Javascript：将逻辑运算符与比较运算符一起使用

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
}

Answer 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()){

}

Answer 2

You may simply use:

if(key) {
   // to do here
}

Taking care of whitespace:

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

Answer 3

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
}

Answer 4

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.

Answer 5

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.