Matching special characters and letters in regex

Add them to the allowed characters, but you'll need to escape some of them, such as -]/\

var pattern = /^[a-zA-Z0-9!@#$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]*$/

That way you can remove any individual character you want to disallow.

Also, you want to include the start and end of string placemarkers ^ and $

Update:

As elclanrs understood (and the rest of us didn't, initially), the only special characters needing to be allowed in the pattern are &-._

/^[\w&.\-]+$/

[\w] is the same as [a-zA-Z0-9_]

Though the dash doesn't need escaping when it's at the start or end of the list, I prefer to do it in case other characters are added. Additionally, the + means you need at least one of the listed characters. If zero is ok (ie an empty value), then replace it with a * instead:

/^[\w&.\-]*$/

Well, why not just add them to your existing character class?

var pattern = /[a-zA-Z0-9&._-]/

If you need to check whether a string consists of nothing but those characters you have to anchor the expression as well:

var pattern = /^[a-zA-Z0-9&._-]+$/

The added ^ and $ match the beginning and end of the string respectively.

Testing for letters, numbers or underscore can be done with \w which shortens your expression:

var pattern = /^[\w&.-]+$/

As mentioned in the comment from Nathan, if you're not using the results from .match() (it returns an array with what has been matched), it's better to use RegExp.test() which returns a simple boolean:

if (pattern.test(qry)) {
    // qry is non-empty and only contains letters, numbers or special characters.
}

Update 2

In case I have misread the question, the below will check if all three separate conditions are met.

if (/[a-zA-Z]/.test(qry) && /[0-9]/.test(qry) && /[&._-]/.test(qry)) {
   // qry contains at least one letter, one number and one special character
}

Try this regex:

/^[\w&.-]+$/

Also you can use test.

if ( pattern.test( qry ) ) {
  // valid
}