反射 - 获取属性名称

Question

我想将属性名称传递给函数而不使用魔法字符串。

是这样的：

Get<ObjectType>(x=>x.Property1); 

其中Property1是类型对象类型的属性。

该方法的实现是什么样的？

Answer 1

这可以使用表达式来实现：

// requires object instance, but you can skip specifying T 
static string GetPropertyName<T>(Expression<Func<T>> exp) 
{ 
    return (((MemberExpression)(exp.Body)).Member).Name; 
} 

// requires explicit specification of both object type and property type 
static string GetPropertyName<TObject, TResult>(Expression<Func<TObject, TResult>> exp) 
{ 
    // extract property name 
    return (((MemberExpression)(exp.Body)).Member).Name; 
} 

// requires explicit specification of object type 
static string GetPropertyName<TObject>(Expression<Func<TObject, object>> exp) 
{ 
    var body = exp.Body; 
    var convertExpression = body as UnaryExpression; 
    if(convertExpression != null) 
    { 
     if(convertExpression.NodeType != ExpressionType.Convert) 
     { 
      throw new ArgumentException("Invalid property expression.", "exp"); 
     } 
     body = convertExpression.Operand; 
    } 
    return ((MemberExpression)body).Member.Name; 
} 

用法：

var x = new ObjectType(); 
// note that in this case we don't need to specify types of x and Property1 
var propName1 = GetPropertyName(() => x.Property1); 
// assumes Property2 is an int property 
var propName2 = GetPropertyName<ObjectType, int>(y => y.Property2); 
// requires only object type 
var propName3 = GetPropertyName<ObjectType>(y => y.Property3); 

更新：固定GetPropertyName<TObject>(Expression<Func<TObject, object>> exp)为属性返回值类型。

Answer 2

class Foo 
{ 
    public string Bar { get; set; } 
} 

class Program 
{ 
    static void Main() 
    { 
     var result = Get<Foo, string>(x => x.Bar); 
     Console.WriteLine(result); 
    } 

    static string Get<T, TResult>(Expression<Func<T, TResult>> expression) 
    { 
     var me = expression.Body as MemberExpression; 
     if (me != null) 
     { 
      return me.Member.Name; 
     } 
     return null; 
    } 
}