在c#中有简单的方法来创建序数吗?

在c#中有一个简单的方法来创建一个数字的序数吗?例如:

1返回第1位 2返回第2 3返回第3 等

这是否可以通过String.Format()来完成，或者是否有可用的函数来完成?

当前回答

简单、干净、快捷

private static string GetOrdinalSuffix(int num)
{
    string number = num.ToString();
    if (number.EndsWith("11")) return "th";
    if (number.EndsWith("12")) return "th";
    if (number.EndsWith("13")) return "th";
    if (number.EndsWith("1")) return "st";
    if (number.EndsWith("2")) return "nd";
    if (number.EndsWith("3")) return "rd";
    return "th";
}

或者更好的是，作为一个扩展方法

public static class IntegerExtensions
{
    public static string DisplayWithSuffix(this int num)
    {
        string number = num.ToString();
        if (number.EndsWith("11")) return number + "th";
        if (number.EndsWith("12")) return number + "th";
        if (number.EndsWith("13")) return number + "th";
        if (number.EndsWith("1")) return number + "st";
        if (number.EndsWith("2")) return number + "nd";
        if (number.EndsWith("3")) return number + "rd";
        return number + "th";
    }
}

现在你可以打电话了

int a = 1;
a.DisplayWithSuffix();

甚至直接到

1.DisplayWithSuffix();

2014-10-17 21:12:10

其他回答

我使用这个扩展类:

public static class Int32Extensions
{
    public static string ToOrdinal(this int i)
    {
        return (i + "th")
            .Replace("1th", "1st")
            .Replace("2th", "2nd")
            .Replace("3th", "3rd");
    }
}

2016-09-23 16:15:29

public static string OrdinalSuffix(int ordinal)
{
    //Because negatives won't work with modular division as expected:
    var abs = Math.Abs(ordinal); 

    var lastdigit = abs % 10; 

    return 
        //Catch 60% of cases (to infinity) in the first conditional:
        lastdigit > 3 || lastdigit == 0 || (abs % 100) - lastdigit == 10 ? "th" 
            : lastdigit == 1 ? "st" 
            : lastdigit == 2 ? "nd" 
            : "rd";
}

2013-12-16 12:02:49

虽然这里有很多很好的答案，但我想还有另一个答案的空间，这一次是基于模式匹配，如果不是为了其他任何东西，那么至少是为了有争议的可读性

public static string Ordinals1(this int number)
{
    switch (number)
    {
        case int p when p % 100 == 11:
        case int q when q % 100 == 12:
        case int r when r % 100 == 13:
            return $"{number}th";
        case int p when p % 10 == 1:
            return $"{number}st";
        case int p when p % 10 == 2:
            return $"{number}nd";
        case int p when p % 10 == 3:
            return $"{number}rd";
        default:
            return $"{number}th";
    }
}

这个溶液有什么特别之处呢?我只是为各种其他解决方案添加了一些性能考虑因素

坦率地说，我怀疑性能对于这种特定的场景真的很重要(谁真的需要数百万个数字的序数呢)，但至少它提供了一些可供考虑的比较……

100万件供参考(当然，根据机器规格，您的米粒可能会有所不同) 使用模式匹配和划分(这个答案) ~ 622毫秒使用模式匹配和字符串(这个答案) ~ 1967毫秒有两个开关和划分(接受答案) ~ 637毫秒用一个开关和除法(另一个答案) ~ 725毫秒

void Main()
{
    var timer = new Stopwatch();
    var numbers = Enumerable.Range(1, 1000000).ToList();

    // 1
    timer.Reset();
    timer.Start();
    var results1 = numbers.Select(p => p.Ordinals1()).ToList();
    timer.Stop();
    timer.Elapsed.TotalMilliseconds.Dump("with pattern matching and divisions");

    // 2
    timer.Reset();
    timer.Start();
    var results2 = numbers.Select(p => p.Ordinals2()).ToList();
    timer.Stop();
    timer.Elapsed.TotalMilliseconds.Dump("with pattern matching and strings");

    // 3
    timer.Reset();
    timer.Start();
    var results3 = numbers.Select(p => p.Ordinals3()).ToList();
    timer.Stop();
    timer.Elapsed.TotalMilliseconds.Dump("with two switches and divisons");
    
    // 4
    timer.Reset();
    timer.Start();
    var results4 = numbers.Select(p => p.Ordinals4()).ToList();
    timer.Stop();
    timer.Elapsed.TotalMilliseconds.Dump("with one switche and divisons");
}

public static class Extensions
{
    public static string Ordinals1(this int number)
    {
        switch (number)
        {
            case int p when p % 100 == 11:
            case int q when q % 100 == 12:
            case int r when r % 100 == 13:
                return $"{number}th";
            case int p when p % 10 == 1:
                return $"{number}st";
            case int p when p % 10 == 2:
                return $"{number}nd";
            case int p when p % 10 == 3:
                return $"{number}rd";
            default:
                return $"{number}th";
        }
    }

    public static string Ordinals2(this int number)
    {
        var text = number.ToString();
        switch (text)
        {
            case string p when p.EndsWith("11"):
                return $"{number}th";
            case string p when p.EndsWith("12"):
                return $"{number}th";
            case string p when p.EndsWith("13"):
                return $"{number}th";
            case string p when p.EndsWith("1"):
                return $"{number}st";
            case string p when p.EndsWith("2"):
                return $"{number}nd";
            case string p when p.EndsWith("3"):
                return $"{number}rd";
            default:
                return $"{number}th";
        }
    }

    public static string Ordinals3(this int number)
    {
        switch (number % 100)
        {
            case 11:
            case 12:
            case 13:
                return $"{number}th";
        }

        switch (number % 10)
        {
            case 1:
                return $"{number}st";
            case 2:
                return $"{number}nd";
            case 3:
                return $"{number}rd";
            default:
                return $"{number}th";
        }
    }

    public static string Ordinals4(this int number)
    {
        var ones = number % 10;
        var tens = Math.Floor(number / 10f) % 10;
        if (tens == 1)
        {
            return $"{number}th";
        }

        switch (ones)
        {
            case 1:
                return $"{number}th";
            case 2:
                return $"{number}nd";
            case 3:
                return $"{number}rd";
            default:
                return $"{number}th";
        }
    }
}

2019-10-14 14:03:47

简单、干净、快捷

private static string GetOrdinalSuffix(int num)
{
    string number = num.ToString();
    if (number.EndsWith("11")) return "th";
    if (number.EndsWith("12")) return "th";
    if (number.EndsWith("13")) return "th";
    if (number.EndsWith("1")) return "st";
    if (number.EndsWith("2")) return "nd";
    if (number.EndsWith("3")) return "rd";
    return "th";
}

或者更好的是，作为一个扩展方法

public static class IntegerExtensions
{
    public static string DisplayWithSuffix(this int num)
    {
        string number = num.ToString();
        if (number.EndsWith("11")) return number + "th";
        if (number.EndsWith("12")) return number + "th";
        if (number.EndsWith("13")) return number + "th";
        if (number.EndsWith("1")) return number + "st";
        if (number.EndsWith("2")) return number + "nd";
        if (number.EndsWith("3")) return number + "rd";
        return number + "th";
    }
}

现在你可以打电话了

int a = 1;
a.DisplayWithSuffix();

甚至直接到

1.DisplayWithSuffix();

2014-10-17 21:12:10

类似于Ryan的解决方案，但更基本，我只是使用一个普通数组，并使用日期来查找正确的序数:

private string[] ordinals = new string[] {"","st","nd","rd","th","th","th","th","th","th","th","th","th","th","th","th","th","th","th","th","th","st","nd","rd","th","th","th","th","th","th","th","st" };
DateTime D = DateTime.Now;
String date = "Today's day is: "+ D.Day.ToString() + ordinals[D.Day];

我没有这个需要，但是我假设如果您想要多语言支持，您可以使用多维数组。

根据我在大学的记忆，这种方法只需要服务器做最少的工作。

2013-05-02 13:48:06

在c#中有简单的方法来创建序数吗?

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