Javascript - 按日期排序对象数组,然后按时间排序

Question

我有以下数组:

[{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"}]

我怎样才能将阵列从最早到最新排序,反之亦然？

我尝试按日期排序,但按时间排序将记录ID 4的顺序交换为id 3,因为它具有比记录3更早的时间值,但在技术上按照定义,稍后.

给定此数组和json结构,如何对数组进行排序以考虑两个字段(date和time)？

Answer 1

按dates中的差异排序,如果没有差异time,则在单个.sort函数中按s 的差异排序:

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);

将返回日期的差异,除非它们是相同的,在这种情况下localCompare将出现0,并且将返回时间差异.

要改为降序,只需切换as和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);