使用DateTime.ToString()时获取日期后缀

时间:2010-01-12 17:10:22

标签: c# .net date

使用DateTime.ToString()格式化日期时是否可以包含日后缀?

例如,我想以下列格式打印日期 - 2009年7月27日星期一。但是,我可以使用DateTime.ToString()查找的最接近的示例是2009年7月27日星期一。

我可以使用DateTime.ToString()执行此操作,还是必须回退到我自己的代码?

18 个答案:

答案 0 :(得分:219)

使用开关的另一个选项:

string GetDaySuffix(int day)
{
    switch (day)
    {
        case 1:
        case 21:
        case 31:
            return "st";
        case 2:
        case 22:
            return "nd";
        case 3:
        case 23:
            return "rd";
        default:
            return "th";
    }
}

答案 1 :(得分:53)

作为参考,我总是使用/参考SteveX String Formatting 并且在任何可用变量中似乎没有任何“th”,但您可以使用

轻松构建字符串
string.Format("{0:dddd dd}{1} {0:MMMM yyyy}", DateTime.Now, (?));

然后你必须为1提供“st”,为2提供“nd”,为3提供“rd”,为所有其他提供“th”,并且可以用“?:”语句提供。 / p>

var now = DateTime.Now;
(now.Day % 10 == 1 && now.Day != 11) ? "st"
: (now.Day % 10 == 2 && now.Day != 12) ? "nd"
: (now.Day % 10 == 3 && now.Day != 13) ? "rd"
: "th"

答案 2 :(得分:28)

使用几种扩展方法:

namespace System
{
    public static class IntegerExtensions
    {
        public static string ToOccurrenceSuffix(this int integer)
        {
            switch (integer % 100)
            {
                case 11:
                case 12:
                case 13:
                    return "th";
            }
            switch (integer % 10)
            {
                case 1:
                    return "st";
                case 2:
                    return "nd";
                case 3:
                    return "rd";
                default:
                    return "th";
            }
        }
    }   

    public static class DateTimeExtensions
    {
        public static string ToString(this DateTime dateTime, string format, bool useExtendedSpecifiers)
        {
            return dateTime.ToString(format)
                .Replace("nn", dateTime.Day.ToOccurrenceSuffix().ToLower())
                .Replace("NN", dateTime.Day.ToOccurrenceSuffix().ToUpper());
        } 
    }
}

用法:

return DateTime.Now.ToString("dddd, dnn MMMM yyyy", useExtendedSpecifiers: true);
// Friday, 7th March 2014

注意:整数扩展方法可用于任何数字,而不仅仅是1到31.例如

return 332211.ToOccurrenceSuffix();
// th

答案 3 :(得分:12)

另一种选择是使用Modulo Operator

public string CreateDateSuffix(DateTime date)
{
    // Get day...
    var day = date.Day;

    // Get day modulo...
    var dayModulo = day%10;

    // Convert day to string...
    var suffix = day.ToString(CultureInfo.InvariantCulture);

    // Combine day with correct suffix...
    suffix += (day == 11 || day == 12 || day == 13) ? "th" :
        (dayModulo == 1) ? "st" :
        (dayModulo == 2) ? "nd" :
        (dayModulo == 3) ? "rd" :
        "th";

    // Return result...
    return suffix;
}

然后,您可以通过传入 DateTime 对象作为参数来调用上述方法,例如:

// Get date suffix for 'October 8th, 2019':
var suffix = CreateDateSuffix(new DateTime(2019, 10, 8));

有关 DateTime 构造函数的详细信息,请参阅Microsoft Docs Page

答案 4 :(得分:7)

这是包括第11,12和13位的扩展版本:

DateTime dt = DateTime.Now;
string d2d = dt.ToString("dd").Substring(1);
string daySuffix =
    (dt.Day == 11 || dt.Day == 12 || dt.Day == 13) ? "th"
    : (d2d == "1") ? "st"
    : (d2d == "2") ? "nd"
    : (d2d == "3") ? "rd"
    : "th";

答案 5 :(得分:5)

更新

NuGet包装:
https://www.nuget.org/packages/DateTimeToStringWithSuffix

例:
https://dotnetfiddle.net/zXQX7y

支持:
.NET Core 1.0及更高版本 .NET Framework 4.5和高级


这是一种扩展方法(因为每个人都喜欢扩展方法),以Lazlow的答案为基础(选择Lazlow的方法很容易阅读)。

ToString()上的常规DateTime方法类似,但如果格式包含ddd,则会自动添加后缀。< / p>

/// <summary>
/// Return a DateTime string with suffix e.g. "st", "nd", "rd", "th"
/// So a format "dd-MMM-yyyy" could return "16th-Jan-2014"
/// </summary>
public static string ToStringWithSuffix(this DateTime dateTime, string format, string suffixPlaceHolder = "$") {
    if(format.LastIndexOf("d", StringComparison.Ordinal) == -1 || format.Count(x => x == 'd') > 2) {
        return dateTime.ToString(format);
    }

    string suffix;
    switch(dateTime.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;
    }

    var formatWithSuffix = format.Insert(format.LastIndexOf("d", StringComparison.InvariantCultureIgnoreCase) + 1, suffixPlaceHolder);
    var date = dateTime.ToString(formatWithSuffix);

    return date.Replace(suffixPlaceHolder, suffix);
}

答案 6 :(得分:5)

以@ Lazlow对完整解决方案的回答,以下是完全可重用的扩展方法,并附带示例用法;

internal static string HumanisedDate(this DateTime date)
{
    string ordinal;

    switch (date.Day)
    {
        case 1:
        case 21:
        case 31:
            ordinal = "st";
            break;
        case 2:
        case 22:
            ordinal = "nd";
            break;
        case 3:
        case 23:
            ordinal = "rd";
            break;
        default:
            ordinal = "th";
            break;
    }

    return string.Format("{0:dddd dd}{1} {0:MMMM yyyy}", date, ordinal);
} 

要使用它,您只需在DateTime对象上调用它;

var myDate = DateTime.Now();
var myDateString = myDate.HumanisedFormat()

哪个会给你:

  

2016年6月17日星期五

答案 7 :(得分:2)

我认为这是一个很好的解决方案,涵盖第111等数字:

private string daySuffix(int day)
{
    if (day > 0)
    {
        if (day % 10 == 1 && day % 100 != 11)
            return "st";
        else if (day % 10 == 2 && day % 100 != 12)
            return "nd";
        else if (day % 10 == 3 && day % 100 != 13)
            return "rd";
        else
            return "th";
    }
    else
        return string.Empty;
}

答案 8 :(得分:0)

这里的价值在于我使用以下答案的最终解决方案

     DateTime dt = DateTime.Now;
        string d2d = dt.ToString("dd").Substring(1); 

        string suffix =
       (dt.Day == 11 || dt.Day == 12 || dt.Day == 13) ? "th"
       : (d2d == "1") ? "st"
       : (d2d == "2") ? "nd"
       : (d2d == "3") ? "rd"
       : "th";


        Date.Text = DateTime.Today.ToString("dddd d") + suffix + " " + DateTime.Today.ToString("MMMM") + DateTime.Today.ToString(" yyyy"); 

答案 9 :(得分:0)

string datestring;    
// datestring = DateTime.Now.ToString("dd MMMM yyyy"); // 16 January 2021

    // code to add 'st' ,'nd', 'rd' and 'th' with day of month 
    // DateTime todaysDate = DateTime.Now.Date; // enable this line for current date 
    DateTime todaysDate = DateTime.Parse("01-13-2021"); // custom date to verify code // 13th January 2021
    int day = todaysDate.Day;
    string dateSuffix;

    if(day==1 || day==21 || day==31){
        dateSuffix= "st";
    }else if(day==2 || day==22 ){
        dateSuffix= "nd";
    }else if(day==3 || day==23 ){
        dateSuffix= "rd";
    }else{
        dateSuffix= "th";
    }
    datestring= day+dateSuffix+" "+todaysDate.ToString("MMMM")+" "+todaysDate.ToString("yyyy");
    Console.WriteLine(datestring);

答案 10 :(得分:0)

查看humanizr: https://github.com/Humanizr/Humanizer#date-time-to-ordinal-words

new DateTime(2015, 1, 1).ToOrdinalWords() => "1st January 2015"
new DateTime(2015, 2, 12).ToOrdinalWords() => "12th February 2015"
new DateTime(2015, 3, 22).ToOrdinalWords() => "22nd March 2015"
// for English US locale
new DateTime(2015, 1, 1).ToOrdinalWords() => "January 1st, 2015"
new DateTime(2015, 2, 12).ToOrdinalWords() => "February 12th, 2015"
new DateTime(2015, 3, 22).ToOrdinalWords() => "March 22nd, 2015"

发布此消息后,我立即意识到@Gumzle提出了同样的建议,但我错过了他的帖子,因为它被埋在了代码片段中。因此,这是他的答案,其中包含足够的代码,使得像我这样的人可以快速滚动浏览该代码。

答案 11 :(得分:0)

对于那些乐于使用外部依赖关系的人(在这种情况下,使用奇妙的Humanizr .net),就像

一样简单

dateVar.Day.Ordinalize(); \\ 1st, 4th etc depending on the value of dateVar

答案 12 :(得分:0)

获取日期后缀。 (静态功能)

public static string GetSuffix(this string day)
{
    string suffix = "th";

    if (int.Parse(day) < 11 || int.Parse(day) > 20)
    {
        day = day.ToCharArray()[day.ToCharArray().Length - 1].ToString();
        switch (day)
        {
            case "1":
                suffix = "st";
                break;
            case "2":
                suffix = "nd";
                break;
            case "3":
                suffix = "rd";
                break;
        }
    }

    return suffix;
}

参考:https://www.aspsnippets.com/Articles/Display-st-nd-rd-and-th-suffix-after-day-numbers-in-Formatted-Dates-using-C-and-VBNet.aspx

答案 13 :(得分:0)

public static String SuffixDate(DateTime date)       {          string ordinal;

     switch (date.Day)
     {
        case 1:
        case 21:
        case 31:
           ordinal = "st";
           break;
        case 2:
        case 22:
           ordinal = "nd";
           break;
        case 3:
        case 23:
           ordinal = "rd";
           break;
        default:
           ordinal = "th";
           break;
     }
     if (date.Day < 10)
        return string.Format("{0:d}{2} {1:MMMM yyyy}", date.Day, date, ordinal);
     else
        return string.Format("{0:dd}{1} {0:MMMM yyyy}", date, ordinal);
  }

答案 14 :(得分:0)

我是这样做的,它解决了其他例子中给出的一些问题。

    public static string TwoLetterSuffix(this DateTime @this)
    {
        var dayMod10 = @this.Day % 10;

        if (dayMod10 > 3 || dayMod10 == 0 || (@this.Day >= 10 && @this.Day <= 19))
        {
            return "th";
        }
        else if(dayMod10 == 1)
        {
            return "st";
        }
        else if (dayMod10 == 2)
        {
            return "nd";
        }
        else
        {
            return "rd";
        }
    }

答案 15 :(得分:0)

廉价而开朗的VB解决方案:

litDate.Text = DatePart("dd", Now) & GetDateSuffix(DatePart("dd", Now))

Function GetDateSuffix(ByVal dateIn As Integer) As String

    '// returns formatted date suffix

    Dim dateSuffix As String = ""
    Select Case dateIn
        Case 1, 21, 31
            dateSuffix = "st"
        Case 2, 22
            dateSuffix = "nd"
        Case 3, 23
            dateSuffix = "rd"
        Case Else
            dateSuffix = "th"
    End Select

    Return dateSuffix

End Function

答案 16 :(得分:-2)

使用最后一个字符串字符的另一个选项:

public static string getDayWithSuffix(int day) {
 string d = day.ToString();
 if (day < 11 || day > 13) {
  if (d.EndsWith("1")) {
   d += "st";
  } else if (d.EndsWith("2")) {
   d += "nd";
  } else if (d.EndsWith("3")) {
   d += "rd";
  } else {
   d += "th";
 } else {
  d += "th";
 }
 return d;
}

答案 17 :(得分:-2)

在MSDN文档中,没有提到可以将17转换为17的文化。所以你应该通过代码隐藏手动完成它。或者构建一个...你可以构建一个能够做到这一点的函数。

public string CustomToString(this DateTime date)
    {
        string dateAsString = string.empty;
        <here wright your code to convert 17 to 17th>
        return dateAsString;
    }