如何快速比较两个日期的大小？

Question

          let time = {
                endDate: "2019-02-23",
                endTime: "17:00:35",
                startDate: "2019-02-23",
                startTime: "17:50:35"
            };
            如何比较上面两对时间endDate+endTime与startDate+startTime的大小？
            我现在解决方案是拼接然后转时间戳比较大小，有没有更快速的方法？

Answer 1

let&nbsp;time&nbsp;=&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;endDate:&nbsp;"2019-02-23", &nbsp;&nbsp;&nbsp;&nbsp;endTime:&nbsp;"17:00:35", &nbsp;&nbsp;&nbsp;&nbsp;startDate:&nbsp;"2019-02-23", &nbsp;&nbsp;&nbsp;&nbsp;startTime:&nbsp;"17:50:35"};//&nbsp;转时间戳比较稳妥。+new&nbsp;Date(time.endDate+time.endTime)&nbsp;<&nbsp;+new&nbsp;Date(time.startDate+time.startTime);//&nbsp;这种方法可以写到`utils`工具方法里去呀。//&nbsp;调用一下就可以了。比如：export&nbsp;const&nbsp;compareDate&nbsp;=&nbsp;(date1,&nbsp;date2)&nbsp;=>&nbsp;{&nbsp;&nbsp;&nbsp;&nbsp;return&nbsp;new&nbsp;Date(date1).getTime()&nbsp;>=&nbsp;(new&nbsp;Date(date2).getTime());

Answer 2

'2019-02-02 17:00:35' > '2019-02-23 17:50:35' //false&nbsp;字符串比较会逐个用&nbsp;Unicode&nbsp;值来比较首先要保证两个比较的字符长度相等或格式相同：'22'>'3' // false