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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.

Difficulty finding where to insert "word exclusion" in a regex

I know the regex for excluding words, roughly anyway, It would be (!?wordToIgnore|wordToIgnore2|wordToIgnore3)
But I have an existing, complicated regex that I need to add this to, and I am a bit confused about how to go about that. I'm still pretty new to regex, and it took me a very long time to make this particular one, but I'm not sure where to insert it or how ...
The regex I have is ...
^(?!.*[ ]{2})(?!.*[']{2})(?!.*[-]{2})(?:[a-zA-Z0-9 \:/\p{L}'-]{1,64}$)$
This should only allow the person typing to insert between 1 and 64 letters that match that pattern, cannot start with a space, quote, double quote, special character, a dash, an escape character, etc, and only allows a-z both upper and lowercase, can include a space, ":", a dash, and a quote anywhere but the beginning.
But I want to forbid them from using certain words, so I have this list of words that I want to be forbidden, I just cannot figure out how to get that to fit into here.. I tried just pasting the whole .. "block" in, and that didn't work.
?!the|and|or|a|given|some|that|this|then|than
Has anyone encountered this before?

ciel, first off, congratulations for getting this far trying to build your regex rule. If you want to read something detailed about all kinds of exclusions, I suggest you have a look at Match (or replace) a pattern except in situations s1, s2, s3 etc
Next, in your particular situation, here is how we could approach your regex.
For consision, let's make all the negative lookarounds more compact, replacing them with a single (?!.*(?: |-|'){2})
In your character class, the \: just escapes the colon, needlessly so as : is enough. I assume you wanted to add a backslash character, and if so we need to use \\
\p{L} includes [a-zA-Z], so you can drop [a-zA-Z]. But are you sure you want to match all letters in any script? (Thai etc). If so, remember to set the u flag after the regex string.
For your "bad word exclusion" applying to the whole string, place it at the same position as the other lookarounds, i.e., at the head of the string, but using the .* as in your other exclusions: (?!.*(?:wordToIgnore|wordToIgnore2|wordToIgnore3)) It does not matter which lookahead comes first because lookarounds do not change your position in the string. For more on this, see Mastering Lookahead and Lookbehind
This gives us this glorious regex (I added the case-insensitive flag):
^(?i)(?!.*(?:wordToIgnore|wordToIgnore2|wordToIgnore3))(?!.*(?: |-|'){2})(?:[\\0-9 :/\p{L}'-]{1,64}$)$
Of course if you don't want unicode letters, replace \p{L} with a-z
Also, if you want to make sure that the wordToIgnore is a real word, as opposed to an embedded string (for instance you don't want cat but you are okay with catalog), add boundaries to the lookahead rule: (?!.*\b(?:wordToIgnore|wordToIgnore2|wordToIgnore3)\b)

use this:
^(?!.*(the|and|or|a|given|some|that|this|then|than))(?!.*[ ]{2})(?!.*[']{2})(?!.*[-]{2})(?:[a-zA-Z0-9 \:\p{L}'-]{1,64}$)$
see demo

What does .* do in regex?

After extensive search, I am unable to find an explanation for the need to use .* in regex. For example, MSDN suggests a password regex of
#\"(?=.{6,})(?=(.*\d){1,})(?=(.*\W){1,})"
for length >= 6, 1+ digit and 1+ special character.
Why can't I just use:
#\"(?=.{6,})(?=(\d){1,})(?=(\W){1,})"

.* just means "0 or more of any character"
It's broken down into two parts:
. - a "dot" indicates any character
* - means "0 or more instances of the preceding regex token"
In your example above, this is important, since they want to force the password to contain a special character and a number, while still allowing all other characters. If you used \d instead of .*, for example, then that would restrict that portion of the regex to only match decimal characters (\d is shorthand for [0-9], meaning any decimal). Similarly, \W instead of .*\W would cause that portion to only match non-word characters.
A good reference containing many of these tokens for .NET can be found on the MSDN here: Regular Expression Language - Quick Reference
Also, if you're really looking to delve into regex, take a look at http://www.regular-expressions.info/. While it can sometimes be difficult to find what you're looking for on that site, it's one of the most complete and begginner-friendly regex references I've seen online.

Just FYI, that regex doesn't do what they say it does, and the way it's written is needlessly verbose and confusing. They say it's supposed to match more than seven characters, but it really matches as few as six. And while the other two lookaheads correctly match at least one each of the required character types, they can be written much more simply.
Finally, the string you copied isn't just a regex, it's an XML attribute value (including the enclosing quotes) that seems to represent a C# string literal (except the closing quote is missing). I've never used a Membership object, but I'm pretty sure that syntax is faulty. In any case, the actual regex is:
(?=.{6,})(?=(.*\d){1,})(?=(.*\W){1,})
..but it should be:
(?=.{8,})(?=.*\d)(?=.*\W)
The first lookahead tries to match eight or more of any characters. If it succeeds, the match position (or cursor, if you prefer) is reset to the beginning and the second lookahead scans for a digit. If it finds one, the cursor is reset again and the third lookahead scans for a special character. (Which, by the way, includes whitespace, control characters, and a boatload of other esoteric characters; probably not what the author intended.)
If you left the .* out of the latter two lookaheads, you would have (?=\d) asserting that the first character is a digit, and (?=\W) asserting that it's not a digit. (Digits are classed as word characters, and \W matches anything that's not a word character.) The .* in each lookahead causes it to initially gobble up the whole string, then backtrack, giving back one character at a time until it reaches a spot where the \d or \W can match. That's how they can match the digit and the special character anywhere in the string.

The .* portion just allows for literally any combination of characters to be entered. It's essentially allowing for the user to add any level of extra information to the password on top of the data you are requiring
Note: I don't think that MSDN page is actually suggesting that as a password validator. It is just providing an example of a possible one.

Regular expression to replace a string

I'm working on some code inherited from someone else and trying to understand some regular expression code in C#:
Regex.Replace(query, #"""[^""~]+""([^~]|$)",
m => string.Format(field + "_exact:{0}", m.Value))
What is the above regular expression doing? This is in relation to input from a user performing a search. It's doing a replace of the query string using the pattern provided in the second argument, with the value of the third. But what is that regular expression? For the life of me, it doesn't make sense. Thanks.

As far as I can see, xanatos' answer is correct. I tried to understand the regex, so here it comes:
"[^"~]+"([^~]|$)
You can test our regex and play with the single parts for better understanding at http://www.regexpal.com/
1.) a single character
"
The first pattern is a literal character. Since there is no statement of relative position, it can occur everywhere.
2.) a character class
[^"~]
The next expression is the []-bracket. This is a character set. It defines a quantity of characters, which maybe follow next. It is a placeholder for one single character... So lets see inside, which content is allowed:
^"~
The definition of the character class begins with an caret (^), which is a special character. Typing a caret after the opening square bracket will negate the character class. So it's "upside down": everything following, which does not match the class expression, matches and is a valid character.
In this case, every literal character is possible, except the two excluded ones: " or ~.
3.) a special character
+
The next expression, a plus, tells the engine to attempt to match the preceding token once or more.
So the defined character class should one or multiple times repeated to match the given expression.
4.) a single character
"
To match, the expression should contain furthermore one further apostrophe, which will be the corresponding apostrophe to the first one in 1.) since the character class in (2.) hence (3.) does not permit an apostrophe.
5.) a lookaround
([^~]|$)
The first structure here to examine is the ()-bracket. This is called a "Lookaround".
It is is a special kind of group. Lookaround matches a position. It does not expand the regex match.
So this means this part does not try to find any certain characters inside of an expression
rather then to localize them.
The localisation demands has two conditions, which are connected by a logical OR by the pipeline symbol: |
So the next character of the matched expression could either be
[^~] one single character out of the class everything excluding the character ~
or
$ the end of the line (or word, if multiline-mode is not used in regex engine)
I'll try to edit my answer to a better format, since this is my first post, I first have to check out how this is working.. :)
Update:
to "detect" a Asterisk/star in front/end of the line, you have to do following:
First it's a special character, so you have to escape it with an backslash: *
To define the position, you can use:
^ to look at the beginning of the line,
$ end of the line
The overall expression would be:
^* in front of the expression to search for an * at the beginning of
the line $* at the end of the regex to demand an * at the end.
.... in your case you can add the * in the last character class to detect an * in the end:
([^~]|$|$*)
and to force an * in the end, delete the other conditions:
($*)
PS:
(somehow my regex is swallowed up by formating engine, so my update is wrong...)

The # makes it necessary to escape all the " with a second ", so "". Without it to escape the " you would have used \", but I consider it better to always use # in regexes, because the \ is used quite often, and it's boring and unreadable to always have to escape it to \\.
Let's see what the regex really is:
Console.WriteLine(#"""[^""~]+""([^~]|$)");
is
"[^"~]+"([^~]|$)
So now we can look at the "real" regex.
It looks for a " followed by one or more non-" and non-~ followed by another " followed by a non-~ or the end of the string. Note that the match could start after the start of the string and it could end before the end of the string (with a non-~)
For example in
car"hello"help
it would match "hello"h

Regex expression not filtering special symbols

I'm currently using the following line of code:
Regex Regex_Alpha = new Regex(#"[a-zA-Z]+('[a-zA-Z])?[a-zA-Z]*");
What I want to do is filter the input of text fields with the condition that input should only be letters and the apostrophe symbol (actually, I still want to add more, but I'm trying to resolve this first).
Right now, it is accepting ALL characters, even numbers.
With my understanding of Regex, I tried to formulate my own expression in the line of:
Regex Regex_Alpha = new Regex(#"^[a-zA-Z'-"+$);
It filters numbers, but doesn't accept the apostrophe symbol. Tried to remove the # sign and filter the apostrophe with the backslash escape character, but still no use.
What should be the best approach to filter the input so that it only accepts letters and apostrophe? (I'll do the rest of the symbols once I understand how this one should work)

As I've commented, your first regular expression is a pretty good shot at "letters, with a single apostrophe not at either end". However, it matchs any string with even a single letter because a regular expression looks for any match in the input, not for whether the entire input matches.
You can fix this by doing what you've done in your second regular expression - just put a ^ at the start and a $ at the end. This means the start and end of the expression have to match the start and end of the input, so it ensures the whole input is only made up of letters and a possible apostrophe.
Regarding your second regular expression, you have a few of problems.
If you want a double-quote in a #"..." string literal, you need to put two double quotes. (I think this might just be a typing mistake in your question, as what you currently have wouldn't even compile.)
You need to close your character class with a ], otherwise the [ and everything inside just get treated as a sequence of characters to match, one after the other.
If you want a hyphen in a character class, it has to go at the start or end, or it gets mistaken for a "between" hyphen (as in A-Z).
The expression #"^[a-zA-Z'""-]+$" should match "any string entirely made of letters, apostrophes, quotes or hyphens".