Question

我一直在做一个扑扑的项目，并在发现以下行为时尝试按特定日期过滤结果。

main() {
  String v = '2020-08-03';
  int ms0 = 1596249000000;
  int ms1 = 1596368040000; // 
  int ms2 = 1596465736799;
  int ms3 = 1596472120778; //
  int ms4 = 1596623965588;
  
  print(DateTime.fromMillisecondsSinceEpoch(ms0));
  print(DateTime.fromMillisecondsSinceEpoch(ms1));
  print(DateTime.fromMillisecondsSinceEpoch(ms2));
  print(DateTime.fromMillisecondsSinceEpoch(ms3));
  print(DateTime.fromMillisecondsSinceEpoch(ms4));

 print(DateTime.parse(v).difference(DateTime.fromMillisecondsSinceEpoch(ms0)).inDays); 
 print(DateTime.parse(v).difference(DateTime.fromMillisecondsSinceEpoch(ms1)).inDays);
 print(DateTime.parse(v).difference(DateTime.fromMillisecondsSinceEpoch(ms2)).inDays);
 print(DateTime.parse(v).difference(DateTime.fromMillisecondsSinceEpoch(ms3)).inDays);
 print(DateTime.parse(v).difference(DateTime.fromMillisecondsSinceEpoch(ms4)).inDays);
}

上面的代码产生以下结果

2020-08-01 10:30:00.000
2020-08-02 19:34:00.000
2020-08-03 22:42:16.799
2020-08-04 00:28:40.778
2020-08-05 18:39:25.588
1
0
0
-1
-2

我不明白变量“ ms1”和“ ms2”将如何产生相同的差异“ inDays”。有人可以帮忙提供一些指示吗？谢谢大家。

Answer 1

一天的总时长为24小时，整个小时为60分钟

如果差异少于24小时1秒或1微秒，依此类推-则不计算整天

回到您的示例

2020-08-03表示一天的开始00:00:00.000

并且不同于

2020-08-02 19:34:00.000仅4h，26m

2020-08-03 22:42:16.799仅停留22h，42m，16s，799ms

此测试说明

  test('datetime difference in days', () {
    final date = DateTime.parse('2020-08-03');
    final date1 = DateTime.fromMillisecondsSinceEpoch(1596368040000);
    final date2 = DateTime.fromMillisecondsSinceEpoch(1596465736799);

    print(date);  // 2020-08-03 00:00:00.000
    print(date1); // 2020-08-02 17:34:00.000
    print(date2); // 2020-08-03 20:42:16.799

    expect(date1.difference(date).inDays, equals(0));
    expect(date1.difference(date).inHours, equals(-6));

    expect(date2.difference(date).inDays, equals(0));
    expect(date2.difference(date).inHours, equals(20));
  });

飞镖-两个不同的日期在天中产生相同的差异

1 个答案: