给定一个类型ExpressionType.MemberAccess，我如何获取字段值？ [英] Given a type ExpressionType.MemberAccess, how do i get the field value?

问题描述

我解析表达式树。鉴于ExpressionType.MemberAccess的一个nodeType，我如何获取字段的值？

从C＃MSDN文档：
MemberAccess是表示从字段或属性读取节点。

一个代码片段将是令人难以置信，难以置信的帮助。在此先感谢！

我的代码看起来是这样的：

 公共静态列表< T>过滤器（表达式来; Func键< T，BOOL>> filterExp）
 {
 //表达确实是在这种情况下，
 BinaryExpression expBody = filterExp.Body作为BinaryExpression一个二进制表达式; 
 
如果（expBody.Left.NodeType == ExpressionType.MemberAccess）
 //做（（MemberExpressionexpBody.Left）.Name点
 
的东西//右手一边是确实的成员访问。事实上，价值来自//aspdroplist.selectedvalue 
如果（expBody.Right.NodeType == ExpressionType.MemberAccess）
 {
 //我如何获得aspdroplist.selected值??音符的价值：它的非静态
} 
 
 //返回一个列表
} 
  

解决方案

[更新为清晰起见]

首先，投在表达到 MemberExpression 。

A MemberExpression 有感兴趣的两件事情：

• 。成员 - 在的PropertyInfo / 字段信息的成员


• .Expression - 评估得到表达OBJ为。成员

也就是说，如果你可以评估 .Expression 来目标文件，以及。成员是字段信息，然后你就可以用 .GetValue（OBJ）在字段信息（和的PropertyInfo 很相似）。

的问题是，评估 .Expression 是非常棘手的;-p

显然，你很幸运，如果它原来是一个常量表达式 - 但在大多数情况下，它不是;它可能是一个 ParameterExpression （在这种情况下，你需要知道你要计算的实际参数值），或的任何其他组合表达秒。

在很多情况下，一个简单的（也许懒惰）选项是使用 .Compile（）来获得.NET框架做繁重;那么你可以评估的lambda作为一个类型代表（传入的拉姆达需要的任何参数）。这并不总是一个选项，但是

要展示如何复杂，这是;考虑这个简单的例子（其中，我的每一步硬编码，而不是测试等）：

 使用系统;使用System.Linq.Expressions 
; 
使用的System.Reflection; 
类Foo 
 {
公共字符串酒吧{搞定;组; } 
} 
 
静态类节目
 {
静态无效的主要（）
 {
富富=新富{吧=ABC }; 
表达式来; Func键<串GT;> FUNC =（）=> foo.Bar; 
 
 MemberExpression outerMember =（MemberExpression）func.Body; 
的PropertyInfo outerProp =（的PropertyInfo）outerMember.Member; 
 MemberExpression innerMember =（MemberExpression）outerMember.Expression; 
字段信息innerField =（字段信息）innerMember.Member; 
常量表达式CE =（常量表达式）innerMember.Expression; 
对象innerObj = ce.Value; 
对象outerObj = innerField.GetValue（innerObj）; 
字符串值=（字符串）outerProp.GetValue（outerObj，NULL）; 
} 
 
} 
  

I am parsing an Expression Tree. Given a NodeType of ExpressionType.MemberAccess, how do I get the value of that Field?

From C# MSDN docs: MemberAccess is A node that represents reading from a field or property.

A code snippet would be incredibly, incredibly helpful. Thanks in advance!!!

My code looks something like this:

public static List<T> Filter(Expression<Func<T, bool>> filterExp) 
{
//the expression is indeed a binary expression in this case
BinaryExpression expBody = filterExp.Body as BinaryExpression;

if (expBody.Left.NodeType == ExpressionType.MemberAccess) 
  //do something with ((MemberExpressionexpBody.Left).Name

//right hand side is indeed member access. in fact, the value comes from //aspdroplist.selectedvalue            
if (expBody.Right.NodeType == ExpressionType.MemberAccess)
{
   //how do i get the value of aspdroplist.selected value?? note: it's non-static                        
}

//return a list
}

解决方案

[updated for clarity]

First; cast the Expression to a MemberExpression.

A MemberExpression has two things of interest:

• .Member - the PropertyInfo / FieldInfo to the member
• .Expression - the expression to evaluate to get the "obj" for the .Member

i.e. if you can evaluate the .Expression to "obj", and the .Member is a FieldInfo, then you can get the actual value via .GetValue(obj) on the FieldInfo (and PropertyInfo is very similar).

The problem is that evaluating the .Expression is very tricky ;-p

Obviously you get lucky if it turns out to be a ConstantExpression - but in most cases it isn't; it could be a ParameterExpression (in which case you'll need to know the actual parameter value that you want to evaluate), or any other combination of Expressions.

In many cases, a simple (perhaps lazy) option is to use .Compile() to get the .NET framework to do the heavy lifting; you can then evaluate the lambda as a typed delegate (passing in any parameters that the lambda requires). This isn't always an option, however.

To show how complex this is; consider this trivial example (where I've hard-coded at every step, rather than testing etc):

using System;
using System.Linq.Expressions;
using System.Reflection;
class Foo
{
    public string Bar { get; set; }
}

static class Program
{
    static void Main()
    {
        Foo foo = new Foo {Bar = "abc"};
        Expression<Func<string>> func = () => foo.Bar;

        MemberExpression outerMember = (MemberExpression)func.Body;
        PropertyInfo outerProp = (PropertyInfo) outerMember.Member;
        MemberExpression innerMember = (MemberExpression)outerMember.Expression;
        FieldInfo innerField = (FieldInfo)innerMember.Member;
        ConstantExpression ce = (ConstantExpression) innerMember.Expression;
        object innerObj = ce.Value;
        object outerObj = innerField.GetValue(innerObj);
        string value = (string) outerProp.GetValue(outerObj, null);    
    }

}

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