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Regex pattern generator

I'm trying to do regex pattern which will match to this:
Name[0]/Something
or
Name/Something
Verbs Name and Something will be always known.
I did for Name[0]/Something, but I want make pattern for this verb in one regex
I've tried to sign [0] as optional but it didn't work :
var regexPattern = "Name" + #"\([\d*\]?)/" + "Something"
Do you know some generator where I will input some verbs and it will make pattern for me?

Use this:
Name(\[\d+\])?\/Something
\d+ allows one or more digits
\[\d+\] allows one or more digits inside [ and ]. So it will allow [0], [12] etc but reject []
(\[\d+\])? allows digit with brackets to be present either zero times or once
\/ indicates a slash (only one)
Name and Something are string literals
Regex 101 Demo

You were close, the regex Name(\[\d+\])?\/Something will do.

The problem is with first '\' in your pattern before '('.
Here is what you need:
var str = "Name[0]/Something or Name/Something";
Regex rg = new Regex(#"Name(\[\d+\])?/Something");
var matches = rg.Matches(str);
foreach(Match a in matches)
{
Console.WriteLine(a.Value);
}

var string = 'Name[0]/Something';
var regex = /^(Name)(\[\d*\])?\/Something$/;
console.log(regex.test(string));
string = 'Name/Something';
console.log(regex.test(string));
You've tried wrong with this pattern: \([\d*\]?)/
No need to use \ before ( (in this case)
? after ] mean: character ] zero or one time
So, if you want the pattern [...] displays zero or one time, you can try: (\[\d*\])?
Hope this helps!

i think this is what you are looking for:
Name(\[\d+\])?\/Something
Name litteral
([\d+])? a number (1 or more digits) between brackets optional 1 or 0 times
/Something Something litteral
https://regex101.com/r/G8tIHC/1

How to find a string with missing fragments?

I'm building a chatbot in C# using AIML files, at the moment I've this code to process:
<aiml>
<category>
<pattern>a * is a *</pattern>
<template>when a <star index="1"/> is not a <star index="2"/>?</template>
</category>
</aiml>
I would like to do something like:
if (user_string == pattern_string) return template_string;
but I don't know how to tell the computer that the star character can be anything, and expecially that can be more than one word!
I was thinking to do it with regular expressions, but I don't have enough experience with it. Can somebody help me? :)

Using Regex
static bool TryParse(string pattern, string text, out string[] wildcardValues)
{
// ^ and $ means that whole string must be matched
// Regex.Escape (http://msdn.microsoft.com/en-us/library/system.text.regularexpressions.regex.escape(v=vs.110).aspx)
// (.+) means capture at least one character and place it in match.Groups
var regexPattern = string.Format("^{0}$", Regex.Escape(pattern).Replace(#"\*", "(.+)"));
var match = Regex.Match(text, regexPattern, RegexOptions.Singleline);
if (!match.Success)
{
wildcardValues = null;
return false;
}
//skip the first one since it is the whole text
wildcardValues = match.Groups.Cast<Group>().Skip(1).Select(i => i.Value).ToArray();
return true;
}
Sample usage
string[] wildcardValues;
if(TryParse("Hello *. * * to *", "Hello World. Happy holidays to all", out wildcardValues))
{
//it's a match
//wildcardValues contains the values of the wildcard which is
//['World','Happy','holidays','all'] in this sample
}
By the way, you don't really need Regex for this, it's overkill. Just implement your own algorithm by splitting the pattern into tokens using string.Split then finding each token using string.IndexOf. Although using Regex does result in shorter code

Do you think this should work for you?
Match match = Regex.Match(pattern_string, #"<pattern>a [^<]+ is a [^<]+</pattern>");
if (match.Success)
{
// do something...
}
Here [^<]+ represents for one or more characters which is/are not <
If you think you may have < character in your *, then you can simply use .+ instead of [^<]+
But this will be risky as .+ means any characters having one or multiple times.

Why isn't this C# regular expression working?

I tried to write an expression to validate the following pattern:
digit[0-9] at 1 time exactly
"dot"
digit[0-9] 1-2 times
"dot"
digit[0-9] 1-3 times
"dot"
digit[0-9] 1-3 times or “hyphen”
For example these are legal numbers:
1.10.23.5
1.10.23.-
these aren't:
10.10.23.5
1.254.25.3
I used RegexBuddy to write the next pattern:
[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.[0-9]{1,3}|[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.-
In RegexBuddy all seems perfect but in my code I am getting true about illegal numbers (like 10.1.1.1)
I wrote the next method for validating this pattern:
public static bool IsVaildEc(string ec)
{
try
{
if (String.IsNullOrEmpty(ec))
return false;
string pattern = #"[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.[0-9]{1,3}|[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.-";
Regex check = new Regex(pattern);
return check.IsMatch(ec);
}
catch (Exception ex)
{
//logger
}
}
What am I doing wrong?

You regex isn't anchored to the start and end of the string, therefore it also matches a substring (e. g. 0.1.1.1 in the string 10.1.1.1).
As you can see, RegexBuddy matches a substring in the first "illegal" number. It correctly fails to match the second number because the three digits in the second octet can't be matched at all:
string pattern = #"^(?:[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.[0-9]{1,3}|[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.-)$";
will fix that problem.
Then, your regex is needlessly complicated. The following does the same but simpler:
string pattern = #"^[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.(?:[0-9]{1,3}|-)$";

try:
#"^[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.[0-9]{1,3}|[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.-"
you are not starting from the beggining of the text

If you match against the "10.1.1.1" the "0.1.1.1" part of your string would be a correct number and therefor return true.
Matching against
#"^[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.[0-9]{1,3}|[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.-"
with the ^ sign at the beginning means that you want to match from the beginning.

You are missing the ^ char in the start of the regex.
Try this regex:
^[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.[0-9]{1,3}|[0-9]\.[0-9]{1,2}\.[0-9]{1,3}\.-
This C# Regex Cheat Sheet can be handy

Regular Expression start-end position

I need to make a validation in a string with positional information using regex.
Example:
020005254877841100557810AAAAAA841158891BBBB
I need to get match position 5 until 10 and has to be only numbers.
How can I do this using RegEx?

hmm does it really have to be regex?
I would have done like this.
var myString = "020005254877841100557810AAAAAA841158891BBBB";
var isValid = myString.Substring(4, 5).All(Char.IsDigit);

If you absolutely HAVE to use regex, here it is.... (Though I think you should go with something like Jonas W's answer).
Match m = Regex.Match(myString, "^.{4}\d{5}.*");
if(m.Success){
//do stuff
}
The regex means, "from the beginning of the string (^), match 4 of any character (.{4}), then five digits, (\d{5}), then however many of any other characters (.*)"

If you really feel you must use a regex this will be the one:
"^....\d{5}"
You can also do multiple checks like this:
"^....(\d{5}|\D{5})
That one will match all numbers or all non-digit characters, but not a mix or anything with whitespace.

To build on what FailedDev mentioned.
string myString = "020005254877841100557810AAAAAA841158891BBBB";
myString.Substring(5, 5); //will get the substring at index 5 to 10.
double Num;
bool isNum = double.TryParse(myString, out Num); //returns true of all numbers
Hope that helps,

Trouble creating a Regex expression

I'm trying to create a regex expression what will accept a certain format of command. The pattern is as follows:
Can start with a $ and have two following value 0-9,A-F,a-f (ie: $00 - $FF)
or
Can be any value except for "&<>'/"
*if the value start with $ the next two values after need to be a valid hex value from 00-ff
So far I have this
Regex correctValue = new Regex("($[0-9a-fA-F][0-9a-fA-F])");
Any help will be greatly appreciated!

You just need to add "\" symbol before your "$" and it works:
string input = "$00";
Match m = Regex.Match(input, #"^\$[0-9a-fA-F][0-9a-fA-F]$");
if (m.Success)
{
foreach (Group g in m.Groups)
Console.WriteLine(g.Value);
}
else
Console.WriteLine("Didn't match");

If I'm following you correctly, the net result you're looking for is any value that is not in the list "&<>'/", since any combination of $ and two alphanumeric characters would also not be in that list. Thus you could make your expression:
Regex correctValue = new Regex("[^&<>'/]");
Update: But just in case you do need to know how to properly match the $00 - $FF, this would do the trick:
Regex correctValue = new Regex("\$[0-9A-Fa-f]{2}");

In Regular Expression $ use for Anchor assertion, and means:
The match must occur at the end of the string or before \n at the end of the line or string.
try using [$] (Character Class for single character) or \$ (Character Escape) instead.