如何从字符串为深属性创建表达式树/ lambda

听起来你是用常规反射排序的,但是对于info,构建嵌套属性的表达式的代码与这个order-by代码非常相似.

请注意,要设置值,您需要GetSetMethod()在属性上使用并调用它 - 在构造之后没有用于分配值的内置表达式(尽管4.0中支持).

(编辑)像这样:

using System;
using System.Linq;
using System.Linq.Expressions;
using System.Reflection;
class Foo
{
    public Foo() { Bar = new Bar(); }
    public Bar Bar { get; private set; }
}
class Bar
{
    public string Name {get;set;}
}
static class Program
{
    static void Main()
    {
        Foo foo = new Foo();
        var setValue = BuildSet("Bar.Name");
        var getValue = BuildGet("Bar.Name");
        setValue(foo, "abc");
        Console.WriteLine(getValue(foo));        
    }
    static Action BuildSet(string property)
    {
        string[] props = property.Split('.');
        Type type = typeof(T);
        ParameterExpression arg = Expression.Parameter(type, "x");
        ParameterExpression valArg = Expression.Parameter(typeof(TValue), "val");
        Expression expr = arg;
        foreach (string prop in props.Take(props.Length - 1))
        {
            // use reflection (not ComponentModel) to mirror LINQ 
            PropertyInfo pi = type.GetProperty(prop);
            expr = Expression.Property(expr, pi);
            type = pi.PropertyType;
        }
        // final property set...
        PropertyInfo finalProp = type.GetProperty(props.Last());
        MethodInfo setter = finalProp.GetSetMethod();
        expr = Expression.Call(expr, setter, valArg);
        return Expression.Lambda>(expr, arg, valArg).Compile();        

    }
    static Func BuildGet(string property)
    {
        string[] props = property.Split('.');
        Type type = typeof(T);
        ParameterExpression arg = Expression.Parameter(type, "x");
        Expression expr = arg;
        foreach (string prop in props)
        {
            // use reflection (not ComponentModel) to mirror LINQ 
            PropertyInfo pi = type.GetProperty(prop);
            expr = Expression.Property(expr, pi);
            type = pi.PropertyType;
        }
        return Expression.Lambda>(expr, arg).Compile();
    }
}