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"}]
如何从最早到最新对数组进行排序，反之亦然？
我尝试按日期排序，但按时间排序会将记录 id 4 的顺序交换为 id 3，因为它的时间值比记录 3 早，但从定义上讲，它更晚。
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 (？
按
date s 中的差值排序，如果没有差值，则按time s 中的差值排序，在单个 .sort 函数中：
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，除非它们相同，在这种情况下，localCompare将出来0，并且将返回时间的差异。
要改为降序排序，只需切换a s 和 b s：
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);