如何从未经申报类型类和属性名和值 [英] How to get class and property names and values from undeclared type

问题描述

如果我有这两个类：

public class A
{
    public int Id { get; set; }
}

public class B
{
    public string Name { get; set; }
}

我可以使用一个通用的方法是这样的：

Can I use a generic method like this:

public void InitMethod(object classProperty)

要传递的数据是这样的：

To pass in data like this:

var a = new A() { Id = 1 };
var b = new B() { Name = "John" };

InitMethod(a.Id);
InitMethod(b.Name);

和从方法中获得以下信息：

And get the following information from within the method:

• 类名（例如：A，B）


• 属性名称（例如：ID，姓名）


• 属性值（例如：1，约翰）

推荐答案

排序的，尽管它可能是更多的麻烦比它的价值。

Sort of, although it may be more trouble than it is worth.

ASP.Net MVC经常使用前pressions获得在一个强类型的时尚属性信息。这位前pression不一定得到评估;相反，它被解析为它的元数据。

ASP.Net MVC frequently uses expressions to get at property info in a strongly-typed fashion. The expression doesn't necessarily get evaluated; instead, it is parsed for its metadata.

这是不特定于MVC;我提到它举在Microsoft框架中的既定模式。

This isn't specific to MVC; I mention it to cite an established pattern in a Microsoft framework.

下面是得到一个属性名称和值从EX pression一个例子：

Here's a sample that gets a property name and value from an expression:

// the type being evaluated
public class Foo
{
    public string Bar {
        get;
        set;
    }
}

// method in an evaluator class
public TProperty EvaluateProperty<TProperty>( Expression<Func<Foo, TProperty>> expression ) {
    string propertyToGetName = ( (MemberExpression)expression.Body ).Member.Name;

    // do something with the property name

    // and/or evaluate the expression and get the value of the property
    return expression.Compile()( null );
}

您这样称呼它（注意传递的前pressions）：

You call it like this (note the expressions being passed):

var foo = new Foo { Bar = "baz" };
string val = EvaluateProperty( o => foo.Bar );

foo = new Foo { Bar = "123456" };
val = EvaluateProperty( o => foo.Bar );

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