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?
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
杰西版本的斯图和萨姆贾德森版本的我的版本:)
包含单元测试,以显示接受的答案是不正确的,当数字< 1
/// <summary>
/// Get the ordinal value of positive integers.
/// </summary>
/// <remarks>
/// Only works for english-based cultures.
/// Code from: http://stackoverflow.com/questions/20156/is-there-a-quick-way-to-create-ordinals-in-c/31066#31066
/// With help: http://www.wisegeek.com/what-is-an-ordinal-number.htm
/// </remarks>
/// <param name="number">The number.</param>
/// <returns>Ordinal value of positive integers, or <see cref="int.ToString"/> if less than 1.</returns>
public static string Ordinal(this int number)
{
const string TH = "th";
string s = number.ToString();
// Negative and zero have no ordinal representation
if (number < 1)
{
return s;
}
number %= 100;
if ((number >= 11) && (number <= 13))
{
return s + TH;
}
switch (number % 10)
{
case 1: return s + "st";
case 2: return s + "nd";
case 3: return s + "rd";
default: return s + TH;
}
}
[Test]
public void Ordinal_ReturnsExpectedResults()
{
Assert.AreEqual("-1", (1-2).Ordinal());
Assert.AreEqual("0", 0.Ordinal());
Assert.AreEqual("1st", 1.Ordinal());
Assert.AreEqual("2nd", 2.Ordinal());
Assert.AreEqual("3rd", 3.Ordinal());
Assert.AreEqual("4th", 4.Ordinal());
Assert.AreEqual("5th", 5.Ordinal());
Assert.AreEqual("6th", 6.Ordinal());
Assert.AreEqual("7th", 7.Ordinal());
Assert.AreEqual("8th", 8.Ordinal());
Assert.AreEqual("9th", 9.Ordinal());
Assert.AreEqual("10th", 10.Ordinal());
Assert.AreEqual("11th", 11.Ordinal());
Assert.AreEqual("12th", 12.Ordinal());
Assert.AreEqual("13th", 13.Ordinal());
Assert.AreEqual("14th", 14.Ordinal());
Assert.AreEqual("20th", 20.Ordinal());
Assert.AreEqual("21st", 21.Ordinal());
Assert.AreEqual("22nd", 22.Ordinal());
Assert.AreEqual("23rd", 23.Ordinal());
Assert.AreEqual("24th", 24.Ordinal());
Assert.AreEqual("100th", 100.Ordinal());
Assert.AreEqual("101st", 101.Ordinal());
Assert.AreEqual("102nd", 102.Ordinal());
Assert.AreEqual("103rd", 103.Ordinal());
Assert.AreEqual("104th", 104.Ordinal());
Assert.AreEqual("110th", 110.Ordinal());
Assert.AreEqual("111th", 111.Ordinal());
Assert.AreEqual("112th", 112.Ordinal());
Assert.AreEqual("113th", 113.Ordinal());
Assert.AreEqual("114th", 114.Ordinal());
Assert.AreEqual("120th", 120.Ordinal());
Assert.AreEqual("121st", 121.Ordinal());
Assert.AreEqual("122nd", 122.Ordinal());
Assert.AreEqual("123rd", 123.Ordinal());
Assert.AreEqual("124th", 124.Ordinal());
}
其他回答
这里是DateTime扩展类。复制,粘贴和享受
public static class DateTimeExtensions
{
public static string ToStringWithOrdinal(this DateTime d)
{
var result = "";
bool bReturn = false;
switch (d.Day % 100)
{
case 11:
case 12:
case 13:
result = d.ToString("dd'th' MMMM yyyy");
bReturn = true;
break;
}
if (!bReturn)
{
switch (d.Day % 10)
{
case 1:
result = d.ToString("dd'st' MMMM yyyy");
break;
case 2:
result = d.ToString("dd'nd' MMMM yyyy");
break;
case 3:
result = d.ToString("dd'rd' MMMM yyyy");
break;
default:
result = d.ToString("dd'th' MMMM yyyy");
break;
}
}
if (result.StartsWith("0")) result = result.Substring(1);
return result;
}
}
结果:
2014年10月9日
你得自己动手了。在我的脑海中:
public static string Ordinal(this int number)
{
var work = number.ToString();
if ((number % 100) == 11 || (number % 100) == 12 || (number % 100) == 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;
}
你可以这样做
Console.WriteLine(432.Ordinal());
针对11/12/13例外进行了编辑。我确实从我的头顶说过:-)
为1011编辑-其他人已经修复了这个问题,只是想确保其他人不会抓取这个错误的版本。
这是dart中的实现,可以根据语言进行修改。
String getOrdinalSuffix(int num){
if (num.toString().endsWith("11")) return "th";
if (num.toString().endsWith("12")) return "th";
if (num.toString().endsWith("13")) return "th";
if (num.toString().endsWith("1")) return "st";
if (num.toString().endsWith("2")) return "nd";
if (num.toString().endsWith("3")) return "rd";
return "th";
}
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();
c# 8和9中接受的带有开关表达式和模式匹配的答案。
没有不必要的字符串转换或分配。
string.Concat(number, number < 0 ? "" : (number % 100) switch
{
11 or 12 or 13 => "th",
int n => (n % 10) switch
{
1 => "st",
2 => "nd",
3 => "rd",
_ => "th",
}
})
或者是不友好的一句话:
$"{number}{(number < 0 ? "" : (number % 100) switch { 11 or 12 or 13 => "th", int n => (n % 10) switch { 1 => "st", 2 => "nd", 3 => "rd", _ => "th" }})}"