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"}]

如何对数组进行从最早到最新的排序，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，因为它的时间值早于记录3，但从技术上讲，它是较晚的。

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

Answer 1

按date中的差异进行排序，如果没有差异，请通过单个time函数按.sort中的差异进行排序：

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);

除非日期相同，否则将返回日期差。在这种情况下，localCompare将出现在0中，而时间差将被返回。

>

要排序为降序，只需切换a和b：

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);

Javascript-按日期然后按时间对对象数组进行排序

1 个答案: