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copied!<p>You'll need to get the member name from the lambda expression, and use reflection to get that member off of the type you've been given:</p> <pre><code>public static PropertyInfo PropInfo<TContainer, TMember>( Expression<Func<TContainer, TMember>> memberGetter) { var memberName = GetExpressionMemberName(memberGetter); return typeof(TContainer).GetProperty(memberName); } public static string GetExpressionMemberName<TContainer, TMember>( Expression<Func<TContainer, TMember>> memberGetter) { var expressionType = memberGetter.Body.NodeType; switch (expressionType) { case ExpressionType.MemberAccess: { var memberExpr = (MemberExpression) memberGetter.Body; return memberExpr.Member.Name; } case ExpressionType.Convert: { var convertExpr = (UnaryExpression) memberGetter.Body; var memberExpr = (MemberExpression) convertExpr.Operand; return memberExpr.Member.Name; } default: throw new InvalidOperationException("Expression {0} does not represent a simple member access."); } } </code></pre> <p>Here's proof that it works:</p> <pre><code>void Main() { Console.WriteLine( MyGenericMethod<Employee>() .GetGetMethod() .Invoke( new Employee {Name = "Bill"}, new object[] {})); } public class Employee : IEmployee { public string Name {get;set;} string IEmployee.Name { get { throw new Exception(); } } } public interface IEmployee {string Name {get;}} public PropertyInfo MyGenericMethod<T>() where T : IEmployee { return PropInfo((T e) => e.Name); } </code></pre> <p>Console output:</p> <pre><code>Bill </code></pre>

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