DateTime.Now.ToString 中的参数问题

最新推荐文章于 2024-01-29 13:28:01 发布

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最新推荐文章于 2024-01-29 13:28:01 发布

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文章标签： c#

本文介绍C#中DateTime.Now.ToString方法的不同格式参数，展示如何根据需求输出日期时间的多种格式，包括短日期、长日期、完整时间等。

DateTime.Now.ToString根据参数不同输出不同内容，功能很强的，总结一下。

[C#]
using System;
using System.Globalization;

public class MainClass {
   public static void Main(string[] args) {
       DateTime dt = DateTime.Now;
       String[] format = {
           "d", "D",
           "f", "F",
           "g", "G",
           "m",
           "r",
           "s",
           "t", "T",
           "u", "U",
           "y",
           "dddd, MMMM dd yyyy",
           "ddd, MMM d \"'\"yy",
           "dddd, MMMM dd",
           "M/yy",
           "dd-MM-yy",
       };
       String date;
       for (int i = 0; i < format.Length; i++) {
           date = dt.ToString(format[i], DateTimeFormatInfo.InvariantInfo);
           Console.WriteLine(String.Concat(format[i], " :" , date));
       }
/** Output.
   *
   * d :08/17/2000
   * D :Thursday, August 17, 2000
   * f :Thursday, August 17, 2000 16:32
   * F :Thursday, August 17, 2000 16:32:32
   * g :08/17/2000 16:32
   * G :08/17/2000 16:32:32
   * m :August 17
   * r :Thu, 17 Aug 2000 23:32:32 GMT
   * s :2000-08-17T16:32:32
   * t :16:32
   * T :16:32:32
   * u :2000-08-17 23:32:32Z
   * U :Thursday, August 17, 2000 23:32:32
   * y :August, 2000
   * dddd, MMMM dd yyyy :Thursday, August 17 2000
   * ddd, MMM d "'"yy :Thu, Aug 17 '00
   * dddd, MMMM dd :Thursday, August 17
   * M/yy :8/00
   * dd-MM-yy :17-08-00
   */
   }
}