MemberExpression，从类构建 Expression.Property

Question

下面的表达式将属性 NAME 与值 PETER 进行比较。

            ParameterExpression pe = Expression.Parameter(typeof(T), "x");
            MemberExpression member = Expression.Property(pe, "name");
            ConstantExpression value = Expression.Constant("Peter");
            exp = Expression.Equal(member, value);

如果该属性是一个类怎么办：

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

那么表达式将类似于以下内容：

            MemberExpression member = Expression.Property(pe, "Address.Name");
            ConstantExpression value = Expression.Constant("Peter");
            exp = Expression.Equal(member, value);

这会失败，因为成员类型与值类型不匹配。

所以，问题是：如何构建一个可以使用上述类示例工作的表达式？

我在 NHibernate.Linq 查询中使用此表达式：

        var q = from f in data //of type IQueryable<T>
            select f;
        if (filter != null) //filter of type Expression<Func<T, bool>>
            q = q.Where(filter);
        etc....

谢谢。

彼得更新：

根据 xanatos 的代码（下一篇文章），我创建了以下测试来了解它是如何工作的。它与 xanatos 所做的并没有太大不同，但一开始我无法让它工作，所以我决定在一个简单的测试中重新编写它，然后就成功了。感谢 xanatos：

    [Test]
    public void FilterWithDeepProperties()
    {
        //Arrange
        IGenericGridRepository repository = ObjectFactory.GetInstance<IGenericGridRepository>();

        FilterDescriptor filter = new FilterDescriptor("AgreementId.Name", FilterOperator.IsEqualTo, "a name");
        string[] properties = filter.Member.Split('.');
        ParameterExpression pe = Expression.Parameter(typeof(SampleDomain), "x");

        //Act
        Expression lastMember = pe;
        for (int i = 0; i < properties.Length; i++)
        {
            MemberExpression member = Expression.Property(lastMember, properties[i]);
            lastMember = member;
        }
        ConstantExpression valueExpression = Expression.Constant(filter.Value);
        Expression equalityExpression = Expression.Equal(lastMember, valueExpression);
        Expression<Func<SampleDomain, bool>> where = Expression.Lambda<Func<SampleDomain, bool>>(equalityExpression, pe);
        var result = repository.GetObjects<SampleDomain>(filter: where);

        //Assert
        result.Count().Should().BeGreaterThan(0, "because there are many schedule items equals to " + filter.Value);
    }

Answer 1

你可能想要这样的东西：

public static Expression<Func<TSource, bool>> GetEquality<TSource>(object value, params string[] properties)
{
    ParameterExpression pe = Expression.Parameter(typeof(TSource), "source");

    Expression lastMember = pe;

    for (int i = 0; i < properties.Length; i++)
    {
        MemberExpression member = Expression.Property(lastMember, properties[i]);
        lastMember = member;
    }

    Expression valueExpression = Expression.Constant(value);
    Expression equalityExpression = Expression.Equal(lastMember, valueExpression);
    Expression<Func<TSource, bool>> lambda = Expression.Lambda<Func<TSource, bool>>(equalityExpression, pe);
    return lambda;
}

像这样使用它：

Expression exp = GetEquality<Person>("Foo", "Address", "Name");

whereFoo是 your Peter（因此必须比较的值），而Address和Name是属性“链”的名称。例如我正在使用

public class Person
{
    public Address Address { get; set; }
}

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

所以生成的表达式是

source.Address.Name == "Foo"

如果你想使用类似的东西Address.Name，你可以使用类似的方法

Expression exp = GetEquality<Person>("Foo", "Address.Name".Split('.'));