JS sort() empty to end

Question

I have a JS sort like this:

records.sort(function(a, b) {
    if (a < b) return -1;
    if (a > b) return 1;
    return 0;
});

This works, but some of my records are "" or null.

The empty records are listed at the begin but I want them at the end.

I think there are better ways to do this than:

if (a == "") a = "zzzz";

But how can I do this?

stackoverflow.com/questions/29829205/…

Ofir Baruch
– Ofir Baruch

2015-10-08 12:45:02 +00:00
Commented Oct 8, 2015 at 12:45 — Ofir Baruch
– Ofir Baruch, Commented Oct 8, 2015 at 12:45

Eric Hotinger · Accepted Answer · 2016-05-18 14:47:52Z

44

Maybe something like this:

records.sort(function(a, b) {
    if(a === "" || a === null) return 1;
    if(b === "" || b === null) return -1;
    if(a === b) return 0;
    return a < b ? -1 : 1;
});

edited May 18, 2016 at 14:47

answered Oct 8, 2015 at 12:44

Eric Hotinger

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2 Comments

Usman Iqbal Over a year ago

care to explain with a diagram ?

ow3n Over a year ago

I improved this by adding !a || and !b || to the first two conditionals, which accounts for undefined (and short circuits the others as well).

Joey Carlisle · Accepted Answer · 2021-09-23 22:10:46Z

5

records.sort((a,b) => a ? b ? a.localeCompare(b) : -1 : 1);

A ES6 one-liner version of Eric Hotinger's answer

answered Sep 23, 2021 at 22:10

Joey Carlisle

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Comments

Nino Filiu · Accepted Answer · 2019-04-19 10:02:22Z

2

Shorter way:

['a', '!', 'z','Z', null, undefined, '']
.sort((a, b) => !a ? 1 : (!b ? -1 : (a.localeCompare(b))) )

edited Apr 19, 2019 at 10:02

Nino Filiu

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answered Apr 19, 2019 at 8:42

Serhii Kimlyk

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