$";
This means "any alphanumeric 1 or more times followed by an optional space, this whole thing one or more times". I don't know how to limit the whole string to 40 characters when you don't know the size of the first expression in advance. Maybe this can be achieved with a "look behind" expression, but I'm not sure. You might have to do it in two steps.
Ok sorry this might seem like a dumb question but I cannot figure this thing out :
I am trying to parse a string and simply want to check whether it only contains the following characters : '0123456789dD+ '
I have tried many things but just can't get to figure out the right regex to use!
Regex oReg = new Regex(#"[\d dD+]+");
oReg.IsMatch("e4");
will return true even though e is not allowed...
I've tried many strings, including Regex("[1234567890 dD+]+")...
It always works on Regex Pal but not in C#...
Please advise and again i apologize this seems like a very silly question
Try this:
#"^[0-9dD+ ]+$"
The ^ and $ at the beginning and end signify the beginning and end of the input string respectively. Thus between the beginning and then end only the stated characters are allowed. In your example, the regex matches if the string contains one of the characters even if it contains other characters as well.
#comments: Thanks, I fixed the missing + and space.
Oops, you forgot the boundaries, try:
Regex oReg = new Regex(#"^[0-9dD +]+$");
oReg.IsMatch("e4");
^ matches the begining of the text stream, $ matches the end.
It is matching the 4; you need ^ and $ to terminate the regex if you want a full match for the entire string - i.e.
Regex re = new Regex(#"^[\d dD+]+$");
Console.WriteLine(re.IsMatch("e4"));
Console.WriteLine(re.IsMatch("4"));
This is because regular expressions can also match parts of the input, in this case it just matches the "4" of "e4". If you want to match a whole line, you have to surround the regex with "^" (matches line start) and "$" (matches line end).
So to make your example work, you have to write is as follows:
Regex oReg = new Regex(#"^[\d dD+]+$");
oReg.IsMatch("e4");
I believe it's returning True because it's finding the 4. Nothing in the regex excludes the letter e from the results.
Another option is to invert everything, so it matches on characters you don't want to allow:
Regex oReg = new Regex(#"[^0-9dD+]");
!oReg.IsMatch("e4");