您是否考虑过使用字典查找表?

enum GroupTypes
{
    TheGroup,
    TheOtherGroup
}

Dictionary<string, GroupTypes> GroupTypeLookup = new Dictionary<string, GroupTypes>();
// initialize lookup table:
GroupTypeLookup.Add("OEM", TheGroup);
GroupTypeLookup.Add("CMB", TheOtherGroup);

然后，您可以使用GroupTypeLookup.TryGetValue()在读取字符串时查找它。

在@Even Mien的回答之后，我试着走得更远一点，使它通用，我似乎几乎在那里，但有一种情况仍然抗拒，我可能可以简化我的代码一点。 我张贴在这里，如果有人看到我可以改进，特别是使它工作，因为我不能从字符串赋值

到目前为止，我有以下结果:

        Console.WriteLine(TestEnum.Test1);//displays "TEST1"

        bool test = "TEST1" == TestEnum.Test1; //true

        var test2 = TestEnum.Test1; //is TestEnum and has value

        string test3 = TestEnum.Test1; //test3 = "TEST1"

        var test4 = TestEnum.Test1 == TestEnum.Test2; //false
         EnumType<TestEnum> test5 = "TEST1"; //works fine

        //TestEnum test5 = "string"; DOESN'T compile .... :(:(

奇迹发生的地方:

public abstract  class EnumType<T>  where T : EnumType<T>   
{

    public  string Value { get; set; }

    protected EnumType(string value)
    {
        Value = value;
    }


    public static implicit operator EnumType<T>(string s)
    {
        if (All.Any(dt => dt.Value == s))
        {
            Type t = typeof(T);

            ConstructorInfo ci = t.GetConstructor(BindingFlags.Instance | BindingFlags.NonPublic,null, new Type[] { typeof(string) }, null);

            return (T)ci.Invoke(new object[] {s});
        }
        else
        {
            return null;
        }
    }

    public static implicit operator string(EnumType<T> dt)
    {
        return dt?.Value;
    }


    public static bool operator ==(EnumType<T> ct1, EnumType<T> ct2)
    {
        return (string)ct1 == (string)ct2;
    }

    public static bool operator !=(EnumType<T> ct1, EnumType<T> ct2)
    {
        return !(ct1 == ct2);
    }


    public override bool Equals(object obj)
    {
        try
        {
            return (string)obj == Value;
        }
        catch
        {
            return false;
        }
    }

    public override int GetHashCode()
    {
        return Value.GetHashCode();
    }

    public static IEnumerable<T> All
     => typeof(T).GetProperties()
       .Where(p => p.PropertyType == typeof(T))
       .Select(x => (T)x.GetValue(null, null));



}

我只需要为我的枚举声明这个:

public class TestEnum : EnumType<TestEnum> 
{

    private TestEnum(string value) : base(value)
    {}

    public static TestEnum Test1 { get { return new TestEnum("TEST1"); } }
    public static TestEnum Test2 { get { return new TestEnum("TEST2"); } }
}

我甚至按照@Even(通过类X和公共静态X成员)的建议实现了一些枚举，只是为了后来发现，从。net 4.5开始，有了正确的ToString()方法。

现在我将所有内容重新实现为枚举。

基于https://stackoverflow.com/a/1343517/1818723，我提出了一个枚举与TryParse方法

public class FancyStringEnum
{
    private FancyStringEnum(string value) { Value = value; }

    public string Value { get; private set; }

    private static List<FancyStringEnum> choices = new List<FancyStringEnum>
    {
        new FancyStringEnum("Small") ,
        new FancyStringEnum("Big Thing") ,
        new FancyStringEnum("Value with Spaces")
    };

    public static FancyStringEnum Small { get { return choices[0]; } }
    public static FancyStringEnum BigThing { get { return choices[1]; } }
    public static FancyStringEnum ValueWithSpaces { get { return choices[2]; } }

    public override string ToString()
    {
        return Value;
    }

    public static bool TryParse(string value, bool ignoreCase, out FancyStringEnum result)
    {
        var sc = StringComparison.InvariantCulture;
        if (ignoreCase)
            sc = StringComparison.InvariantCultureIgnoreCase;

        foreach (var choice in choices)
        {

            if (choice.Value.Equals(value, sc))
            {
                result = choice;
                return true;
            }
        }

        result = new FancyStringEnum(null);
        return false;
    }

    public static FancyStringEnum Parse(string value, bool ignoreCase)
    {
        var sc = StringComparison.InvariantCulture;
        if (ignoreCase)
            sc = StringComparison.InvariantCultureIgnoreCase;

        foreach (var choice in choices)
        {

            if (choice.Value.Equals(value, sc))
            {
                return choice;
            }
        }

        return new FancyStringEnum(null);
    }
}

传入类型安全的字符串值作为参数:

public static void Do(string message, FancyStringEnum value)
{
    if (value == FancyStringEnum.Small)
    {
        //do something
    } else if (value == FancyStringEnum.BigThing)
    {
        //do something else
    }
}

TryParse和Parse在行动:

string something = "something"; //substiture with "small" to see it parsed 
if (FancyStringEnum.TryParse(something, true, out var se))
    Console.WriteLine(se.Value);
else
    Console.WriteLine($"unable to parse {something}");

//or    

var v2 = FancyStringEnum.Parse(something, true);
if (v2.Value == null)
    Console.WriteLine($"unable to parse {something}");
else
    Console.WriteLine(v2.Value); //do something with parsed enum

    

是否可以提取基类，用更少的代码创建StringEnums

您是否考虑过使用字典查找表?

enum GroupTypes
{
    TheGroup,
    TheOtherGroup
}

Dictionary<string, GroupTypes> GroupTypeLookup = new Dictionary<string, GroupTypes>();
// initialize lookup table:
GroupTypeLookup.Add("OEM", TheGroup);
GroupTypeLookup.Add("CMB", TheOtherGroup);

然后，您可以使用GroupTypeLookup.TryGetValue()在读取字符串时查找它。

c#不支持枚举字符串，但在大多数情况下，您可以使用List或Dictionary来获得所需的效果。

打印通过/失败结果:

List<string> PassFail = new List<string> { "FAIL", "PASS" };
bool result = true;
Console.WriteLine("Test1: " + PassFail[result.GetHashCode()]);