Alright, let’s assume that we are lazy coders, we are building a lot of Linq queries lately, and it’s getting repetitive. We keep having to remember to add a certain where
clause to every query, couldn’t we just abstract this somehow? Well sure, we can use Expressions!
Let’s first take a look at the System.Linq.Queryable.Where
extension method, here is the method’s signature:
public static IQueryable<TSource> Where<TSource>(
this IQueryable<TSource> source,
Expression<Func<TSource, bool>> predicate);
Notice that it takes a lambda expression, which we can generate dynamically. For examples sake, let’s make a Foo
and a Bar
class, and two methods that generate a list of each:
static IEnumerable<Foo> MakeFoos() {
for (int i = 0; i < 10; i++)
yield return new Foo { A = "Foo " + i, B = i };
}
static IEnumerable<Bar> MakeBars() {
for (int i = 0; i < 10; i++)
yield return new Bar { A = "Bar " + i, B = i };
}
class Foo {
public string A { get; set; }
public int B { get; set; }
}
class Bar {
public string A { get; set; }
public int B { get; set; }
}
So what if we want to get all of the items in a collection with A
and even B
? What we need is an extension method that applies our filter to an IQueryable<>
of any type. First off, let’s look at the method signature:
public static IQueryable<T> FilterB<T>(this IQueryable<T> query)
It’s fairly simple, it is an extension method that takes an IQueryable<T>
and returns an IQueryable<T>
so that we can keep our fluent interface going. The first thing we have to do in our method is create a parameter and property expression.
var param = Expression.Parameter(typeof(T), "x");
var prop = Expression.Property(param, "B");
Then we need a binary expression that %’s our property by 2:
var two = Expression.Constant(2);
var mod = Expression.MakeBinary(ExpressionType.Modulo, prop, two);
The ConstantExpressions
are pretty self explanatory, they just hold the value that they are given. BinaryExpressions
can represent any kind of binary expression, which is any expression that has a left and right side, like a == b
, a != b
, a + b
, and b – a
for example. Next, we need to make another binary expression that checks to see if our first binary expression, mod
, is equal to zero
:
var zero = Expression.Constant(0) ;
var eqZero = Expression.MakeBinary(ExpressionType.Equal, mod, zero);
Notice that we used one binary expression as a parameter for another, MakeBinary
’s second and third parameters are just Expressions which can be any type of expression. All that is left to do is create the lambda expression, where
it and return it:
var lambda = Expression.Lambda<Func<T, bool>>(eqZero, param);
return query.Where(lambda);
We don’t have to compile the lambda here because query providers use it in its raw expression tree form which they translate into whatever they need to get the job done. Here is the entire method:
public static IQueryable<T> FilterB<T>(this IQueryable<T> query) {
var param = Expression.Parameter(typeof(T), "x");
var prop = Expression.Property(param, "B");
var two = Expression.Constant(2);
var mod = Expression.MakeBinary(ExpressionType.Modulo, prop, two);
var zero = Expression.Constant(0);
var eqZero = Expression.MakeBinary(ExpressionType.Equal, mod, zero);
var lambda = Expression.Lambda<Func<T, bool>>(eqZero, param);
return query.Where(lambda);
}
This should work with any query provider including Linq to SQL and the Entity Framework. We haven’t built in any error checking so expect it to throw if you use it on a collection of a type that does not have a B
property. So, given the types we defined above, let’s see how we would use it:
var foos = MakeFoos().AsQueryable();
var bars = MakeBars().AsQueryable();
foreach (var foo in foos.FilterB())
Console.WriteLine(foo.A);
foreach (var bar in bars.FilterB())
Console.WriteLine(bar.A);
We are first getting collections of both Foo
s and Bar
s, and then changing them into IQueryable
collections. We then just iterate over each collection after calling FilterB()
on them. We could chain other calls after the FilterB
call if we wanted. The output will be:
Foo 0
Foo 2
Foo 4
Foo 6
Foo 8
Bar 0
Bar 2
Bar 4
Bar 6
Bar 8
Yay! It works!