Sort Element in Array of Array Javascript

Question

I wish to sort the arrays by the time (which is the 3rd element). However, when I print out the result, the sequence of the array seems to have no difference.

Can I know what's the error? Thank you!

const events =[
  [ 1, '1230000003', '09:00:00' ],
  [ -1, '1230000003', '14:00:00' ],
  [ -1, '1110000002', '12:30:00' ],
  [ 1, '1110000002', '14:00:00' ],
  [ -1, '1110000007', '08:30:00' ],
  [ 1, '1110000007', '11:00:00' ],
  [ -1, '1110000008', '09:00:00' ],
  [ 1, '1110000008', '12:00:00' ]
]

This is my code:

 events.sort((a, b) => a[0][2] - b[0][2]);
 console.log("sorted start time",events);

Hm. a[0][2] is undefined in your case. Same as a b[0][2]. — Vasyl Moskalov
– Vasyl Moskalov, Commented Jul 15, 2020 at 3:01
@VasylMoskalov I see.. but it seems okay when i do it to other elements like a[0][1] and b[0][1] — wink
– wink, Commented Jul 15, 2020 at 3:12

Sven.hig · Accepted Answer · 2020-07-15 03:18:55Z

2

You could use localCompare

const events =[ [ 1, '1230000003', '09:00:00' ], [ -1, '1230000003', '14:00:00' ], [ -1, '1110000002', '12:30:00' ], [ 1, '1110000002', '14:00:00' ], [ -1, '1110000007', '08:30:00' ], [ 1, '1110000007', '11:00:00' ], [ -1, '1110000008', '09:00:00' ], [ 1, '1110000008', '12:00:00' ] ]
events.sort((a, b) => a[2].localeCompare(b[2]));
console.log("sorted start time",events);

edited Jul 15, 2020 at 3:18

answered Jul 15, 2020 at 3:04

Sven.hig

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2 Comments

wink Over a year ago

yes this solution works for my code! However when I run the code, may I know why this will come first [ 1, "1230000003", "09:00:00" ], instead of this [ -1, "1110000008", "09:00:00" ], ?

Sven.hig Over a year ago

that's because since the function compares time, both values in this case are equal, and since the first value in the first array is passed first it will show first in the result does that makes since ?

Harmandeep Singh Kalsi · Accepted Answer · 2020-07-15 03:02:56Z

2

Since it is date is string , you should use the sort function as

const events =[
  [ 1, '1230000003', '09:00:00' ],
  [ -1, '1230000003', '14:00:00' ],
  [ -1, '1110000002', '12:30:00' ],
  [ 1, '1110000002', '14:00:00' ],
  [ -1, '1110000007', '08:30:00' ],
  [ 1, '1110000007', '11:00:00' ],
  [ -1, '1110000008', '09:00:00' ],
  [ 1, '1110000008', '12:00:00' ]
]

events.sort((a, b) => a[2].localeCompare(b[2]));
console.log(events);

answered Jul 15, 2020 at 3:02

Harmandeep Singh Kalsi

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2 Comments

wink Over a year ago

thanks! this actually helps! but can I know why when it comes to same timing like [ 1,'1230000003', '09:00:00' ] and [ -1, '1110000008', '09:00:00' ], the first one will come first instead of second one, what if i want this [ -1, '1110000008', '09:00:00' ] to come first?

Harmandeep Singh Kalsi Over a year ago

you can customize the logic to check if the field you are checking on is same , then compare based on the other one you mentioned

GirkovArpa · Accepted Answer · 2020-07-15 03:32:31Z

1

Change this;

 events.sort((a, b) => a[0][2] - b[0][2]);

To this:

events.sort((a, b) => a[2] > b[2] ? 1 : -1);

Result:

[
  [-1, "1110000007", "08:30:00"],
  [-1, "1110000008", "09:00:00"],
  [1, "1230000003", "09:00:00"],
  [1, "1110000007", "11:00:00"],
  [1, "1110000008", "12:00:00"],
  [-1, "1110000002", "12:30:00"],
  [1, "1110000002", "14:00:00"],
  [-1, "1230000003", "14:00:00"]
]

edited Jul 15, 2020 at 3:32

answered Jul 15, 2020 at 3:01

GirkovArpa

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