Develop Reference

How to match string by using regular expression which will not allow same special character at same time?

I m trying to matching a string which will not allow same special character at same time
my regular expression is:
[RegularExpression(#"^+[a-zA-Z0-9]+[a-zA-Z0-9.&' '-]+[a-zA-Z0-9]$")]
this solve my all requirement except the below two issues
this is my string : bracks
acceptable :
bra-cks, b-r-a-c-ks, b.r.a.c.ks, bra cks (by the way above regular expression solved this)
not acceptable:
issue 1: b.. or bra..cks, b..racks, bra...cks (two or more any special character together),
issue 2: bra cks (two ore more white space together)

You can use a negative lookahead to invalidate strings containing two consecutive special characters:
^(?!.*[.&' -]{2})[a-zA-Z0-9.&' -]+$
Demo: https://regex101.com/r/7j14bu/1

The goal
From what i can tell by your description and pattern, you are trying to match text, which start and end with alphanumeric (due to ^+[a-zA-Z0-9] and [a-zA-Z0-9]$ inyour original pattern), and inside, you just don't want to have any two consecuive (adjacent) special characters, which, again, guessing from the regex, are . & ' -
What was wrong
^+ - i think here you wanted to assure that match starts at the beginning of the line/string, so you don't need + here
[a-zA-Z0-9.&' '-] - in this character class you doubled ' which is totally unnecessary
Solution
Please try pattern
^[a-zA-Z0-9](?:(?![.& '-]{2,})[a-zA-Z0-9.& '-])*[a-zA-Z0-9]$
Pattern explanation
^ - anchor, match the beginning of the string
[a-zA-Z0-9] - character class, match one of the characters inside []
(?:...) - non capturing group
(?!...) - negative lookahead
[.& '-]{2,} - match 2 or more of characters inside character class
[a-zA-Z0-9.& '-] - character class, match one of the characters inside []
* - match zero or more text matching preceeding pattern
$ - anchor, match the end of the string
Regex demo

Some remarks on your current regex:
It looks like you placed the + quantifiers before the pattern you wanted to quantify, instead of after. For instance, ^+ doesn't make much sense, since ^ is just the start of the input, and most regex engines would not even allow that.
The pattern [a-zA-Z0-9.&' '-]+ doesn't distinguish between alphanumerical and other characters, while you want the rules for them to be different. Especially for the other characters you don't want them to repeat, so that + is not desired for those.
In a character class it doesn't make sense to repeat the same character, like you have a repeat of a quote ('). Maybe you wanted to somehow delimit the space, but realise that those quotes are interpreted literally. So probably you should just remove them. Or if you intended to allow for a quote, only list it once.
Here is a correction (add the quote if you still need it):
^[a-zA-Z0-9]+(?:[.& -][a-zA-Z0-9]+)*$
Follow-up
Based on a comment, I suspect you would allow a non-alphanumerical character to be surrounded by single spaces, even if that gives a sequence of more than one non-alphanumerical character. In that case use this:
^[a-zA-Z0-9]+(?:(?:[ ]|[ ]?[.&-][ ]?)[a-zA-Z0-9]+)*$
So here the space gets a different role: it can optionally occur before and after a delimiter (one of ".&-"), or it can occur on its own. The brackets around the spaces are not needed, but I used them to stress that the space is intended and not a typo.

Asp.Net Regex C# replace function not working [duplicate]

https://regex101.com/r/sB9wW6/1
(?:(?<=\s)|^)#(\S+) <-- the problem in positive lookbehind
Working like this on prod: (?:\s|^)#(\S+), but I need a correct start index (without space).
Here is in JS:
var regex = new RegExp(/(?:(?<=\s)|^)#(\S+)/g);
Error parsing regular expression: Invalid regular expression:
/(?:(?<=\s)|^)#(\S+)/
What am I doing wrong?
UPDATE
Ok, no lookbehind in JS :(
But anyways, I need a regex to get the proper start and end index of my match. Without leading space.

Make sure you always select the right regex engine at regex101.com. See an issue that occurred due to using a JS-only compatible regex with [^] construct in Python.
JS regex - at the time of answering this question - did not support lookbehinds. Now, it becomes more and more adopted after its introduction in ECMAScript 2018. You do not really need it here since you can use capturing groups:
var re = /(?:\s|^)#(\S+)/g;
var str = 's #vln1\n#vln2\n';
var res = [];
while ((m = re.exec(str)) !== null) {
res.push(m[1]);
}
console.log(res);
The (?:\s|^)#(\S+) matches a whitespace or the start of string with (?:\s|^), then matches #, and then matches and captures into Group 1 one or more non-whitespace chars with (\S+).
To get the start/end indices, use
var re = /(\s|^)#\S+/g;
var str = 's #vln1\n#vln2\n';
var pos = [];
while ((m = re.exec(str)) !== null) {
pos.push([m.index+m[1].length, m.index+m[0].length]);
}
console.log(pos);
BONUS
My regex works at regex101.com, but not in...
First of all, have you checked the Code Generator link in the Tools pane on the left?
All languages - "Literal string" vs. "String literal" alert - Make sure you test against the same text used in code, literal string, at the regex tester. A common scenario is copy/pasting a string literal value directly into the test string field, with all string escape sequences like \n (line feed char), \r (carriage return), \t (tab char). See Regex_search c++, for example. Mind that they must be replaced with their literal counterparts. So, if you have in Python text = "Text\n\n abc", you must use Text, two line breaks, abc in the regex tester text field. Text.*?abc will never match it although you might think it "works". Yes, . does not always match line break chars, see How do I match any character across multiple lines in a regular expression?
All languages - Backslash alert - Make sure you correctly use a backslash in your string literal, in most languages, in regular string literals, use double backslash, i.e. \d used at regex101.com must written as \\d. In raw string literals, use a single backslash, same as at regex101. Escaping word boundary is very important, since, in many languages (C#, Python, Java, JavaScript, Ruby, etc.), "\b" is used to define a BACKSPACE char, i.e. it is a valid string escape sequence. PHP does not support \b string escape sequence, so "/\b/" = '/\b/' there.
All languages - Default flags - Global and Multiline - Note that by default m and g flags are enabled at regex101.com. So, if you use ^ and $, they will match at the start and end of lines correspondingly. If you need the same behavior in your code check how multiline mode is implemented and either use a specific flag, or - if supported - use an inline (?m) embedded (inline) modifier. The g flag enables multiple occurrence matching, it is often implemented using specific functions/methods. Check your language reference to find the appropriate one.
line-breaks - Line endings at regex101.com are LF only, you can't test strings with CRLF endings, see regex101.com VS myserver - different results. Solutions can be different for each regex library: either use \R (PCRE, Java, Ruby) or some kind of \v (Boost, PCRE), \r?\n, (?:\r\n?|\n)/(?>\r\n?|\n) (good for .NET) or [\r\n]+ in other libraries (see answers for C#, PHP). Another issue related to the fact that you test your regex against a multiline string (not a list of standalone strings/lines) is that your patterns may consume the end of line, \n, char with negated character classes, see an issue like that. \D matched the end of line char, and in order to avoid it, [^\d\n] could be used, or other alternatives.
php - You are dealing with Unicode strings, or want shorthand character classes to match Unicode characters, too (e.g. \w+ to match Стрибижев or Stribiżew, or \s+ to match hard spaces), then you need to use u modifier, see preg_match() returns 0 although regex testers work - To match all occurrences, use preg_match_all, not preg_match with /...pattern.../g, see PHP preg_match to find multiple occurrences and "Unknown modifier 'g' in..." when using preg_match in PHP?- Your regex with inline backreference like \1 refuses to work? Are you using a double quoted string literal? Use a single-quoted one, see Backreference does not work in PHP
phplaravel - Mind you need the regex delimiters around the pattern, see https://stackoverflow.com/questions/22430529
python - Note that re.search, re.match, re.fullmatch, re.findall and re.finditer accept the regex as the first argument, and the input string as the second argument. Not re.findall("test 200 300", r"\d+"), but re.findall(r"\d+", "test 200 300"). If you test at regex101.com, please check the "Code Generator" page. - You used re.match that only searches for a match at the start of the string, use re.search: Regex works fine on Pythex, but not in Python - If the regex contains capturing group(s), re.findall returns a list of captures/capture tuples. Either use non-capturing groups, or re.finditer, or remove redundant capturing groups, see re.findall behaves weird - If you used ^ in the pattern to denote start of a line, not start of the whole string, or used $ to denote the end of a line and not a string, pass re.M or re.MULTILINE flag to re method, see Using ^ to match beginning of line in Python regex
- If you try to match some text across multiple lines, and use re.DOTALL or re.S, or [\s\S]* / [\s\S]*?, and still nothing works, check if you read the file line by line, say, with for line in file:. You must pass the whole file contents as the input to the regex method, see Getting Everything Between Two Characters Across New Lines. - Having trouble adding flags to regex and trying something like pattern = r"/abc/gi"? See How to add modifers to regex in python?
c#, .net - .NET regex does not support possessive quantifiers like ++, *+, ??, {1,10}?, see .NET regex matching digits between optional text with possessive quantifer is not working - When you match against a multiline string and use RegexOptions.Multiline option (or inline (?m) modifier) with an $ anchor in the pattern to match entire lines, and get no match in code, you need to add \r? before $, see .Net regex matching $ with the end of the string and not of line, even with multiline enabled - To get multiple matches, use Regex.Matches, not Regex.Match, see RegEx Match multiple times in string - Similar case as above: splitting a string into paragraphs, by a double line break sequence - C# / Regex Pattern works in online testing, but not at runtime - You should remove regex delimiters, i.e. #"/\d+/" must actually look like #"\d+", see Simple and tested online regex containing regex delimiters does not work in C# code - If you unnecessarily used Regex.Escape to escape all characters in a regular expression (like Regex.Escape(#"\d+\.\d+")) you need to remove Regex.Escape, see Regular Expression working in regex tester, but not in c#
dartflutter - Use raw string literal, RegExp(r"\d"), or double backslashes (RegExp("\\d")) - https://stackoverflow.com/questions/59085824
javascript - Double escape backslashes in a RegExp("\\d"): Why do regex constructors need to be double escaped?
- (Negative) lookbehinds unsupported by most browsers: Regex works on browser but not in Node.js - Strings are immutable, assign the .replace result to a var - The .replace() method does change the string in place - Retrieve all matches with str.match(/pat/g) - Regex101 and Js regex search showing different results or, with RegExp#exec, RegEx to extract all matches from string using RegExp.exec- Replace all pattern matches in string: Why does javascript replace only first instance when using replace?
javascriptangular - Double the backslashes if you define a regex with a string literal, or just use a regex literal notation, see https://stackoverflow.com/questions/56097782
java - Word boundary not working? Make sure you use double backslashes, "\\b", see Regex \b word boundary not works - Getting invalid escape sequence exception? Same thing, double backslashes - Java doesn't work with regex \s, says: invalid escape sequence - No match found is bugging you? Run Matcher.find() / Matcher.matches() - Why does my regex work on RegexPlanet and regex101 but not in my code? - .matches() requires a full string match, use .find(): Java Regex pattern that matches in any online tester but doesn't in Eclipse - Access groups using matcher.group(x): Regex not working in Java while working otherwise - Inside a character class, both [ and ] must be escaped - Using square brackets inside character class in Java regex - You should not run matcher.matches() and matcher.find() consecutively, use only if (matcher.matches()) {...} to check if the pattern matches the whole string and then act accordingly, or use if (matcher.find()) to check if there is a single match or while (matcher.find()) to find multiple matches (or Matcher#results()). See Why does my regex work on RegexPlanet and regex101 but not in my code?
scala - Your regex attempts to match several lines, but you read the file line by line (e.g. use for (line <- fSource.getLines))? Read it into a single variable (see matching new line in Scala regex, when reading from file)
kotlin - You have Regex("/^\\d+$/")? Remove the outer slashes, they are regex delimiter chars that are not part of a pattern. See Find one or more word in string using Regex in Kotlin - You expect a partial string match, but .matchEntire requires a full string match? Use .find, see Regex doesn't match in Kotlin
mongodb - Do not enclose /.../ with single/double quotation marks, see mongodb regex doesn't work
c++ - regex_match requires a full string match, use regex_search to find a partial match - Regex not working as expected with C++ regex_match - regex_search finds the first match only. Use sregex_token_iterator or sregex_iterator to get all matches: see What does std::match_results::size return? - When you read a user-defined string using std::string input; std::cin >> input;, note that cin will only get to the first whitespace, to read the whole line properly, use std::getline(std::cin, input); - C++ Regex to match '+' quantifier - "\d" does not work, you need to use "\\d" or R"(\d)" (a raw string literal) - This regex doesn't work in c++ - Make sure the regex is tested against a literal text, not a string literal, see Regex_search c++
go - Double backslashes or use a raw string literal: Regular expression doesn't work in Go - Go regex does not support lookarounds, select the right option (Go) at regex101.com before testing! Regex expression negated set not working golang
groovy - Return all matches: Regex that works on regex101 does not work in Groovy
r - Double escape backslashes in the string literal: "'\w' is an unrecognized escape" in grep - Use perl=TRUE to PCRE engine ((g)sub/(g)regexpr): Why is this regex using lookbehinds invalid in R?
oracle - Greediness of all quantifiers is set by the first quantifier in the regex, see Regex101 vs Oracle Regex (then, you need to make all the quantifiers as greedy as the first one)] - \b does not work? Oracle regex does not support word boundaries at all, use workarounds as shown in Regex matching works on regex tester but not in oracle
firebase - Double escape backslashes, make sure ^ only appears at the start of the pattern and $ is located only at the end (if any), and note you cannot use more than 9 inline backreferences: Firebase Rules Regex Birthday
firebasegoogle-cloud-firestore - In Firestore security rules, the regular expression needs to be passed as a string, which also means it shouldn't be wrapped in / symbols, i.e. use allow create: if docId.matches("^\\d+$").... See https://stackoverflow.com/questions/63243300
google-data-studio - /pattern/g in REGEXP_REPLACE must contain no / regex delimiters and flags (like g) - see How to use Regex to replace square brackets from date field in Google Data Studio?
google-sheets - If you think REGEXEXTRACT does not return full matches, truncates the results, you should check if you have redundant capturing groups in your regex and remove them, or convert the capturing groups to non-capturing by add ?: after the opening (, see Extract url domain root in Google Sheet
sed - Why does my regular expression work in X but not in Y?
word-boundarypcrephp - [[:<:]] and [[:>:]] do not work in the regex tester, although they are valid constructs in PCRE, see https://stackoverflow.com/questions/48670105
snowflake-cloud-data-platform snowflake-sql - If you are writing a stored procedure, and \\d does not work, you need to double them again and use \\\\d, see REGEX conversion of VARCHAR value to DATE in Snowflake stored procedure using RLIKE not consistent.

The value of regex match groups remain empty [duplicate]

My regex pattern looks something like
<xxxx location="file path/level1/level2" xxxx some="xxx">
I am only interested in the part in quotes assigned to location. Shouldn't it be as easy as below without the greedy switch?
/.*location="(.*)".*/
Does not seem to work.

You need to make your regular expression lazy/non-greedy, because by default, "(.*)" will match all of "file path/level1/level2" xxx some="xxx".
Instead you can make your dot-star non-greedy, which will make it match as few characters as possible:
/location="(.*?)"/
Adding a ? on a quantifier (?, * or +) makes it non-greedy.
Note: this is only available in regex engines which implement the Perl 5 extensions (Java, Ruby, Python, etc) but not in "traditional" regex engines (including Awk, sed, grep without -P, etc.).

location="(.*)" will match from the " after location= until the " after some="xxx unless you make it non-greedy.
So you either need .*? (i.e. make it non-greedy by adding ?) or better replace .* with [^"]*.
[^"] Matches any character except for a " <quotation-mark>
More generic: [^abc] - Matches any character except for an a, b or c

How about
.*location="([^"]*)".*
This avoids the unlimited search with .* and will match exactly to the first quote.

Use non-greedy matching, if your engine supports it. Add the ? inside the capture.
/location="(.*?)"/

Use of Lazy quantifiers ? with no global flag is the answer.
Eg,
If you had global flag /g then, it would have matched all the lowest length matches as below.

Here's another way.
Here's the one you want. This is lazy [\s\S]*?
The first item:
[\s\S]*?(?:location="[^"]*")[\s\S]* Replace with: $1
Explaination: https://regex101.com/r/ZcqcUm/2
For completeness, this gets the last one. This is greedy [\s\S]*
The last item:[\s\S]*(?:location="([^"]*)")[\s\S]*
Replace with: $1
Explaination: https://regex101.com/r/LXSPDp/3
There's only 1 difference between these two regular expressions and that is the ?

The other answers here fail to spell out a full solution for regex versions which don't support non-greedy matching. The greedy quantifiers (.*?, .+? etc) are a Perl 5 extension which isn't supported in traditional regular expressions.
If your stopping condition is a single character, the solution is easy; instead of
a(.*?)b
you can match
a[^ab]*b
i.e specify a character class which excludes the starting and ending delimiiters.
In the more general case, you can painstakingly construct an expression like
start(|[^e]|e(|[^n]|n(|[^d])))end
to capture a match between start and the first occurrence of end. Notice how the subexpression with nested parentheses spells out a number of alternatives which between them allow e only if it isn't followed by nd and so forth, and also take care to cover the empty string as one alternative which doesn't match whatever is disallowed at that particular point.
Of course, the correct approach in most cases is to use a proper parser for the format you are trying to parse, but sometimes, maybe one isn't available, or maybe the specialized tool you are using is insisting on a regular expression and nothing else.

Because you are using quantified subpattern and as descried in Perl Doc,
By default, a quantified subpattern is "greedy", that is, it will
match as many times as possible (given a particular starting location)
while still allowing the rest of the pattern to match. If you want it
to match the minimum number of times possible, follow the quantifier
with a "?" . Note that the meanings don't change, just the
"greediness":
*? //Match 0 or more times, not greedily (minimum matches)
+? //Match 1 or more times, not greedily
Thus, to allow your quantified pattern to make minimum match, follow it by ? :
/location="(.*?)"/

import regex
text = 'ask her to call Mary back when she comes back'
p = r'(?i)(?s)call(.*?)back'
for match in regex.finditer(p, str(text)):
print (match.group(1))
Output:
Mary

Why doesn't finite repetition in lookbehind work in some flavors?

I want to parse the 2 digits in the middle from a date in dd/mm/yy format but also allowing single digits for day and month.
This is what I came up with:
(?<=^[\d]{1,2}\/)[\d]{1,2}
I want a 1 or 2 digit number [\d]{1,2} with a 1 or 2 digit number and slash ^[\d]{1,2}\/ before it.
This doesn't work on many combinations, I have tested 10/10/10, 11/12/13, etc...
But to my surprise (?<=^\d\d\/)[\d]{1,2} worked.
But the [\d]{1,2} should also match if \d\d did, or am I wrong?

On lookbehind support
Major regex flavors have varying supports for lookbehind differently; some imposes certain restrictions, and some doesn't even support it at all.
Javascript: not supported
Python: fixed length only
Java: finite length only
.NET: no restriction
References
regular-expressions.info/Flavor comparison
Python
In Python, where only fixed length lookbehind is supported, your original pattern raises an error because \d{1,2} obviously does not have a fixed length. You can "fix" this by alternating on two different fixed-length lookbehinds, e.g. something like this:
(?<=^\d\/)\d{1,2}|(?<=^\d\d\/)\d{1,2}
Or perhaps you can put both lookbehinds as alternates of a non-capturing group:
(?:(?<=^\d\/)|(?<=^\d\d\/))\d{1,2}
(note that you can just use \d without the brackets).
That said, it's probably much simpler to use a capturing group instead:
^\d{1,2}\/(\d{1,2})
Note that findall returns what group 1 captures if you only have one group. Capturing group is more widely supported than lookbehind, and often leads to a more readable pattern (such as in this case).
This snippet illustrates all of the above points:
p = re.compile(r'(?:(?<=^\d\/)|(?<=^\d\d\/))\d{1,2}')
print(p.findall("12/34/56")) # "[34]"
print(p.findall("1/23/45")) # "[23]"
p = re.compile(r'^\d{1,2}\/(\d{1,2})')
print(p.findall("12/34/56")) # "[34]"
print(p.findall("1/23/45")) # "[23]"
p = re.compile(r'(?<=^\d{1,2}\/)\d{1,2}')
# raise error("look-behind requires fixed-width pattern")
References
regular-expressions.info/Lookarounds, Character classes, Alternation, Capturing groups
Java
Java supports only finite-length lookbehind, so you can use \d{1,2} like in the original pattern. This is demonstrated by the following snippet:
String text =
"12/34/56 date\n" +
"1/23/45 another date\n";
Pattern p = Pattern.compile("(?m)(?<=^\\d{1,2}/)\\d{1,2}");
Matcher m = p.matcher(text);
while (m.find()) {
System.out.println(m.group());
} // "34", "23"
Note that (?m) is the embedded Pattern.MULTILINE so that ^ matches the start of every line. Note also that since \ is an escape character for string literals, you must write "\\" to get one backslash in Java.
C-Sharp
C# supports full regex on lookbehind. The following snippet shows how you can use + repetition on a lookbehind:
var text = #"
1/23/45
12/34/56
123/45/67
1234/56/78
";
Regex r = new Regex(#"(?m)(?<=^\d+/)\d{1,2}");
foreach (Match m in r.Matches(text)) {
Console.WriteLine(m);
} // "23", "34", "45", "56"
Note that unlike Java, in C# you can use #-quoted string so that you don't have to escape \.
For completeness, here's how you'd use the capturing group option in C#:
Regex r = new Regex(#"(?m)^\d+/(\d{1,2})");
foreach (Match m in r.Matches(text)) {
Console.WriteLine("Matched [" + m + "]; month = " + m.Groups[1]);
}
Given the previous text, this prints:
Matched [1/23]; month = 23
Matched [12/34]; month = 34
Matched [123/45]; month = 45
Matched [1234/56]; month = 56
Related questions
How can I match on, but exclude a regex pattern?

Unless there's a specific reason for using the lookbehind which isn't noted in the question, how about simply matching the whole thing and only capturing the bit you're interested in instead?
JavaScript example:
>>> /^\d{1,2}\/(\d{1,2})\/\d{1,2}$/.exec("12/12/12")[1]
"12"

To quote regular-expressions.info:
The bad news is that most regex
flavors do not allow you to use just
any regex inside a lookbehind, because
they cannot apply a regular expression
backwards. Therefore, the regular
expression engine needs to be able to
figure out how many steps to step back
before checking the lookbehind.
Therefore, many regex flavors,
including those used by Perl and
Python, only allow fixed-length
strings. You can use any regex of
which the length of the match can be
predetermined. This means you can use
literal text and character classes.
You cannot use repetition or optional
items. You can use alternation, but
only if all options in the alternation
have the same length.
In other words your regex does not work because you're using a variable-width expression inside a lookbehind and your regex engine does not support that.

In addition to those listed by #polygenelubricants, there are two more exceptions to the "fixed length only" rule. In PCRE (the regex engine for PHP, Apache, et al) and Oniguruma (Ruby 1.9, Textmate), a lookbehind may consist of an alternation in which each alternative may match a different number of characters, as long as the length of each alternative is fixed. For example:
(?<=\b\d\d/|\b\d/)\d{1,2}(?=/\d{2}\b)
Note that the alternation has to be at the top level of the lookbehind subexpression. You might, like me, be tempted to factor out the common elements, like this:
(?<=\b(?:\d\d/|\d)/)\d{1,2}(?=/\d{2}\b)
...but it wouldn't work; at the top level, the subexpression now consists of a single alternative with a non-fixed length.
The second exception is much more useful: \K, supported by Perl and PCRE. It effectively means "pretend the match really started here." Whatever appears before it in the regex is treated as a positive lookbehind. As with .NET lookbehinds, there are no restrictions; whatever can appear in a normal regex can be used before the \K.
\b\d{1,2}/\K\d{1,2}(?=/\d{2}\b)
But most of the time, when someone has a problem with lookbehinds, it turns out they shouldn't even be using them. As #insin pointed out, this problem can be solved much more easily by using a capturing group.
EDIT: Almost forgot JGSoft, the regex flavor used by EditPad Pro and PowerGrep; like .NET, it has completely unrestricted lookbehinds, positive and negative.

How do I include a hyphen in a hyperlink Regex?

I am trying to find links in user entered text and convert them to link automatically.
I am using current Regex as following, which good to find hyperlinks from text.
Regex regexResolveUrl = new Regex("((http://|www\\.)([A-Z0-9.-:]{1,})\\.[0-9A-Z?;~&#=\\-_\\./]{2,})", RegexOptions.Compiled | RegexOptions.IgnoreCase);
It is working good for almost all links so far i came across but it is giving problem when i want to detect links with hypen.
i.e. www.abc-xyz.com will not work, with above regex, can anyone help me with this?

If you want - to mean dash literally in a character class definition, you need to put it as the last (or first) character. So [abc-] is a character class containing 4 characters, a, b, c, -. On the other hand, [ab-c] only contains 3 characters, not including the -, because - is a range definition.
So, something like this (from your pattern):
[A-Z0-9.-:]
Defines 3 ranges, from A to Z, from 0 to 9, and from . (ASCII 46) to : (ASCII 58). You want instead:
[A-Z0-9.:-]
References
regular-expressions.info/Character Class
Note on repetition
I noticed that you used {1,} in your pattern to denote "one-or-more of".
.NET regex (like most other flavors) support these shorthands:
?: "zero-or-one" {0,1}
*: "zero-or-more" {0,}
+: "one-or-more" {1,}
They may take some getting used to, but they're also pretty standard.
References
regular-expressions.info/Repetition with Star and Plus
Related questions
Using explicitly numbered repetition instead of question mark, star and plus
Note on C# #-quoted string literals
While doubling the slashes in string literals for regex pattern is the norm in e.g. Java (out of necessity), in C# you actually have an option to use #-quoted string literals.
That is, these pairs of strings are identical:
"(http://|www\\.)"
#"(http://|www\.)"
"c:\\Docs\\Source\\a.txt"
#"c:\Docs\Source\a.txt"
Using # can lead to more readable regex patterns because a literal slash don't have to be doubled (although on the other hand, a double quote must now in turn be doubled).
References
MSDN / C# Programmer's Reference / string

Escape the hyphen:
Regex("((http://|www\\.)([A-Z0-9.\-:]{1,})\\.[0-9A-Z?;~&#=\\-_\\./]{2,})", RegexOptions.Compiled | RegexOptions.IgnoreCase);

Add the hyphen as the first or last character in the character class.