Matching special characters and letters in regex

JavascriptRegexSpecial Characters

Javascript Problem Overview


I am trying to validate a string, that should contain letters numbers and special characters &-._ only. For that I tried with a regular expression.

var pattern = /[a-zA-Z0-9&_\.-]/
var qry = 'abc&*';
if(qry.match(pattern)) {
    alert('valid');
}
else{
    alert('invalid');
}

While using the above code, the string abc&* is valid. But my requirement is to show this invalid. ie Whenever a character other than a letter, a number or special characters &-._ comes, the string should evaluate as invalid. How can I do that with a regex?

Javascript Solutions


Solution 1 - Javascript

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&.\-]*$/

Solution 2 - Javascript

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
}

Solution 3 - Javascript

Try this regex:

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

Also you can use test.

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

Solution 4 - Javascript

let pattern = /^(?=.*[0-9])(?=.*[!@#$%^&*])(?=.*[a-z])(?=.*[A-Z])[a-zA-Z0-9!@#$%^&*]{6,16}$/;

//following will give you the result as true(if the password contains Capital, small letter, number and special character) or false based on the string format

let reee =pattern .test("helLo123@");   //true as it contains all the above

Solution 5 - Javascript

I tried a bunch of these but none of them worked for all of my tests. So I found this:

^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[^a-zA-Z0-9])(?!.*\s).{8,15}$

from this source: https://www.w3resource.com/javascript/form/password-validation.php

Solution 6 - Javascript

Try this RegEx: Matching special charecters which we use in paragraphs and alphabets

   Javascript : /^[a-zA-Z]+(([\'\,\.\-_ \/)(:][a-zA-Z_ ])?[a-zA-Z_ .]*)*$/.test(str)

                .test(str) returns boolean value if matched true and not matched false

            c# :  ^[a-zA-Z]+(([\'\,\.\-_ \/)(:][a-zA-Z_ ])?[a-zA-Z_ .]*)*$

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionQuicksilverView Question on Stackoverflow
Solution 1 - JavascriptHighly IrregularView Answer on Stackoverflow
Solution 2 - JavascriptJa͢ckView Answer on Stackoverflow
Solution 3 - JavascriptelclanrsView Answer on Stackoverflow
Solution 4 - JavascriptAmrutha VSView Answer on Stackoverflow
Solution 5 - JavascriptAaronView Answer on Stackoverflow
Solution 6 - Javascriptchadalavada harishView Answer on Stackoverflow