Replacing empty values with nulls in JavaScript string

Question

Given the following string:

var str = "[,,,1,2,,,3,4,,,,,,5,6]";

I want to replace all "empty" values with nulls. In other words, I want this result:

"[null,null,null,1,2,null,null,3,4,null,null,null,null,null,5,6]"

This almost works, but it misses the first empty value:

var str = "[,,,1,2,,,3,4,,,,,,5,6]";
str.split(',').map(function(x) { return x ? x : 'null' }).join(',')
// Gives [,null,null,1,2,null,null,3,4,null,null,null,null,null,5,6]

Likewise, I noticed if I have trailing empty values, it misses the last one there too:

var str = "[,,,1,2,,,3,4,,,,,,5,6,,]";
str.split(',').map(function(x) { return x ? x : 'null' }).join(',')
// Gives [,null,null,1,2,null,null,3,4,null,null,null,null,null,5,6,null,]

Any ideas how I can make sure that first and last empty values are also replaced?

Thanks!

everything is working as expected you have indeed removed all empty values if you just use str.split(',').map(function(x) { return x ? x : 'null' }) — Shanon Jackson
– Shanon Jackson, Commented Feb 13, 2018 at 3:14

gavgrif · Accepted Answer · 2018-02-13 03:18:15Z

1

As mentioned - the [ and ] are altering your return - this removes them and using the same function as your existing function - but on the shortened string and returns the correct value.

var str = "[,,,1,2,,,3,4,,,,,,5,6]";
let newStr = str.substring(1,str.length-1).split(',').map(function(x) { return x ? x : 'null' }).join(',')

console.log(newStr);
// Gives [null,null,null,1,2,null,null,3,4,null,null,null,null,null,5,6]


var str = "[,,,1,2,,,3,4,,,,,,5,6,,]";

newStr = str.substring(1,str.length-1).split(',').map(function(x) { return x ? x : 'null' }).join(',');
console.log(newStr);
// Gives null,null,null,1,2,null,null,3,4,null,null,null,null,null,5,6,null,null

answered Feb 13, 2018 at 3:18

gavgrif

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1 Comment

JoshG Over a year ago

Ahhh, that makes sense. Doh! Thank you!

Obsidian Age · Accepted Answer · 2018-02-13 03:16:55Z

1

The secret is that your brackets are getting combined with your commas / numbers at the start and end of the string when it is split. In order to resolve this, you need to remove the brackets before splitting.

This can be done with .replace() on /[\][]/g:

var str = "[,,,1,2,,,3,4,,,,,,5,6,]";
console.log(
  str.replace(/[\][]/g, '').split(',').map(function(x) {
    return x ? x : 'null'
  }).join(',')
);

answered Feb 13, 2018 at 3:16

Obsidian Age

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Comments

user9274775 · Accepted Answer · 2018-02-13 03:17:20Z

1

var str = "[,,,1,2,,,3,4,,,,,,5,6]";
var a = `[${str.slice(1, -1).split(",").map(n => n || "null")}]`
  
console.log(a);

answered Feb 13, 2018 at 3:17

user9274775

Comments

xianshenglu · Accepted Answer · 2018-02-13 03:25:35Z

1

this is going to work:

	 var str = "[,,,1,2,,,3,4,,,,,,5,6]";
	 var result = '[' + str.replace(/^\[|\]$/g, '').split(/[,]/).map(function(x) { return x ? x : 'null' }).join(',') + ']';
	 console.log(result);

edited Feb 13, 2018 at 3:25

answered Feb 13, 2018 at 3:15

xianshenglu

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Comments

Shanon Jackson · Accepted Answer · 2018-02-13 03:16:17Z

0

You have removed all the empty values. "[" and "]" are not nothing.

answered Feb 13, 2018 at 3:16

Shanon Jackson

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Comments

Ari Singh · Accepted Answer · 2018-02-13 03:49:18Z

0

String.prototype.repEmptyWithNull = 
function() {return this
  .replace(/,,/g, ',null,')
  .replace(/,,/g, ',null,')
  .replace('[,', '[null,')
  .replace(',]', ',null]')
}

console.log ("[,,,1,2,,,3,4,,,,,,5,6,]".repEmptyWithNull())
console.log ("[,,1,2,,,3,4,,,,,,5,6]".repEmptyWithNull())
console.log ("[,1,2,,,3,4,,,,,,5,6,,]".repEmptyWithNull())
console.log ("[,1,2,,,3,4,,,,,,5,6]".repEmptyWithNull())
console.log ("[1,2,,,3,4,,,,,,5,6,]".repEmptyWithNull())
console.log ("[1,2,,,3,4,,,,,,5,6]".repEmptyWithNull())

answered Feb 13, 2018 at 3:49

Ari Singh

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