我如何格式化07/03/2012到2012年3月7日在c＃
任何人请帮助我需要显示日期03/03/2012作为2012年3月3日等
您可以创建自己的自定义格式提供程序来执行此操作：
public class MyCustomDateProvider: IFormatProvider, ICustomFormatter { public object GetFormat(Type formatType) { if (formatType == typeof(ICustomFormatter)) return this; return null; } public string Format(string format, object arg, IFormatProvider formatProvider) { if (!(arg is DateTime)) throw new NotSupportedException(); var dt = (DateTime) arg; string suffix; if (new[] {11, 12, 13}.Contains(dt.Day)) { suffix = "th"; } else if (dt.Day % 10 == 1) { suffix = "st"; } else if (dt.Day % 10 == 2) { suffix = "nd"; } else if (dt.Day % 10 == 3) { suffix = "rd"; } else { suffix = "th"; } return string.Format("{0:MMMM} {1}{2}, {0:yyyy}", arg, dt.Day, suffix); } } 然后可以这样调用：
var formattedDate = string.Format(new MyCustomDateProvider(), "{0}", date);
导致（例如）：
2012年3月3日
Humanizer满足您操作和显示字符串，枚举，日期，时间，时间跨度，数字和数量的所有.NET需求
要安装Humanizer，请在程序包管理器控制台中运行以下命令
PM> Install-Package Humanizer
Ordinalize将数字转换为序数字符串，用于表示有序序列中的位置，如第1，第2，第3，第4：
1.Ordinalize() => "1st" 5.Ordinalize() => "5th"
然后你可以使用：
String.Format("{0} {1:MMMM yyyy}", date.Day.Ordinalize(), date)
自定义日期和时间格式字符串
date.ToString("MMMM d, yyyy")
或者如果你也需要“rd”：
string.Format("{0} {1}, {2}", date.ToString("MMMM"), date.Day.Ordinal(), date.ToString("yyyy"))
-
Ordinal()方法可以在这里找到
不，string.Format（）中没有任何内容可以为您提供序数（第1，第2，第3，第4等）。
您可以将其他答案中建议的日期格式与您自己的序数结合起来，如本答案中所示
有没有一种简单的方法可以在C＃中创建序数？
string Format(DateTime date) { int dayNo = date.Day; return string.Format("{0} {1}{2}, {3}", date.ToString("MMMM"), dayNo, AddOrdinal(dayNo), date.Year); }
public static class IntegerExtensions { /// /// converts an integer to its ordinal representation /// public static String AsOrdinal(this Int32 number) { if (number < 0) throw new ArgumentOutOfRangeException("number"); var work = number.ToString("n0"); var modOf100 = number % 100; if (modOf100 == 11 || modOf100 == 12 || modOf100 == 13) return work + "th"; switch (number % 10) { case 1: work += "st"; break; case 2: work += "nd"; break; case 3: work += "rd"; break; default: work += "th"; break; } return work; } }
certificate：
[TestFixture] class IntegerExtensionTests { [Test] public void TestCases_1s_10s_100s_1000s() { Assert.AreEqual("1st", 1.AsOrdinal()); Assert.AreEqual("2nd", 2.AsOrdinal()); Assert.AreEqual("3rd", 3.AsOrdinal()); foreach (var integer in Enumerable.Range(4, 6)) Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal()); Assert.AreEqual("11th", 11.AsOrdinal()); Assert.AreEqual("12th", 12.AsOrdinal()); Assert.AreEqual("13th", 13.AsOrdinal()); foreach (var integer in Enumerable.Range(14, 6)) Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal()); Assert.AreEqual("21st", 21.AsOrdinal()); Assert.AreEqual("22nd", 22.AsOrdinal()); Assert.AreEqual("23rd", 23.AsOrdinal()); foreach (var integer in Enumerable.Range(24, 6)) Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal()); Assert.AreEqual("31st", 31.AsOrdinal()); Assert.AreEqual("32nd", 32.AsOrdinal()); Assert.AreEqual("33rd", 33.AsOrdinal()); //then just jump to 100 Assert.AreEqual("101st", 101.AsOrdinal()); Assert.AreEqual("102nd", 102.AsOrdinal()); Assert.AreEqual("103rd", 103.AsOrdinal()); foreach (var integer in Enumerable.Range(104, 6)) Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal()); Assert.AreEqual("111th", 111.AsOrdinal()); Assert.AreEqual("112th", 112.AsOrdinal()); Assert.AreEqual("113th", 113.AsOrdinal()); foreach (var integer in Enumerable.Range(114, 6)) Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal()); Assert.AreEqual("121st", 121.AsOrdinal()); Assert.AreEqual("122nd", 122.AsOrdinal()); Assert.AreEqual("123rd", 123.AsOrdinal()); foreach (var integer in Enumerable.Range(124, 6)) Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal()); //then just jump to 1000 Assert.AreEqual("1,001st", 1001.AsOrdinal()); Assert.AreEqual("1,002nd", 1002.AsOrdinal()); Assert.AreEqual("1,003rd", 1003.AsOrdinal()); foreach (var integer in Enumerable.Range(1004, 6)) Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal()); Assert.AreEqual("1,011th", 1011.AsOrdinal()); Assert.AreEqual("1,012th", 1012.AsOrdinal()); Assert.AreEqual("1,013th", 1013.AsOrdinal()); foreach (var integer in Enumerable.Range(1014, 6)) Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal()); Assert.AreEqual("1,021st", 1021.AsOrdinal()); Assert.AreEqual("1,022nd", 1022.AsOrdinal()); Assert.AreEqual("1,023rd", 1023.AsOrdinal()); foreach (var integer in Enumerable.Range(1024, 6)) Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal()); } }
基于Rob Levine的回答和对该答案的评论…我已经在DateTime上作为扩展方法进行了调整，因此您可以调用：
var formattedDate = date.Friendly();
这是扩展方法：
public static class DateFormatter { public static string Friendly(this DateTime dt) { string suffix; switch (dt.Day) { case 1: case 21: case 31: suffix = "st"; break; case 2: case 22: suffix = "nd"; break; case 3: case 23: suffix = "rd"; break; default: suffix = "th"; break; } return string.Format("{0:MMMM} {1}{2}, {0:yyyy}", dt, dt.Day, suffix); } }
这是Ordinalize（）扩展的另一个版本，short和sweet：
public static string Ordinalize(this int x) { return x.ToString() + ((x % 10 == 1 && x != 11) ? "st" : (x % 10 == 2 && x != 12) ? "nd" : (x % 10 == 3 && x != 13) ? "rd" : "th"); }
然后像这样调用该扩展名
myDate.Day.Ordinalize()
要么
myAnyNumber.Ordinalize()
DateTime dt = new DateTime(args); String.Format("{0:ddd, MMM d, yyyy}", dt);
//“Sun，2008年3月9日”
使用以下代码：
DateTime thisDate1 = new DateTime(2011, 6, 10); Console.WriteLine("Today is " + thisDate1.ToString("MMMM dd, yyyy") + "."); DateTimeOffset thisDate2 = new DateTimeOffset(2011, 6, 10, 15, 24, 16, TimeSpan.Zero); Console.WriteLine("The current date and time: {0:MM/dd/yy H:mm:ss zzz}", thisDate2); // The example displays the following output: // Today is June 10, 2011. // The current date and time: 06/10/11 15:24:16 +00:00