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Regex only letters except set of numbers

I'm using Replace(#"[^a-zA-Z]+", "");
leave only letters, but I have a set of numbers or characters that I want to keep as well, ex: 122456 and 112466. But I'm having trouble leaving it only if it's this sequence:
ex input:
abc 1239 asm122456000
I want to:
abscasm122456
tried this: ([^a-zA-Z])+|(?!122456)

My answer doesn't applying Replace(), but achieves a similar result:
(?:[a-zA-Z]+|\d{6})
which captures the group (non-capturing group) with the alphabetic character(s) or a set of digits with 6 occurrences.
Regex 101 & Test Result
Join all the matching values into a single string.
using System.Linq;
Regex regex = new Regex("(?:[a-zA-Z]+|\\d{6})");
string input = "abc 1239 asm12245600";
string output = "";
var matches = regex.Matches(input);
if (matches.Count > 0)
output = String.Join("", matches.Select(x => x.Value));
Sample .NET Fiddle

Alternate way,
using .Split() and .All(),
string input = "abc 1239 asm122456000";
string output = string.Join("", input.Split().Where(x => !x.All(char.IsDigit)));
.NET Fiddle

It is very simple: you need to match and capture what you need to keep, and just match what you need to remove, and then utilize a backreference to the captured group value in the replacement pattern to put it back into the resulting string.
Here is the regex:
(122456|112466)|[^a-zA-Z]
See the regex demo. Details:
(122456|112466) - Capturing group with ID 1: either of the two alternatives
| - or
[^a-zA-Z] - a char other than an ASCII letter (use \P{L} if you need to match any char other than any Unicode letter).
Note the removed + quantifier as [^A-Za-z] also matches digits.
You need to use $1 in the replacement:
var result = Regex.Replace(text, #"(122456|112466)|[^a-zA-Z]", "$1");

Regex to get first 6 and last 4 characters of a string

I would like to use regex instead of string.replace() to get the first 6 chars of a string and the last 4 chars of the same string and substitute it with another character: & for example. The string is always with 16 chars. Im doing some research but i never worked with regex before. Thanks

If you prefer to use regular expression, you could use the following. The dot . will match any character except a newline sequence, so you can specify {n} to match exactly n times and use beginning/end of string anchors.
String r = Regex.Replace("123456foobar7890", #"^.{6}|.{4}$",
m => new string('&', m.ToString().Length));
Console.WriteLine(r); //=> "&&&&&&foobar&&&&"
If you want to invert the logic, replacing the middle portion of your string you can use Positive Lookbehind.
String r = Regex.Replace("123456foobar7890", #"(?<=^.{6}).{6}",
m => new string('&', m.ToString().Length));
Console.WriteLine(r); //=> "123456&&&&&&7890"

Regex problems with equal sign?

In C# I'm trying to validate a string that looks like:
I#paramname='test'
or
O#paramname=2827
Here is my code:
string t1 = "I#parameter='test'";
string r = #"^([Ii]|[Oo])#\w=\w";
var re = new Regex(r);
If I take the "=\w" off the end or variable r I get True. If I add an "=\w" after the \w it's False. I want the characters between # and = to be able to be any alphanumeric value. Anything after the = sign can have alphanumeric and ' (single quotes). What am I doing wrong here. I very rarely have used regular expressions and normally can find example, this is custom format though and even with cheatsheets I'm having issues.

^([Ii]|[Oo])#\w+=(?<q>'?)[\w\d]+\k<q>$
Regular expression:
^ start of line
([Ii]|[Oo]) either (I or i) or (O or o)
\w+ 1 or more word characters
= equals sign
(?<q>'?) capture 0 or 1 quotes in named group q
[\w\d]+ 1 or more word or digit characters
\k<q> repeat of what was captured in named group q
$ end of line

use \w+ instead of \w to one character or more. Or \w* to get zero or more:
Try this: Live demo
^([Ii]|[Oo])#\w+=\'*\w+\'*

If you are being a bit more strict with using paramname:
^([Ii]|[Oo])#paramname=[']?[\w]+[']?
Here is a demo

You could try something like this:
Regex rx = new Regex( #"^([IO])#(\w+)=(.*)$" , RegexOptions.IgnoreCase ) ;
Match group 1 will give you the value of I or O (the parameter direction?)
Match group 2 will give you the name of the parameter
Match group 3 will give you the value of the parameter
You could be stricter about the 3rd group and match it as
(([^']+)|('(('')|([^']+))*'))
The first alternative matches 1 or more non quoted character; the second alternative match a quoted string literal with any internal (embedded) quotes escape by doubling them, so it would match things like
'' (the empty string
'foo bar'
'That''s All, Folks!'

Find words in text by start and stop characters

I need to find and replace all words in text.
Format of these words :
start with (long), end with ;
example
(long)Row["Id"];
whats is the regexp pattern for this format ? I tried some but dont works for me.
Thanks.

\(long\)(.*?);
(.*?) generally tries to capture as many as necessary to find the ; at the end. And as for the (long) you will need to escape the parentheses.

Try the following:
var input = "(long)Row["Id"];";
var result = Regex.Replace(input, #"\(long\)([^;]+)", "$1.ToLong()");
The following expression: \(long\)([^;]+):
\(: Matches an open parentheses (.
long: Matches the word long literally.
\): Matches a closed parentheses ).
([^;]+): Matches one or more non-semicolon characters and puts them into capturing group 1.

As an alternative to regex, you can use String.StartsWith and String.EndsWith methods.
For example;
string[] lines = File.ReadAllLines(#"C:\Users\Public\TestFolder\Text.txt");
foreach(string word in lines)
{
if (word.StartsWith("(long)", StringComparison.InvariantCulture) && word.EndsWith(';', StringComparison.InvariantCulture))
{
//Replace your string here.
}
}

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.