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Regular expression that works on dots

I have this regular expression :
string[] values = Regex
.Matches(mystring4, #"([\w-[\d]][\w\s-[\d]]+)|([0-9]+)")
.OfType<Match>()
.Select(match => match.Value.Trim())
.ToArray();
This regular expression turns this string :
MY LIMITED COMPANY (52100000 / 58447000)";
To these strings :
MY LIMITED COMPANY - 52100000 - 58447000
This also works on non-English characters.
But there is one problem, when I have this string : MY. LIMITED. COMPANY. , it splits that too. I don't want that. I don't want that regular expression to work on dots. How can I do that? Thanks.

You may add the dot after each \w in your pattern, and I also suggest removing unnecessary ( and ):
string[] values = Regex
.Matches("MY. LIMITED. COMPANY. (52100000 / 58447000)", #"[\w.-[\d]][\w.\s-[\d]]+|[0-9]+")
.OfType<Match>()
.Select(match => match.Value.Trim())
.ToArray();
foreach (var s in values)
Console.WriteLine(s);
See the C# demo
Pattern:
[\w.-[\d]] - one Unicode letter or underscore ([\w-[\d]]) or a dot (.)
[\w.\s-[\d]]+ - 1 or more (due to + quantifier at the end) characters that are either Unicode letters or underscore, ., or whitespace (\s)
| - or
[0-9]+ - one or more ASCII-only digits

I'd simplify the expression. What if the names in the front include numbers? Not that my solution doesn't exactly mimic the original expression. It will allow numbers in the name part.
Let's start from the beginning:
To match words all you need is a sequence of word characters:
\w+
This will match any alphanumerical characters including underscores (_).
Considering you want the possibility of the word ending with a dot, you can add it and make it optional (one or zero matches):
\w+\.?
Note the escape to make it an actual character rather than a character class "any character".
To match another potential word following, we now simply duplicate this match, add a white space before, and once again make it optional using the * quantifier:
\w+\.?(?:\w+\.?)*
In case you haven't seen a group starting with ?: is a non-matching group. In essence this works like a usual group, but won't save a matching group in your results.
And that's it already. This pattern will split your demo string as expected. Of course there could be other possible characters not being covered by this.
You can see the results of this matching online here and also play around with it.
To test your regular expressions (and to learn them), I'd really recommend you using a tool such as http://regex101.com
It has an input mask allowing you to provide your pattern and your target string. On the right hand side it will first explain the pattern to you (to see if it's indeed what you had in mind) and below it will show all the groups matched. Just keep in mind it actually uses slightly different flavors of regular expressions, but this shouldn't matter for such simple patterns. (I'm not affiliated with that site, just consider it really useful.)
As an alternative, to directly use C#'s regex parser, you can also try this Regex Tester. This works in a similar way, although doesn't include any explanations, which might be not as ideal for someone just getting started.

Regex lookbeaind only when contains colon

Today I use c# Regex.IsMatch function to matching key:value format.
I have some code that checking if string format is: key:value (like: H:15).
The Regex pattern that I am using today is: [D,H,M,S]:[1-9]+\d?
I what to add the option for default key, when the input is 15, I would like to consider it like: H:15
So, I need to improve my Regex to support key:value or only value (without colon), H:15 is good and 15 is also good
I tried to use the or regex condition (|) something like : ([D,H,M,S]:[1-9]+\d?)|([1-9]+\d?)
But now it match more thinks like :1 and H:01 that are bad input for me.
I try to use also lookbehind regex without success
Any help would be greatly appreciated,
Nadav.

This should do the trick:
\b(?:[DHMS]:|(?<!:))[1-9][0-9]*\b
Demo
So, either match [DHMS]: or a word boundary not preceded by :.
Also, [1-9]+\d? looks very suspicious to me, so I replaced it with [1-9][0-9]*. Note that in .NET \d is not equivalent to [0-9] because it includes Unicode digits as well.

Looks like Avinash just beat me to it, but I added word boundaries with this expression, which works well in tests.
\b(?<=[DHMS]:)?[1-9]\d*\b

Seems like you wants something like this,
#"^(?:[DHMS]:)?[1-9]\d*$"
[DHMS] matches a single character from the given list. ? after the non-capturing group will turn the key part to an optional one. \d* matches zero or more digit characters.

Why is this regex not allowing this text?

I have a username validator IsValidUsername, and I am testing "baconman" but it is failing, could someone please help me out with this regex?
if(!Regex.IsMatch(str, #"^[a-zA-Z]\\w+|[0-9][0-9_]*[a-zA-Z]+\\w*$")) {
isValid = false;
}
I want the restrictions to be: (It's very close)
Be between 5 & 17 characters long
contain at least one letter
no spaces
no special characters

You're escaping unnecessarily: if you write your regex as starting with # outside the string, you don't need both \ - just one is fine.
Either:
#"\w"
or
"\\w"
Edit: I didn't make this clear: right now due to the double escaping, you're looking for a \ in your regex and a w. So your match would need [some character]\w to match (example: "a\w" or "a\wwwwww" would match.

Your requirements are best taken care of in normal C#. They don't map well to a regular expression. Just code them up using LINQ which works on strings like it would on an IEnumerable<char>.
Also, understanding a query of a string is much easier than understanding a Regex with the requirements that you have.

It is possible to do everything as part of a Regex, however it is not pretty :-)
^(\w(?=\w*[a-zA-Z])|[a-zA-Z]|\w(?<=[a-zA-Z]\w*)){5,17}$
It does 3 checks that always results in 1 character being matched (so we can perform the length check in the end)
Either the character is any word character \w which is before [a-zA-Z]
Or it is [a-zA-Z]
Or it is any word character \w which is after [a-zA-Z]

Simple Regex Question

I am new to regex (15 minutes of experience) so I can't figure this one out. I just want something that will match an alphanumeric string with no spaces in it. For example:
"ThisIsMyName" should match, but
"This Is My Name" should not match.

^[a-zA-Z0-9]+$ will match any letters and any numbers with no spaces (or any punctuation) in the string. It will also require at least one alphanumeric character. This uses a character class for the matching. Breakdown:
^ #Match the beginning of the string
[ #Start of a character class
a-z #The range of lowercase letters
A-Z #The range of uppercase letters
0-9 #The digits 0-9
] #End of the character class
+ #Repeat the previous one or more times
$ #End of string
Further, if you want to "capture" the match so that it can be referenced later, you can surround the regex in parens (a capture group), like so:
^([a-zA-Z0-9]+)$
Even further: since you tagged this with C#, MSDN has a little howto for using regular expressions in .NET. It can be found here. You can also note the fact that if you run the regex with the RegexOptions.IgnoreCase flag then you can simplify it to:
^([a-z0-9])+$

this will match any sequence of non-space characters:
\S+

Take a look at this link for a good basic Regex information source: http://regexlib.com/CheatSheet.aspx
They also have a handy testing tool that I use quite a bit: http://regexlib.com/RETester.aspx
That said, #eldarerathis' or #Nicolas Bottarini's answers should work for you.

I have just written a blog entry about regex, maybe it's something you may find useful:)
http://blogs.appframe.com/erikv/2010-09-23-Regular-Expression

Try using this regex to see if it works: (\w+)

Regex search and replace where the replacement is a mod of the search term

i'm having a hard time finding a solution to this and am pretty sure that regex supports it. i just can't recall the name of the concept in the world of regex.
i need to search and replace a string for a specific pattern but the patterns can be different and the replacement needs to "remember" what it's replacing.
For example, say i have an arbitrary string: 134kshflskj9809hkj
and i want to surround the numbers with parentheses,
so the result would be: (134)kshflskj(9809)hkj
Finding numbers is simple enough, but how to surround them?
Can anyone provide a sample or point me in the right direction?

In some various langauges:
// C#:
string result = Regex.Replace(input, #"(\d+)", "($1)");
// JavaScript:
thestring.replace(/(\d+)/g, '($1)');
// Perl:
s/(\d+)/($1)/g;
// PHP:
$result = preg_replace("/(\d+)/", '($1)', $input);
The parentheses around (\d+) make it a "group" specifically the first (and only in this case) group which can be backreferenced in the replacement string. The g flag is required in some implementations to make it match multiple times in a single string). The replacement string is fairly similar although some languages will use \1 instead of $1 and some will allow both.

Most regex replacement functions allow you to reference capture groups specified in the regex (a.k.a. backreferences), when defining your replacement string. For instance, using preg_replace() from PHP:
$var = "134kshflskj9809hkj";
$result = preg_replace('/(\d+)/', '(\1)', $var);
// $result now equals "(134)kshflskj(9809)hkj"
where \1 means "the first capture group in the regex".

Another somewhat generic solution is this:
search : /([\d]+)([^\d]*)/g
replace: ($1)$2
([\d]+): match a set of one or more digits and retain them in a group
([^\d]*): match a set of non-digits, and retain them as well. \D could work here, too.
g: indicate this is a global expression, to work multiple times on the input.
($1): in the replace block, parens have no special meaning, so output the first group, surrounding it with parens.
$2: output the second group
I used a pretty good online regex tool to test out my expression. The next step would be to apply it to the language that you are using, as each has its own implemention nuance.

Backreferences (grouping) are not necessary if you're just looking to search for numbers and replace with the found regex surrounded by parens. It is simpler to use the whole regex match in the replacement string.
e.g for perl
$text =~ s/\d+/($&)/g;
This searches for 1 or more digits and replaces with parens surrounding the match (specified by $&), with trailing g to find and replace all occurrences.
see http://www.regular-expressions.info/refreplace.html for the correct syntax for your regex language.

Depending on your language, you're looking to match groups.
So typically you'll make a pattern in the form of
([0-9]{1,})|([a-zA-Z]{1,})
Then, you'll iterate over the resulting groups in (specific to your language).