(见下面我用我接受的答案创建的解决方案)

我试图提高一些涉及反射的代码的可维护性。该应用程序有一个。net Remoting接口,其中包括一个名为Execute的方法,用于访问应用程序中未包含在其发布的远程接口中的部分。

下面是应用程序如何指定属性(本例中的静态属性),这意味着可以通过Execute访问:

RemoteMgr.ExposeProperty("SomeSecret", typeof(SomeClass), "SomeProperty");

因此远程用户可以调用:

string response = remoteObject.Execute("SomeSecret");

应用程序将使用反射来查找SomeClass。SomeProperty并以字符串形式返回它的值。

不幸的是,如果有人重命名SomeProperty并且忘记更改ExposeProperty()的第3个参数,就会破坏这个机制。

我需要的是:

SomeClass.SomeProperty.GetTheNameOfThisPropertyAsAString()

作为ExposeProperty中的第三个parm,这样重构工具就会照顾到重命名。

有办法做到这一点吗?

好吧,这是我最终创建的(基于我选择的答案和他引用的问题):

// <summary>
// Get the name of a static or instance property from a property access lambda.
// </summary>
// <typeparam name="T">Type of the property</typeparam>
// <param name="propertyLambda">lambda expression of the form: '() => Class.Property' or '() => object.Property'</param>
// <returns>The name of the property</returns>
public string GetPropertyName<T>(Expression<Func<T>> propertyLambda)
{
    var me = propertyLambda.Body as MemberExpression;

    if (me == null)
    {
        throw new ArgumentException("You must pass a lambda of the form: '() => Class.Property' or '() => object.Property'");
    }

    return me.Member.Name;
 }

用法:

// Static Property
string name = GetPropertyName(() => SomeClass.SomeProperty);

// Instance Property
string name = GetPropertyName(() => someObject.SomeProperty);

现在有了这个很酷的功能,是时候简化ExposeProperty方法了。擦门把手是危险的工作。


当前回答

有一个著名的黑客从lambda表达式中提取它(这是来自PropertyObserver类,由Josh Smith在他的MVVM基金会):

    private static string GetPropertyName<TPropertySource>
        (Expression<Func<TPropertySource, object>> expression)
    {
        var lambda = expression as LambdaExpression;
        MemberExpression memberExpression;
        if (lambda.Body is UnaryExpression)
        {
            var unaryExpression = lambda.Body as UnaryExpression;
            memberExpression = unaryExpression.Operand as MemberExpression;
        }
        else
        {
            memberExpression = lambda.Body as MemberExpression;
        }

        Debug.Assert(memberExpression != null, 
           "Please provide a lambda expression like 'n => n.PropertyName'");

        if (memberExpression != null)
        {
            var propertyInfo = memberExpression.Member as PropertyInfo;

            return propertyInfo.Name;
        }

        return null;
    }

抱歉,这里缺少一些上下文。这是一个更大的类的一部分,其中TPropertySource是包含属性的类。您可以将TPropertySource中的函数设为泛型,以便从类中提取它。我建议查看来自MVVM基金会的完整代码。

其他回答

基于已经在问题和本文中的答案:https://handcraftsman.wordpress.com/2008/11/11/how-to-get-c-property-names-without-magic-strings/我提出了我对这个问题的解决方案:

public static class PropertyNameHelper
{
    /// <summary>
    /// A static method to get the Propertyname String of a Property
    /// It eliminates the need for "Magic Strings" and assures type safety when renaming properties.
    /// See: http://stackoverflow.com/questions/2820660/get-name-of-property-as-a-string
    /// </summary>
    /// <example>
    /// // Static Property
    /// string name = PropertyNameHelper.GetPropertyName(() => SomeClass.SomeProperty);
    /// // Instance Property
    /// string name = PropertyNameHelper.GetPropertyName(() => someObject.SomeProperty);
    /// </example>
    /// <typeparam name="T"></typeparam>
    /// <param name="propertyLambda"></param>
    /// <returns></returns>
    public static string GetPropertyName<T>(Expression<Func<T>> propertyLambda)
    {
        var me = propertyLambda.Body as MemberExpression;

        if (me == null)
        {
            throw new ArgumentException("You must pass a lambda of the form: '() => Class.Property' or '() => object.Property'");
        }

        return me.Member.Name;
    }
    /// <summary>
    /// Another way to get Instance Property names as strings.
    /// With this method you don't need to create a instance first.
    /// See the example.
    /// See: https://handcraftsman.wordpress.com/2008/11/11/how-to-get-c-property-names-without-magic-strings/
    /// </summary>
    /// <example>
    /// string name = PropertyNameHelper((Firma f) => f.Firmenumsatz_Waehrung);
    /// </example>
    /// <typeparam name="T"></typeparam>
    /// <typeparam name="TReturn"></typeparam>
    /// <param name="expression"></param>
    /// <returns></returns>
    public static string GetPropertyName<T, TReturn>(Expression<Func<T, TReturn>> expression)
    {
        MemberExpression body = (MemberExpression)expression.Body;
        return body.Member.Name;
    }
}

还有一个Test,它也显示了实例和静态属性的用法:

[TestClass]
public class PropertyNameHelperTest
{
    private class TestClass
    {
        public static string StaticString { get; set; }
        public string InstanceString { get; set; }
    }

    [TestMethod]
    public void TestGetPropertyName()
    {
        Assert.AreEqual("StaticString", PropertyNameHelper.GetPropertyName(() => TestClass.StaticString));

        Assert.AreEqual("InstanceString", PropertyNameHelper.GetPropertyName((TestClass t) => t.InstanceString));
    }
}

我修改了您的解决方案,以链接多个属性:

public static string GetPropertyName<T>(Expression<Func<T>> propertyLambda)
{
    MemberExpression me = propertyLambda.Body as MemberExpression;
    if (me == null)
    {
        throw new ArgumentException("You must pass a lambda of the form: '() => Class.Property' or '() => object.Property'");
    }

    string result = string.Empty;
    do
    {
        result = me.Member.Name + "." + result;
        me = me.Expression as MemberExpression;
    } while (me != null);

    result = result.Remove(result.Length - 1); // remove the trailing "."
    return result;
}

用法:

string name = GetPropertyName(() => someObject.SomeProperty.SomeOtherProperty);
// returns "SomeProperty.SomeOtherProperty"

我在使用已经为我的特定用例建议的解决方案时遇到了一些困难,但最终解决了这个问题。我不认为我的具体情况值得一个新的问题,所以我在这里张贴我的解决方案供参考。(这与问题密切相关,并为其他与我有类似情况的人提供了解决方案)。

我最终得到的代码是这样的:

public class HideableControl<T>: Control where T: class
{
    private string _propertyName;
    private PropertyInfo _propertyInfo;

    public string PropertyName
    {
        get { return _propertyName; }
        set
        {
            _propertyName = value;
            _propertyInfo = typeof(T).GetProperty(value);
        }
    }

    protected override bool GetIsVisible(IRenderContext context)
    {
        if (_propertyInfo == null)
            return false;

        var model = context.Get<T>();

        if (model == null)
            return false;

        return (bool)_propertyInfo.GetValue(model, null);
    }

    protected void SetIsVisibleProperty(Expression<Func<T, bool>> propertyLambda)
    {
        var expression = propertyLambda.Body as MemberExpression;
        if (expression == null)
            throw new ArgumentException("You must pass a lambda of the form: 'vm => vm.Property'");

        PropertyName = expression.Member.Name;
    }
}

public interface ICompanyViewModel
{
    string CompanyName { get; }
    bool IsVisible { get; }
}

public class CompanyControl: HideableControl<ICompanyViewModel>
{
    public CompanyControl()
    {
        SetIsVisibleProperty(vm => vm.IsVisible);
    }
}

对我来说重要的部分是,在CompanyControl类编译器只允许我选择一个布尔属性的ICompanyViewModel,这使得其他开发人员更容易得到它的权利。

我的解决方案与公认答案之间的主要区别是,我的类是泛型的,我只想匹配泛型类型中的布尔型属性。

从lambda表达式中检索属性名,你可以这样做:

RemoteMgr.ExposeProperty(() => SomeClass.SomeProperty)

public class SomeClass
{
    public static string SomeProperty
    {
        get { return "Foo"; }
    }
}

public class RemoteMgr
{
    public static void ExposeProperty<T>(Expression<Func<T>> property)
    {
        var expression = GetMemberInfo(property);
        string path = string.Concat(expression.Member.DeclaringType.FullName,
            ".", expression.Member.Name);
        // Do ExposeProperty work here...
    }
}

public class Program
{
    public static void Main()
    {
        RemoteMgr.ExposeProperty("SomeSecret", () => SomeClass.SomeProperty);
    }
}

如果我理解正确的话,PropertyInfo类应该帮助您实现这一点。

Type.GetProperties()方法 PropertyInfo[] propInfos = typeof(ReflectedType).GetProperties(); propInfos.ToList()。ForEach (p = > Console.WriteLine(字符串。格式("属性名:{0}",p.Name));

这是你需要的吗?