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Regex not able to parse last group [duplicate]

This question already has answers here:
What is the difference between .*? and .* regular expressions?
(3 answers)
Closed 3 years ago.
This is the test. I expect the last group to be ".png", but this pattern returns "" instead.
var inputStr = #"C:\path\to\dir\[yyyy-MM-dd_HH-mm].png";
var pattern = #"(.*?)\[(.*?)\](.*?)";
var regex = new Regex(pattern);
var match = regex.Match(inputStr);
var thirdGroupValue = match.Groups[3].Value;
// ✓ EXPECTED: ".png"
// ✗ CURRENT: ""
The 1st and 2nd groups work fine.

This is because you made the * in Group 3 lazy:
(.*?)\[(.*?)\](.*?)
^
here
This means it will match as little as possible. What's the least .* can match? An empty string!
You can learn more about lazy vs greedy here.
You can fix this either by removing ?, making it greedy, or put a $ at the end, telling it to match until the end of the string:
(.*?)\[(.*?)\](.*)
or
(.*?)\[(.*?)\](.*?)$

Easy Regex capture [duplicate]

This question already has answers here:
Regular Expression Groups in C#
(5 answers)
Closed 6 years ago.
New to using C# Regex, I am trying to capture two comma separated integers from a string into two variables.
Example: 13,567
I tried variations on
Regex regex = new Regex(#"(\d+),(\d+)");
var matches = regex.Matches("12,345");
foreach (var itemMatch in matches)
Debug.Print(itemMatch.Value);
This just captures 1 variable, which is the entire string. I did workaround this by changing the capture pattern to "(\d+)", but that then ignores the middle comma entirely and I would get a match if there were any text between the integers.
How do I get it to extract both integers and ensure it also sees a comma between.

Can do this with String.Split
Why not just use a split and parse?
var results = "123,456".Split(',').Select(int.Parse).ToArray();
var left = results[0];
var right = results[1];
Alternatively, you can use a loop and use int.TryParse to handle failures but for what you're looking for this should cover it
If you're really committed to a Regex
You can do this with a Regex too, just need to use groups of the match
Regex r = new Regex(#"(\d+)\,(\d+)", RegexOptions.Compiled);
var r1 = r.Match("123,456");
//first is total match
Console.WriteLine(r1.Groups[0].Value);
//Then first and second groups
var left = int.Parse(r1.Groups[1].Value);
var right = int.Parse(r1.Groups[2].Value);
Console.WriteLine("Left "+ left);
Console.WriteLine("Right "+right);
Made a dotnetfiddle you can test the solution in as well

With Regex, you can use this:
Regex regex = new Regex(#"\d+(?=,)|(?<=,)\d+");
var matches = regex.Matches("12,345");
foreach (Match itemMatch in matches)
Console.WriteLine(itemMatch.Value);
prints:
12
345
Actually this is doing a look-ahead and look-behind a , :
\d+(?=,) <---- // Match numbers followed by a ,
| <---- // OR
(?<=,)\d+ <---- // Match numbers preceeded by a ,

Replace text place holders with Regular Expression [duplicate]

This question already has answers here:
Extract string between braces using RegEx, ie {{content}}
(3 answers)
Closed 6 years ago.
I have a text template that has text variables wrapped with {{ and }}.
I need a regular expression to gives me all the matches that "Include {{ and }}".
For example if I have {{FirstName}} in my text I want to get {{FirstName}} back as a match to be able to replace it with the actual variable.
I already found a regular expression that probably gives me what is INSIDE { and } but I don't know how can I modify it to return what I want.
/\{([^)]+)\}/

This pattern should do the trick:
string str = "{{FirstName}} {{LastName}}";
Regex rgx = new Regex("{{.*?}}");
foreach (var match in rgx.Matches(str))
{
// {{FirstName}}
// {{LastName}}
}

Maybe:
alert(/^\{{2}[\w|\s]+\}{2}$/.test('{{FirstName}}'))
^: In the beginning.
$: In the end.
\{{2}: Character { 2 times.
[\w|\s]+: Alphabet characters or whitespace 1 or more times.
\}{2}: Character } 2 times.
UPDATE:
alert(/(^\{{2})?[\w|\s]+(\}{2})?$/.test('FirstName'))

Regex only checks first character in string C# [duplicate]

This question already has answers here:
What special characters must be escaped in regular expressions?
(13 answers)
Closed 4 years ago.
Why does the following method only check the first character in the supplied string?
public static bool IsUnicodeSms(string message)
{
var strMap = new Regex(#"^[#£$¥èéùìòÇØøÅå_ÆæßÉ!""#%&'()*+,-./0123456789:;<=>? ¡ABCDEFGHIJKLMNOPQRSTUVWXYZÄÖÑܧ¿abcdefghijklmnopqrstuvwxyzäöñüà^{}\[~]|€]+$");
return !strMap.IsMatch(message);
}
So for example the following string returns false: "abcლ" but "ლabc" returns true.

You have to escape ] with \] and also put the - at the end:
Change this:
var strMap = new Regex(#"^[#£$¥èéùìòÇØøÅå_ÆæßÉ!""#%&'()*+,-./0123456789:;<=>? ¡ABCDEFGHIJKLMNOPQRSTUVWXYZÄÖÑܧ¿abcdefghijklmnopqrstuvwxyzäöñüà^{}\[~]|€]+$");
To this:
var strMap = new Regex(#"^[#£$¥èéùìòÇØøÅå_ÆæßÉ!""#%&'()*+,./0123456789:;<=>? ¡ABCDEFGHIJKLMNOPQRSTUVWXYZÄÖÑܧ¿abcdefghijklmnopqrstuvwxyzäöñüà^{}\[~\]|€-]+$");
Btw, you can improve your regex and use:
var strMap = new Regex(#"^[#£$¥èéùìòÇØøÅå_ÆæßÉ!"#%&'()*+,./\w:;<=>? ¡ÄÖÑܧ¿äöñüà^{}\[~\]|€-]+$");
And not sure if using the ignore case flag might help you to shorten it a little more like this:
var strMap = new Regex(#"(?i)^[#£$¥èéùìòÇøå_Ææß!"#%&'()*+,./\w:;<=>? ¡§¿äöñüà^{}\[~\]|€-]+$");

You copied the code from here.
It's very flawed. It needs more escaping. From Regexp Tutorial - Character Classes or Character Sets:
the only special characters or metacharacters inside a character class are the closing bracket (]), the backslash (\), the caret (^), and the hyphen (-)
So, it needs to be:
new Regex(#"^[#£$¥èéùìòÇØøÅå_ÆæßÉ!""#%&'()*+,\-./0123456789:;<=>? ¡ABCDEFGHIJKLMNOPQRSTUVWXYZÄÖÑܧ¿abcdefghijklmnopqrstuvwxyzäöñüà^{}\[~\]|€]+$");
You can of course improve the regex even further like #Fede demonstrates.

C# RegEx to create string array split on spaces and phrases (in quotes) [duplicate]

This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Regular Expression to split on spaces unless in quotes
I am dealing with various strings that I need to split into an array wherever there is a space, except for if that space exists within "quotes".
So for example, I would like this:
this is "a simple" test
..to become:
[0] = this
[1] = is
[2] = "a simple"
[3] = test
Note I would like to retain the quotes surrounding the phrase, not remove them.

The regex:
".*?"|[^\s]+
Usage:
String input = #"this is ""a simple"" test";
String[] matches =
Regex.Matches(input, #""".*?""|[^\s]+").Cast<Match>().Select(m => m.Value).ToArray();