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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"}]

我怎样才能对数组从最早到最新进行排序，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 更早，但从技术上讲，它的定义更晚。

鉴于此数组和 json 结构，我如何对数组进行排序以同时考虑两个字段 (date和time)？

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1 回答 1

6

按dates 中的差异排序，如果没有差异，则按times 中的差异排序，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0返回 ，而将返回时间的差异。

要改为降序排序，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);

于 2018-08-09T02:31:48.813 回答