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Reg Expression to ignore limited special characters javascript

Using Jquery validator plugin in my implementation. Need a regular expression which excludes special characters like , and &.
is there any regular expression for this. also if this special characters are anywhere in the string it should find and throw the error.

You can use regular expressions like this:
[\,\&]
you can add as much as u want to this.
try it out yourself on this site:
http://www.regexr.com/

/[,&]/g
matches , and &.
Demo: https://regex101.com/r/gY0mC3/2#javascript
If you want to search for every special character except letters, numbers and the underscore, use
/\W/g
Demo: https://regex101.com/r/gY0mC3/5#javascript
If you need to include spaces (e.g. a name) use
/[^\w\s]/g
Demo: https://regex101.com/r/gY0mC3/4#javascript

The brackets [] define custom regex classes.
To match a character for only those characters, you can do [\,\&].
To match all except that, you can add a ^, such as [^\,\&].
To match any non-word character, you can use \W (any character not a-z, A-Z, 0-9, or _).
To include an underscore, you can do [\W_].
Keep in mind that whitespaces are represented by \s and that depending on your environment, you may need to escape (add an additional backslash to) your backslashes.

regex to match alphanumeric and hyphen only, strip everything else in javascript

I want to strip everything except alphanumeric and hyphens.
so far i've got this but its not working:
String = String.replace(/^[a-zA-Z0-9-_]+$/ig,'');
any help appreciated?

If you want to remove everything except alphanum, hypen and underscore, then negate the character class, like this
String = String.replace(/[^a-zA-Z0-9-_]+/ig,'');
Also, ^ and $ anchors should not be there.
Apart from that, you have already covered both uppercase and lowercase characters in the character class itself, so i flag is not needed. So, RegEx becomes
String = String.replace(/[^a-zA-Z0-9-_]+/g,'');
There is a special character class, which matches a-zA-Z0-9_, \w. You can make use of it like this
String = String.replace(/[^\w-]+/g,'');
Since \w doesn't cover -, we included that separately.
Quoting from MDN RegExp documentation,
\w
Matches any alphanumeric character from the basic Latin alphabet, including the underscore. Equivalent to [A-Za-z0-9_].
For example, /\w/ matches 'a' in "apple," '5' in "$5.28," and '3' in "3D."

Javascript regex - no white space at beginning + allow space in the middle

I would like to have a regex which matches the string with NO whitespace(s) at the beginning. But the string containing whitespace(s) in the middle CAN match. So far i have tried below
[^-\s][a-zA-Z0-9-_\\s]+$
Debuggex Demo
Above is not allowing whitespace(s) at the beginning, but also not allowing in the middle/end of the string. Please help me.

In your 2nd character class, \\s will match \ and s, and not \s. Thus it doesn't matches a whitespace. You should use just \s there. Also, move the hyphen towards the end, else it will create unintentional range in character class:
^[^-\s][a-zA-Z0-9_\s-]+$

If you plan to match a string of any length (even an empty string) that matches your pattern and does not start with a whitespace, use (?!\s) right after ^:
/^(?!\s)[a-zA-Z0-9_\s-]*$/
^^^^^^
Or, bearing in mind that [A-Za-z0-9_] in JS regex is equal to \w:
/^(?!\s)[\w\s-]*$/
The (?!\s) is a negative lookahead that matches a location in string that is not immediately followed with a whitespace (matched with the \s pattern).
If you want to add more "forbidden" chars at the string start (it looks like you also disallow -) keep using the [\s-] character class in the lookahead:
/^(?![\s-])[\w\s-]*$/
To match at least 1 character, replace * with +:
/^(?![\s-])[\w\s-]+$/
See the regex demo. JS demo:
console.log(/^(?![\s-])[\w\s-]+$/.test("abc def 123 ___ -loc- "));
console.log(/^(?![\s-])[\w\s-]+$/.test(" abc def 123 ___ -loc- "));

You need to use this regex:
^[^-\s][\w\s-]+$
Use start anchor ^
No need to double escape \s
Also important is to use hyphen as the first OR last character in the character class.
\w is same as [a-zA-Z0-9_]

use \S at the beginning
^\S+[a-zA-Z0-9-_\\s]+$

This RegEx will allow neither white-space at the beginning nor at the end of. Your string/word and allows all the special characters.
^[^\s].+[^\s]$
This Regex also works Fine
^[^\s]+(\s+[^\s]+)*$

try this should work
[a-zA-Z0-9_]+.*$

/^[^.\s]/
try this instead it will not allow a user to enter character at first place
^ matches position just before the first character of the string
. matches a single character. Does not matter what character it is, except newline
\s is space

If your field for user name only accept letters and middle of space but not for begining and end
User name: /^[^\s][a-zA-Z\s]+[^\s]$/
If your field for user ID only accept letters,numbers and underscore and no spaces allow
user ID: /^[\w]+$/
If your field for password only accept letters,number and special character no spaces allow
Password: /^[\w##&]+$/
Note: \w content a-zA-Z, number, underscore (_) if you add more character, add you special character after \w.
You can compare with user ID and password field in password field im only add some special character (##&).
India public thoko like 😁

I suggest below regex for this,
^[^\s].*[^\s]$
You can try regex in here

Accented characters and regular expression

I have this regexp:
(\b)(emozioni|gioia|felicità)(\b)
In a string like the one below:
emozioni emozioniamo felicità felicitàs
it should match the first and the third word. Instead it matches the first and the last. I assume it is because of the accented character. I tried this alternative:
(\b)(emozioni|gioia|felicità\s)(\b)
but it matched "felicità" only if there is an other word after it. So for being specific only if it is in this context:
emozioni emozioniamo felicità felicitàs
and not in this other:
emozioni emozioniamo felicitàs felicità
I've found an article about accented characters in French (so at the beginning of the word) here, i followed the second answer. If anyone knows a better solution it is very welcome.

A word boundary \b works only with characters that are in \w character class, i.e [0-9a-zA-Z_], thus you can't put a \b after an accentued character like à.
You can solve the problem in your case using a lookahead:
felicità(?=\s|$)
or shorter:
felicità(?!\S)
(or \W in place of \s as suggested #Sniffer, but you take the risk to match something like :felicitàà)

Try the following alternative:
\b(emozioni|gioia|felicità)(?=\W|$)
This will match any of your listed words, as long as any of those words is followed by either a non-word character \W or end-of-string $.
Regex101 Demo

Regex with special chars

Hi I have this regex.
/^[\w]|[åäöæøÅÄÖÆØ]$/
"tå" is ok but "åå" is not. Why is that? How can I make it accept words starting with åäöæøÅÄÖÆØ?

Note that the \w (and \W, \b, and \B) are English-centric. \w just means [A-Za-z0-9_], where A-Z means only the 26 English letters. Other letters are not considered part of a "word" by JavaScript's built-in character classes.
You'll need to build a character class including all of the letters you want to treat as word characters (then use the negated version of that wherever you "non-word character").
But that's not the only problem. Your regular expression says:
Match one English word character at the beginning of the string, or match one of this list of characters at the end of the string.
The | operator is fairly greedy, in this case it treats ^[\w] and [åäöæøÅÄÖÆØ]$ as the alternatives. I don't get the impression that's what you wanted.
"tå" is ok but "åå" is not.
I guess it depends on what you mean by "ok". Both match the expression:
console.log("tå".match(/^[\w]|[åäöæøÅÄÖÆØ]$/)); // ["t", index: 0, input: "tå"]
console.log("åå".match(/^[\w]|[åäöæøÅÄÖÆØ]$/)); // ["å", index: 1, input: "åå"]
"tå" matches because it matches the ^[\w] alternative. "åå" matches because it matches the [åäöæøÅÄÖÆØ]$ alternative.
How can I make it accept words starting with åäöæøÅÄÖÆØ?
If the goal is to accept only strings containing exactly one word, where "word" includes digits and the underscore (since \w does), then:
/^[A-Za-z0-9_åäöæøÅÄÖÆØ]+$/

Why do you think it fails? I would not put the \w in square brackets but various systems seem to allow that and both the following match the text being tested.
Javascript
var test = 'åå';
if (test.match(/^[\w]|[åäöæøÅÄÖÆØ]$/)) { alert("Match"); }
PHP
echo(preg_match("/^[\w]|[åäöæøÅÄÖÆØ]$/","åå")."</br>");
What are you trying to achieve here?