Check if a string contains special charcaters using Javascript

Question

I have a password field and I need to check using javascript if it has the following characters:

! @ # $ % ^ & *

I tried to do it like this, and it's working as expected:

function ValidarPass()
{
    var Contrasena = document.getElementById('Clave').value;

    if(Contrasena!='' && 
      (Contrasena.indexOf('!')>-1||
       Contrasena.indexOf('@')>-1||
       Contrasena.indexOf('#')>-1||
       Contrasena.indexOf('$')>-1||
       Contrasena.indexOf('%')>-1||
       Contrasena.indexOf('^')>-1||
       Contrasena.indexOf('&')>-1||
       Contrasena.indexOf('*')>-1))
    {
        alert("Exito!");
    }
    else
    {
        alert("Error!");
    }
}

Is there an easier/efficient way to do this?

I'm voting to close this question as off-topic because it is asking for improvements to working code. ask on [codereview.se] — Daniel A. White, Apr 01 '19 at 14:52
Possible duplicate of [Matching special characters and letters in regex](https://stackoverflow.com/questions/13946651/matching-special-characters-and-letters-in-regex) — Cup of Java, Apr 01 '19 at 14:53

Flash · Answer 1 · 2019-04-01T15:00:02.460

2

you can test a string using regex: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions

function isValid(str){
     return !/[!@#$%^&*+=\-\[\]\\';,/{}|\\":<>\?]/g.test(str);
    }

edited Apr 01 '19 at 15:00

answered Apr 01 '19 at 14:51

Flash

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1

That allows more special characters than is asked for. – Ivar Apr 01 '19 at 14:53

score 0 · Answer 2 · answered Apr 01 '19 at 14:54

The Regular Expression that you would use to achieve this would be [!@#$%^&*]

This will check for all the special characters that you have listed. For use in a function you can use the matches function like so:

public boolean isSpecialCharacter(String string) {
    String pattern = "[!@#$%^&*]";
    return string.matches(pattern);
}

score 0 · Answer 3 · answered Apr 01 '19 at 14:58

A regex would be a nice one line solution. But it's actually slower than indexOf (like the code you posted). See some information here: https://jsperf.com/regexp-vs-indexof

Regex docs: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions

function isValid(str){
 return /[!@#$%^&*(),.?":{}|<>]/g.test(str);
}

Also this site is great for testing regexs https://www.regextester.com/

mishu · Answer 4 · 2019-04-01T15:06:07.127

While both answers here do solve the issue, just for demo I will submit one approach that doesn't use RegEx. Maybe we don't always need to use regular expressions, even for the simplest checks. And the list of characters is pretty short here, one other option could be:

let testString = 'abcd';// the string you are checking
let chars = '!@#$%^&*'.split('');// just a way to write your chars as an array
let isValid = chars.some((c) => testString.includes(c));// check if at least one of the chars is included in your string

Did not really benchmark, but I am hoping it's pretty quick, as the some method should return as early as it finds a positive match and includes also.

Maybe you will need some polyfills for some and includes methods or replace the arrow function with a regular function depending on what kind of browsers you want to support, this is just to demo an idea.

Check if a string contains special charcaters using Javascript

4 Answers4