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
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
其他回答
FWIW，对于MS-SQL，这个表达式将完成工作。将第一个WHEN (WHEN num % 100 IN (11,12,13) THEN 'th')作为列表中的第一个，因为这依赖于在其他尝试之前尝试。
CASE
WHEN num % 100 IN (11, 12, 13) THEN 'th' -- must be tried first
WHEN num % 10 = 1 THEN 'st'
WHEN num % 10 = 2 THEN 'nd'
WHEN num % 10 = 3 THEN 'rd'
ELSE 'th'
END AS Ordinal
对于Excel:
=MID("thstndrdth",MIN(9,2*RIGHT(A1)*(MOD(A1-11,100)>2)+1),2)
表达式(MOD(A1- 11100)>2)对于除以11,12,13结尾的任何数字(FALSE = 0)外的所有数字都是TRUE(1)。因此2 * RIGHT(A1) * (MOD(A1- 11100)>2) +1)对于11/12/13最终为1，否则: 1等于3 2点到5点， 3至7点 其他:9 -所需的2个字符从该位置开始的“第thstndrdth”中选择。
如果你真的想把它直接转换成SQL，这对我来说适用于一些测试值:
DECLARE @n as int
SET @n=13
SELECT SubString( 'thstndrdth'
, (SELECT MIN(value) FROM
(SELECT 9 as value UNION
SELECT 1+ (2* (ABS(@n) % 10) * CASE WHEN ((ABS(@n)+89) % 100)>2 THEN 1 ELSE 0 END)
) AS Mins
)
, 2
)
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());
}
我使用的另一种替代方法是基于所有其他建议，但不需要特殊的大小写:
public static string DateSuffix(int day)
{
if (day == 11 | day == 12 | day == 13) return "th";
Math.DivRem(day, 10, out day);
switch (day)
{
case 1:
return "st";
case 2:
return "nd";
case 3:
return "rd";
default:
return "th";
}
}
虽然我还没有对此进行基准测试，但通过避免所有条件case语句，您应该能够获得更好的性能。
这是java，但是移植到c#很简单:
public class NumberUtil {
final static String[] ORDINAL_SUFFIXES = {
"th", "st", "nd", "rd", "th", "th", "th", "th", "th", "th"
};
public static String ordinalSuffix(int value) {
int n = Math.abs(value);
int lastTwoDigits = n % 100;
int lastDigit = n % 10;
int index = (lastTwoDigits >= 11 && lastTwoDigits <= 13) ? 0 : lastDigit;
return ORDINAL_SUFFIXES[index];
}
public static String toOrdinal(int n) {
return new StringBuffer().append(n).append(ordinalSuffix(n)).toString();
}
}
注意，如果在一个紧密循环中生成大量序数，减少条件和使用数组查找应该会提高性能。然而，我也承认这并不像case语句解决方案那样可读。
编辑:正如YM_Industries在评论中指出的那样，samjudson的答案确实适用于超过1000的数字，nickf的评论似乎已经消失了，我不记得我看到的问题是什么。留下这个答案在这里比较时间。
正如nickf在评论中指出的(编辑:现在丢失了)，很多数字> 999都不起作用。
以下是一个基于samjudson的公认答案的修改版本。
public static String GetOrdinal(int i)
{
String res = "";
if (i > 0)
{
int j = (i - ((i / 100) * 100));
if ((j == 11) || (j == 12) || (j == 13))
res = "th";
else
{
int k = i % 10;
if (k == 1)
res = "st";
else if (k == 2)
res = "nd";
else if (k == 3)
res = "rd";
else
res = "th";
}
}
return i.ToString() + res;
}
同样，Shahzad Qureshi使用字符串操作的回答也很好，但它确实有性能损失。为了生成大量这样的类型，LINQPad示例程序使字符串版本比整数版本慢6-7倍(尽管您必须生成很多才会注意到)。
LINQPad例子:
void Main()
{
"Examples:".Dump();
foreach(int i in new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 22, 113, 122, 201, 202, 211, 212, 2013, 1000003, 10000013 })
Stuff.GetOrdinal(i).Dump();
String s;
System.Diagnostics.Stopwatch sw = System.Diagnostics.Stopwatch.StartNew();
for(int iter = 0; iter < 100000; iter++)
foreach(int i in new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 22, 113, 122, 201, 202, 211, 212, 2013, 1000003, 1000013 })
s = Stuff.GetOrdinal(i);
"Integer manipulation".Dump();
sw.Elapsed.Dump();
sw.Restart();
for(int iter = 0; iter < 100000; iter++)
foreach(int i in new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 22, 113, 122, 201, 202, 211, 212, 2013, 1000003, 1000013 })
s = (i.ToString() + Stuff.GetOrdinalSuffix(i));
"String manipulation".Dump();
sw.Elapsed.Dump();
}
public class Stuff
{
// Use integer manipulation
public static String GetOrdinal(int i)
{
String res = "";
if (i > 0)
{
int j = (i - ((i / 100) * 100));
if ((j == 11) || (j == 12) || (j == 13))
res = "th";
else
{
int k = i % 10;
if (k == 1)
res = "st";
else if (k == 2)
res = "nd";
else if (k == 3)
res = "rd";
else
res = "th";
}
}
return i.ToString() + res;
}
// Use string manipulation
public static 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";
}
}
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