下面是我对这个问题的看法，使用c# 9.0语法来保持整洁。我为枚举定义了一个基类:

public class StringEnum
{
    protected StringEnum(string value) { Value = value; }
    public string Value { get; }
    public override string ToString() => Value;
}

创建新的枚举样式类型是简单而紧凑的:

public class GroupTypes : StringEnum
{ 
    private GroupTypes(string value) : base(value) {}

    public static readonly GroupTypes TheGroup = new("OEM");
    public static readonly GroupTypes TheOtherGroup = new("CMB");
}

像这样使用它:

void Example(GroupTypes groupType)
{
    Console.WriteLine(groupType); // Will print "OEM" or "CMB"
    if (groupType == GroupTypes.TheGroup) { ... }
}

你也可以给StringEnum添加更多的功能，这样你的所有子类都可以使用(例如，实现IComparable和重写Equals和GetHashCode)

这是一种将它用作强类型参数或字符串的方法:

public class ClassLikeEnum
{
    public string Value
    {
        get;
        private set;
    }

    ClassLikeEnum(string value) 
    {
        Value = value;
    }

    public static implicit operator string(ClassLikeEnum c)
    {
        return c.Value;
    }

    public static readonly ClassLikeEnum C1 = new ClassLikeEnum("RandomString1");
    public static readonly ClassLikeEnum C2 = new ClassLikeEnum("RandomString2");
}

我想完全避免使用字符串字面量，而且我也不需要在项目描述中有空间。更重要的是，我想要有一种机制来检查所提供的字符串是否是一个有效的项目，所以我想出了这个解决方案:

public class Seasons
{
    public static string Spring { get; }
    public static string Summer { get; }
    public static string Fall { get; }
    public static string Winter { get; }

    public static bool IsValid(string propertyName)
    {
        if (string.IsNullOrEmpty(propertyName))
        {
            return false;
        }

        try
        {           
            return typeof(Seasons).GetProperty(propertyName) != null;
        }
        catch
        {
            return false;
        }       
    }
}

它是这样工作的:

void Main()
{
    string s = nameof(Seasons.Fall);
    Console.WriteLine($"Fall is valid: {Seasons.IsValid(s)}"); // true

    s = "WrongSeason";
    Console.WriteLine($"WrongSeason is valid: {Seasons.IsValid(s)}"); // false
}

我尝试将IsValid()重构为基类并使用反射来读取类型(MethodBase.GetCurrentMethod(). declaringtype)，但由于我希望它是静态的，所以它返回基类类型，而不是继承的类型。你的补救措施将是非常欢迎的!这就是我想要达到的目标:

public  class Seasons : ConstantStringsBase
{
    // ... same
}

public  class ConstantStringsBase
{
    public static bool IsValid(string propertyName)
    {       
        return MethodBase.GetCurrentMethod().DeclaringType.GetProperty(propertyName) != null;
    }
}

使用类。

编辑:更好的例子

class StarshipType
{
    private string _Name;
    private static List<StarshipType> _StarshipTypes = new List<StarshipType>();

    public static readonly StarshipType Ultralight = new StarshipType("Ultralight");
    public static readonly StarshipType Light = new StarshipType("Light");
    public static readonly StarshipType Mediumweight = new StarshipType("Mediumweight");
    public static readonly StarshipType Heavy = new StarshipType("Heavy");
    public static readonly StarshipType Superheavy = new StarshipType("Superheavy");

    public string Name
    {
        get { return _Name; }
        private set { _Name = value; }
    }

    public static IList<StarshipType> StarshipTypes
    {
        get { return _StarshipTypes; }
    }

    private StarshipType(string name, int systemRatio)
    {
        Name = name;
        _StarshipTypes.Add(this);
    }

    public static StarshipType Parse(string toParse)
    {
        foreach (StarshipType s in StarshipTypes)
        {
            if (toParse == s.Name)
                return s;
        }
        throw new FormatException("Could not parse string.");
    }
}

我要分享一个不同的答案。我想发送一个字符串来返回一个字符串。例如:Airport = A, Railway = R。

public enum LocationType
    {
        AIRPORT = 1,
        RAILWAY = 2,
        DOWNTOWN = 3
    }
    
public enum ShortLocationType
    {
        A = 1,
        R = 2,
        D = 3
    }

((ShortLocationType)(int)Enum.Parse(typeof(LocationType), "AIRPORT")).ToString();

输出“”

((ShortLocationType)(int)Enum.Parse(typeof(LocationType), "DOWNTOWN")).ToString();

输出“D”

我使用了在之前的回答中提到的结构，但去掉了任何复杂性。对我来说，这就像创建一个字符串枚举。它的使用方式与枚举的使用方式相同。

    struct ViewTypes
    {
        public const string View1 = "Whatever string you like";
        public const string View2 = "another string";
    }

使用示例:

   switch( some_string_variable )
   {
      case ViewTypes.View1: /* do something */ break;
      case ViewTypes.View2: /* do something else */ break;
   }