String validation function in JavaScript

Question

I'm trying to create a nice util function to validate strings. Conditions are:

Cannot be typeof "undefined"
Cannot be null
Must be a string i.e. not a number, object, array, etc.
Must have at least one character in it. So a string with value of '' would be invalid.

I found some pointers on this that suggest using RegEx is the way to do it but the following is not working when I give it a numeric value.

Here's what I have so far:

const isValidString = (str1) => {

   if(typeof str1 === "undefined" || str1 === null) return false;

   const validRegEx = /^[^\\\/&]*$/;
   if(str1.match(validRegEx)) {
      return true;
   } else {
      return false;
   }
}

As I said, if send const str1 = 3; to this function, I get an error that reads:

"TypeError: badString.match is not a function

What am I missing?

Eric · Accepted Answer · 2018-03-26 04:50:29Z

2

if (str1 != null) will coerce the str1 value, and will check both against null and undefined.

EDIT: ignore this part (Also, you don't need to use a regex to check if there's at least one character. You can use the string length. str1.length will evaluate to a falsey value if it's 0, otherwise it will evaluate to a truethy value.) You need a strict boolean value, so it should be str1.length > 0.

function isValidString(str1) {
  return str1 != null && typeof str1 === "string" && str1.length > 0;
}

edited Mar 26, 2018 at 4:50

answered Mar 26, 2018 at 4:41

Eric

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6 Comments

Sam Over a year ago

Nice and terse!

Eric Over a year ago

@Sam Glad it helped! Thanks!

Patrick Roberts Over a year ago

If you want the function to strictly return a boolean, you'll need to expand str1.length to str1.length > 0

Eric Over a year ago

@PatrickRoberts actually you're right... I'm editing it.. Thanks a lot!!

Sam Over a year ago

@PatrickRoberts You beat me to it. I was just posting this: return str1 != null && typeof str1 === "string" && str1.length > 0; will always return true or false

|

arbghl · Accepted Answer · 2018-03-26 04:31:20Z

1

Why don't you add a typeof str1 === 'string' as shown below

if(typeof str1 === 'string' && str1.match(validRegEx)) {
  return true;
} else {
  return false;
}

answered Mar 26, 2018 at 4:31

arbghl

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Comments

djfdev · Accepted Answer · 2018-03-26 04:31:03Z

0

That is because the match method is only available on strings, and you are attempting to call it on a number.

I would suggest re-writing the function to ensure that the argument is a string, rather than only checking for null or undefined values:

const isValidString = str1 => {
  if (!str1 || typeof str1 !== 'string') return false

  const validRegEx = /^[^\\\/&]*$/

  if (str1.match(validRegEx)) {
    return true
  } else {
    return false
  }
}

answered Mar 26, 2018 at 4:31

djfdev

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1 Comment

Sam Over a year ago

I think I like this the best. Thank you.

Harsh Makadia · Accepted Answer · 2018-03-26 04:33:14Z

0

Convert number to string perform evaluating it

const isValidString = (str1) => {
str1 = str1.toString();
   if(typeof str1 === "undefined" || str1 === null) return false;

   const validRegEx = /^[^\\\/&]*$/;
   if(str1.match(validRegEx)) {
      return true;
   } else {
      return false;
   }
}

console.log(isValidString(3))

answered Mar 26, 2018 at 4:33

Harsh Makadia

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1 Comment

Sam Over a year ago

This is doing the opposite of what I want. If I enter 3 which is a number, I should get a false. By converting it toString() almost all values become acceptable which is not what I want.

I. Ahmed · Accepted Answer · 2018-03-26 04:33:26Z

0

Replace the if with following:

if(typeof str1 === "undefined" || str1 === null || typeof str1 !== 'string') return false;

answered Mar 26, 2018 at 4:33

I. Ahmed

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Comments

31piy · Accepted Answer · 2018-03-26 04:33:27Z

0

Use RegExp#test instead, because your regular expression will always be available. Also, String#match returns an array (or null) which is not well-suited for boolean expressions and may produce unwanted results.

const isValidString = (str1) => {

  if (typeof str1 === "undefined" || str1 === null) return false;
  const validRegEx = /^[^\\\/&]*$/;
  
  if (validRegEx.test(str1)) {
    return true;
  } else {
    return false;
  }
}

console.log(isValidString(1));
console.log(isValidString('string'));

Note, your regexp is still wrong because it validates the number as true. You need to tweak it more.

answered Mar 26, 2018 at 4:33

31piy

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Comments

Rocky Sims · Accepted Answer · 2018-03-26 05:00:05Z

0

Credit to djfdev on this one, all I've really done is simplify his answer to be more succinct.

const isValidString = str1 => {
    return typeof str1 === 'string' && /^[^\\\/&]*$/.test(str1);
}

edited Mar 26, 2018 at 5:00

answered Mar 26, 2018 at 4:44

Rocky Sims

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