用子lambda表达式扩展lambda表达式[重复]

问题描述 投票:0回答:1

我试图找出是否可以组合2个LambdaExpressions,其中一个lambda表达式使用第一个的Child属性。

鉴于以下2个类:

class MyClass
{
    public string Name { get; set; }
    public SubClass Child { get; set; }
}

class SubClass
{
    public string SubClassName { get; set; }
}

以下表达式:

Expression<Func<MyClass, bool>> Expression1 = c => c.Name == "Test";

Expression<Func<SubClass, bool>> Expression2 = sc => sc.SubClassName == "SubTest";

我想将它组合成以下类型的lambda:

Expression<Func<MyClass, bool>> Combined;

原因是:上面的lambda是内部的,下面的lamba将由方法的'user'传入,它不知道(也不应该)MyClass存在,只知道SubClass。

这是可能的,还是我应该找到另一条路?

c# lambda expression
1个回答
-1
投票

我不知道这是不是你所追求的,但是;

public class Search
{
    private IEnumerable<MyClass> _data;

    public Search(IEnumerable<MyClass> data)
    {
        _data = data;
    }

    public IEnumerable<MyClass> Find(Expression<Func<SubClass, bool>> expr)
    {
        ParameterExpression x = Expression.Parameter(typeof(MyClass), "x");
        PropertyInfo p = typeof(MyClass).GetProperty("Child");

        var propertyExpression = (expr.Body as BinaryExpression).Left as MemberExpression;
        var constant = (expr.Body as BinaryExpression).Right;

        var memberAccess = Expression.MakeMemberAccess(x, p);
        var upperMemberAccess = Expression.MakeMemberAccess(memberAccess, propertyExpression.Member);

        var equals = Expression.Equal(upperMemberAccess, constant);

        var expression = Expression.Lambda<Func<MyClass, bool>>(equals, x);

        return _data.Where(expression.Compile());
    }
}

然后使用它看起来像这样;

var search = new Search(new[] {
        new MyClass { Name = "Test 1", Child = new SubClass { SubClassName = "Bob" }},
        new MyClass { Name = "Test 2", Child = new SubClass { SubClassName = "Subclass"}},
        new MyClass { Name = "Test 3", Child = new SubClass { SubClassName = "Subclass"}}
    });

    search.Find(k => k.SubClassName == "Subclass");

我确信有更好的方法可以做到这一点,因为这感觉很笨,但基本上它会发现MyClass的所有成员都有一个名为'Subclass'的子类。它提取二进制表达式的组件(x => x.SubClassname ==“Subclass”),并使用基于MyClass的equals重建一个新表达式。

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