Javascript: using Logical Operators with Comparison Operators

Question

I am trying to understand how to use logical operators with my code and be as efficient as possible.

Let's say I have this variable:

var key = localStorage.getItem('keyName');

And this is my conditional:

if (key !== null && key !== '' && key !== ' ') { // localStorage key exists and is not empty or a single space
  // Do something with key
}

Is there a way to write it more efficiently? I tried this, but it's not working:

if (key !== null && '' && ' ') {
  // Do something with key
}

@Vamsi that works, unless key is 0 or false. Local Storage can include any type of value: w3.org/TR/2009/WD-webstorage-20090910/#the-storage-interface — nils
– nils, Commented Sep 18, 2015 at 11:41
@nils, thanks for correcting me. I've never checked those conditions. Very helpful. :) I only assumed that the key would be some value, not the 2 options that u have mentioned. — Vamsi
– Vamsi, Commented Sep 18, 2015 at 11:43

digitalfrost · Accepted Answer · 2015-09-18 12:20:53Z

1

if( value ) {
}

will evaluate to true if value is not:

null
undefined
NaN
empty string ("")
0
false

so key !== null && key !== '' can be replaced by key

We can use /\S/.test(key) to test for a string that is just spaces (1 or more)

so how about

if (key && /\S/.test(key) ){

}

or alternatively

if (key && key.trim()){

}

edited Sep 18, 2015 at 12:20

answered Sep 18, 2015 at 12:10

digitalfrost

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1 Comment

digitalfrost Over a year ago

trim() will only work from IE9 and up but this is not an issue in your case

user2575725 · Accepted Answer · 2015-09-18 11:40:35Z

0

You may simply use:

if(key) {
   // to do here
}

Taking care of whitespace:

if(key && key.replace(/\s/g,'')) {
   //
}

answered Sep 18, 2015 at 11:40

user2575725

Comments

user1438038 · Accepted Answer · 2015-09-18 11:41:47Z

0

You can use the trim() method to remove any leading or trailing whitespace characters from a string. There is no shorthand notation as you did in your second code example.

if (key !== null && key.trim() !== '')
  // Do something with key
}

answered Sep 18, 2015 at 11:41

user1438038

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Comments

Matt Morgan · Accepted Answer · 2015-09-18 11:50:07Z

0

try:

if (key && key.trim())

the first check fails if key is null, and the second check fails if key is only whitespace. If either check fails, the whole thing fails.

answered Sep 18, 2015 at 11:50

Matt Morgan

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Comments

Vivek · Accepted Answer · 2015-09-18 12:24:47Z

0

if(!~[null, '', ' '].indexOf(key))

or

if(!~[null, ''].indexOf(key.trim()))

Explanation:

indexOf return int>=0. Int >=0 means the element is found. All result will evaluate to true except 0, that's where ~(invert/negate the result and subtract 1) comes handy.

Rest is self-explanatory, please let me know if any further explanation is required.

edited Sep 18, 2015 at 12:24

answered Sep 18, 2015 at 11:42

Vivek

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