Access the value of a member expression

Question

If i have a product.

var p = new Product { Price = 30 };

and i have the following linq query.

var q = repo.Products().Where         


        

Answer 1

 MemberExpression right = (MemberExpression)((BinaryExpression)p.Body).Right;
 Expression.Lambda(right).Compile().DynamicInvoke();

Answer 2

If you had a class:

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

and an instance of the object:

var myItem = new Item { Id = 7 };

You can get the value of Id using an Expression using the following code:

Expression<Func<Item, int>> exp = x => x.Id;
var me = exp.Body as MemberExpression;
var propInfo = me.Member as PropertyInfo;
var myValue = propInfo.GetValue(myItem, null);

myValue will contain "7"

Answer 3

Can you use the following:

var price = p.Price;
var q = repo.Products().Where(x=>x.Price == price).ToList()

Answer 4

As of 2020

This helper method will gracefully retrieve any expression value, without "compiling hack" :

public static object GetMemberExpressionValue (MemberExpression expression)
{
    // Dependency chain of a MemberExpression is of the form:
    // MemberExpression expression
    //    MemberExpression expression.Expression
    //        ... MemberExpression expression.[...].Expression
    //            ConstantExpression expression.[...].Expression.Expression <- base object
    var dependencyChain = new List<MemberExpression>();
    var pointingExpression = expression;
    while (pointingExpression != null)
    {
        dependencyChain.Add(pointingExpression);
        pointingExpression = pointingExpression.Expression as MemberExpression;
    }

    if (!(dependencyChain.Last().Expression is ConstantExpression baseExpression))
    {
        throw new Exception(
            $"Last expression {dependencyChain.Last().Expression} of dependency chain of {expression} is not a constant." +
            "Thus the expression value cannot be found.");
    }

    var resolvedValue = baseExpression.Value;

    for (var i = dependencyChain.Count; i > 0; i--)
    {
        var expr = dependencyChain[i - 1];
        resolvedValue = new PropOrField(expr.Member).GetValue(resolvedValue);
    }

    return resolvedValue;
}

PropOrField class :

public class PropOrField
{
    public readonly MemberInfo MemberInfo;

    public PropOrField (MemberInfo memberInfo)
    {
        if (!(memberInfo is PropertyInfo) && !(memberInfo is FieldInfo))
        {
            throw new Exception(
                $"{nameof(memberInfo)} must either be {nameof(PropertyInfo)} or {nameof(FieldInfo)}");
        }

        MemberInfo = memberInfo;
    }

    public object GetValue (object source)
    {
        if (MemberInfo is PropertyInfo propertyInfo) return propertyInfo.GetValue(source);
        if (MemberInfo is FieldInfo fieldInfo) return fieldInfo.GetValue(source);

        return null;
    }

    public void SetValue (object target, object source)
    {
        if (MemberInfo is PropertyInfo propertyInfo) propertyInfo.SetValue(target, source);
        if (MemberInfo is FieldInfo fieldInfo) fieldInfo.SetValue(target, source);
    }

    public Type GetMemberType ()
    {
        if (MemberInfo is PropertyInfo propertyInfo) return propertyInfo.PropertyType;
        if (MemberInfo is FieldInfo fieldInfo) return fieldInfo.FieldType;

        return null;
    }
}

Answer 5

q is of type List<Product>. The List doesn't have a Price property - only the individual Products.

The first or last Product will have a price.

q.First().Price
q.Last().Price

If you know there's only one in the collection you can also flatten it using Single

q.Single().Price

Answer 6

You can compile and invoke a lambda expression whose body is the member access:

private object GetValue(MemberExpression member)
{
    var objectMember = Expression.Convert(member, typeof(object));

    var getterLambda = Expression.Lambda<Func<object>>(objectMember);

    var getter = getterLambda.Compile();

    return getter();
}

Local evaluation is a common technique when parsing expression trees. LINQ to SQL does this exact thing in quite a few places.