I have the following array:

[{id: 1, value : "value1", date: "2018-08-08", time: "15:27:17"},
{id: 2, value : "value2", date: "2018-08-09", time: "12:27:17"},
{id: 3, value : "value3", date: "2018-08-10", time: "17:27:17"},
{id: 4, value : "value4", date: "2018-08-11", time: "10:27:17"}]

How can I go about sorting the array from earliest to latest or vice versa?

I tried sorting by date, but sorting it by time swaps the order of record id 4 to id 3, because it has an earlier time value than record 3, but is technically by definition, later.

Given this array and json structure, how do I sort the array to take both fields (date and time) into consideration?

I have the following array:

[{id: 1, value : "value1", date: "2018-08-08", time: "15:27:17"},
{id: 2, value : "value2", date: "2018-08-09", time: "12:27:17"},
{id: 3, value : "value3", date: "2018-08-10", time: "17:27:17"},
{id: 4, value : "value4", date: "2018-08-11", time: "10:27:17"}]

How can I go about sorting the array from earliest to latest or vice versa?

I tried sorting by date, but sorting it by time swaps the order of record id 4 to id 3, because it has an earlier time value than record 3, but is technically by definition, later.

Given this array and json structure, how do I sort the array to take both fields (date and time) into consideration?

• javascript
• arrays
• typescript
• sorting

• Possible duplicate of How to sort date strings (format exmaple: 2014 7 23) in JavaScript? – Blue Commented Aug 9, 2018 at 2:32
• hello, the answer does not seem to consider a 2nd property time that is separate from date. – muffin Commented Aug 9, 2018 at 2:54

1 Answer 1

Sort by the difference in the dates, and if there is no difference, sort by the difference in the times, in a single .sort function:

const arr = [{id: 1, value : "value1", date: "2018-08-08", time: "15:27:17"},
{id: 2, value : "value2", date: "2018-08-09", time: "12:27:17"},
{id: 3, value : "value3", date: "2018-08-10", time: "17:27:17"},
{id: 4, value : "value4", date: "2018-08-10", time: "01:27:17"},
{id: 5, value : "value5", date: "2018-08-10", time: "09:27:17"},
{id: 6, value : "value6", date: "2018-08-10", time: "23:27:17"},
{id: 7, value : "value7", date: "2018-08-10", time: "16:27:17"},
{id: 8, value : "value8", date: "2018-08-11", time: "10:27:17"}
];

arr.sort((a, b) => a.date.localeCompare(b.date) || a.time.localeCompare(b.time));
console.log(arr);

The difference in dates will be returned, except if they're the same, in which case the localCompare will e out to 0, and the difference in times will be returned instead.

To sort to descending instead, just switch the as and bs:

const arr = [{id: 1, value : "value1", date: "2018-08-08", time: "15:27:17"},
{id: 2, value : "value2", date: "2018-08-09", time: "12:27:17"},
{id: 3, value : "value3", date: "2018-08-10", time: "17:27:17"},
{id: 4, value : "value4", date: "2018-08-10", time: "01:27:17"},
{id: 5, value : "value5", date: "2018-08-10", time: "09:27:17"},
{id: 6, value : "value6", date: "2018-08-10", time: "23:27:17"},
{id: 7, value : "value7", date: "2018-08-10", time: "16:27:17"},
{id: 8, value : "value8", date: "2018-08-11", time: "10:27:17"}
];

arr.sort((a, b) => b.date.localeCompare(a.date) || b.time.localeCompare(a.time));
console.log(arr);