最简单的直接方法，无需安装额外的包是

约会:

datetime.toIso8601String().split("T")[0];

获得时间:

datetime.toIso8601String().split("T")[1];

使用这种方法，不需要导入任何库。

DateTime now = DateTime.now();

String convertedDateTime = "${now.year.toString()}-${now.month.toString().padLeft(2,'0')}-${now.day.toString().padLeft(2,'0')} ${now.hour.toString().padLeft(2,'0')}-${now.minute.toString().padLeft(2,'0')}";

输出

2020-12-05 14:57

您也可以使用这种语法。YYYY-MM-JJ HH-MM:

var now = DateTime.now();
var month = now.month.toString().padLeft(2, '0');
var day = now.day.toString().padLeft(2, '0');
var text = '${now.year}-$month-$day ${now.hour}:${now.minute}';

使用字符串分割方法删除:00.000

var formatedTime = currentTime.toString().split(':')
Text(formatedTime[0])

=======或使用以下代码YYYY-MM-DD HH:MM:SS格式，不使用库====

var stringList =  DateTime.now().toIso8601String().split(new RegExp(r"[T\.]"));
var formatedDate = "${stringList[0]} ${stringList[1]}";

为DateTime提供一个简单的扩展方法如何?运行时可能不是很好，因为我们多次迭代字符串并遍历每个%格式选项。可能会浏览一次，贪婪地替换%。

extension DateTimeFormat on DateTime {

  /// Supports the following, inspired by: https://linux.die.net/man/3/strptime
  /// %Y: The year, including century (for example, 1991).
  /// %m: The month number (1-12).
  /// %d: The day of month (1-31).
  /// %H: The hour (0-23).
  /// %M: The minute (0-59).
  /// %S: The second (0-59).
  String format(String formatString) {
    var hourString = hour.toString();
    var dayString = day.toString();
    var monthString = month.toString();
    var minuteString = minute.toString();
    var secondString = second.toString();
    var yearString = year.toString();

    var map = {
      '%H': hourString.padLeft(3 - hourString.length, '0'), // the pad values here are the desired length + 1
      '%d': dayString.padLeft(3 - dayString.length, '0'),
      '%m': monthString.padLeft(3 - monthString.length, '0'),
      '%M': minuteString.padLeft(3 - minuteString.length, '0'),
      '%S': secondString.padLeft(3 - secondString.length, '0'),
      '%Y': yearString.padLeft(5 - yearString.length, '0'),
    };
    return map.entries.fold(formatString, (acc, entry) => acc.replaceAll(entry.key, entry.value));
  }
}

用法:

print(DateTime(2021, 10, 16, 4, 4, 4, 4, 4).format('%Y-%m-%d-%H-%M-%S'));
// '2021-10-16-04-04-04'

请随意提出修改建议。

自从0.16以来有了一些变化这是我做的，

在pubspec.yaml中导入

dependencies:
      flutter:
        sdk: flutter
      intl: ^0.16.1

然后 使用

  txdate= DateTime.now()


  DateFormat.yMMMd().format(txdate)