Remove empty values from array simply - javascript

Question

I have looked at previous Q/A and I did not find much help there. Mainly as I did not understand what was coded.

I am just looking to remove any empty values in my array.

My simple approach - that is not working!

My code is -

var colors = [a, b, c, d, e, f];
var newArray = [];
for (var i = 0; i < colors.length; i++) {
  if (colors[i] !== 'undefined' || colors[i] !== null || colors[i] !== "") {
    newArray.push(colors[i]);
  }
}
console.log(newArray.length); // == 6 
console.log(newArray) //== yellow,blue,red,,,

I would have thought that my if statement would filter all elements with value and push on to my new array. I really need newArray length to equal 3 and just hold vales, no empty string "" should be in the newArray.

Thank you in advance.

"If the color is null or if the color is the empty string, include it in the result". No value can be identical to both null and "" at the same time. — Jon
– Jon, Commented Apr 23, 2013 at 8:42
you use || between conditions, so if colors is not undefined you ll insert it (just example). use && — Bojan Kovacevic
– Bojan Kovacevic, Commented Apr 23, 2013 at 8:42
which values are supposed to be null or empty? what are you expecting? — Oliver Watkins
– Oliver Watkins, Commented Apr 23, 2013 at 8:42
What is with the downvotes? You don't vote on OPs experience, you should vote if the question is legit. He has tried, he has failed, he has shown us the codes. His assumption is wrong, which is why he came here in the first place! — Zlatko
– Zlatko, Commented Apr 23, 2013 at 9:05

Bojan Kovacevic · Accepted Answer · 2013-04-23 08:51:33Z

4

use && instead of ||:

var colors = ["yellow", "","red", "blue", "", ""];
var newArray = [];
for (var i = 0; i < colors.length; i++) {
  if (colors[i] !== undefined && colors[i] !== null && colors[i] !== "") {
    newArray.push(colors[i]);
  }
 }
console.log(newArray.length); // == 3 
console.log(newArray) //== yellow,blue,red,,,

edited Apr 23, 2013 at 8:51

answered Apr 23, 2013 at 8:45

Bojan Kovacevic

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1 Comment

shishirmk Over a year ago

I think you mean colors[i] !== undefined it should not be 'undefined' that will make it a string

shishirmk · Accepted Answer · 2013-04-23 08:54:57Z

use && instead of ||:

var colors = ["yellow", "","red", "blue", "", ""];
var newArray = [];
for (var i = 0; i < colors.length; i++) {
  if (colors[i] !== undefined && colors[i] !== null && colors[i] !== "") {
    newArray.push(colors[i]);
  }
 }
console.log(newArray.length); // == 3 
console.log(newArray) //== yellow,blue,red,,,

For your use case you could also use

for (var i = 0; i < colors.length; i++) {
  if (colors[i]) {
    newArray.push(colors[i]);
  }
 }

This will filter out any falsy values. Falsy values include

false
0
""
null
undefined
NaN

shakib · Accepted Answer · 2013-04-23 09:49:08Z

1

You can simply use colors[i] to existence checking,

var colors = ["yellow", "","red", "blue", "", "", true, 1];
var newArray = [];
for (var i = 0; i < colors.length; i++) {
    if (typeof colors[i] == 'string' && colors[i]) {
        newArray.push(colors[i]);
    }
}
console.log(newArray) //["yellow", "red", "blue"]

relevant resource javascript type conversion

hope this helps.

edited Apr 23, 2013 at 9:49

answered Apr 23, 2013 at 9:02

shakib

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Comments

3y3 · Accepted Answer · 2013-04-23 09:31:44Z

0

If 'false' value is important then:

var colors = [0,1,'a',,'',null,undefined,false,true];
    colors = colors.filter(function(e){
        return (e===undefined||e===null||e==='')?false:~e;
    });

else:

var colors = [0,1,'a',,'',null,undefined,false,true];
        colors = colors.filter(function(e){return e;});

answered Apr 23, 2013 at 9:31

3y3

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Comments

Khalizar · Accepted Answer · 2016-06-01 15:00:25Z

0

var colors = ["yellow", null, "blue", "red", undefined, 0, ""];

// es5:
var newArray = colors.filter(function(e){return !!e;});

// es6:
var newArray = colors.filter((e)=>!!e);

console.log(newArray.length); // ==> 3
console.log(newArray) // ==> ["yellow","blue","red"]

answered Jun 1, 2016 at 15:00

Khalizar

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