6

我正在寻找一种使用 lambda 表达式调用泛型方法的方法,该 lambda 表达式在项目数组中调用 Contains。

在这种情况下,我使用的是实体框架 Where 方法,但该场景可以应用于其他 IEnumerables。

我需要通过Reflection调用上面代码的最后一行,所以我可以使用任何类型和任何属性来传递给Contains方法。

var context = new TestEntities();

var items = new[] {100, 200, 400, 777}; //IN list (will be tested through Contains)
var type = typeof(MyType); 

context.Set(type).Where(e => items.Contains(e.Id)); //**What is equivalent to this line using Reflection?**

在研究中,我注意到我应该使用 GetMethod、MakeGenericType 和 Expression 来实现这一点,但我不知道该怎么做。拥有此示例将非常有帮助,因此我可以了解反射如何与 Lambda 和通用概念一起使用。

基本上,目标是编写一个正确版本的函数,如下所示:

//Return all items from a IEnumerable(target) that has at least one matching Property(propertyName) 
//with its value contained in a IEnumerable(possibleValues)
static IEnumerable GetFilteredList(IEnumerable target, string propertyName, IEnumerable searchValues)
{
    return target.Where(t => searchValues.Contains(t.propertyName));
    //Known the following:
    //1) This function intentionally can't be compiled
    //2) Where function can't be called directly from an untyped IEnumerable
    //3) t is not actually recognized as a Type, so I can't access its property 
    //4) The property "propertyName" in t should be accessed via Linq.Expressions or Reflection
    //5) Contains function can't be called directly from an untyped IEnumerable
}

//Testing environment
static void Main()
{
    var listOfPerson = new List<Person> { new Person {Id = 3}, new Person {Id = 1}, new Person {Id = 5} };
    var searchIds = new int[] { 1, 2, 3, 4 };

    //Requirement: The function must not be generic like GetFilteredList<Person> or have the target parameter IEnumerable<Person>
    //because the I need to pass different IEnumerable types, not known in compile-time
    var searchResult = GetFilteredList(listOfPerson, "Id", searchIds);

    foreach (var person in searchResult)
        Console.Write(" Found {0}", ((Person) person).Id);

    //Should output Found 3 Found 1
}

我不确定其他问题是否解决了这种情况,因为我认为我无法清楚地理解表达式的工作原理。

更新:

我不能使用泛型,因为我只有在运行时要测试的类型和属性(在包含中)。在第一个代码示例中,假设“MyType”在编译时未知。在第二个代码示例中,类型可以作为参数传递给 GetFilteredList 函数,也可以通过反射 (GetGenericArguments) 获取。

谢谢,

4

3 回答 3

14

经过广泛的研究和对表达式的大量研究,我可以自己编写一个解决方案。它当然可以改进,但完全符合我的要求。希望它可以帮助别人。

//Return all items from a IEnumerable(target) that has at least one matching Property(propertyName) 
//with its value contained in a IEnumerable(possibleValues)
static IEnumerable GetFilteredList(IEnumerable target, string propertyName, IEnumerable searchValues)
{
    //Get target's T 
    var targetType = target.GetType().GetGenericArguments().FirstOrDefault();
    if (targetType == null)
        throw new ArgumentException("Should be IEnumerable<T>", "target");

    //Get searchValues's T
    var searchValuesType = searchValues.GetType().GetGenericArguments().FirstOrDefault();
    if (searchValuesType == null)
        throw new ArgumentException("Should be IEnumerable<T>", "searchValues");

    //Create a p parameter with the type T of the items in the -> target IEnumerable<T>
    var containsLambdaParameter = Expression.Parameter(targetType, "p");

    //Create a property accessor using the property name -> p.#propertyName#
    var property = Expression.Property(containsLambdaParameter, targetType, propertyName);

    //Create a constant with the -> IEnumerable<T> searchValues
    var searchValuesAsConstant = Expression.Constant(searchValues, searchValues.GetType());

    //Create a method call -> searchValues.Contains(p.Id)
    var containsBody = Expression.Call(typeof(Enumerable), "Contains", new[] { searchValuesType }, searchValuesAsConstant, property);

    //Create a lambda expression with the parameter p -> p => searchValues.Contains(p.Id)
    var containsLambda = Expression.Lambda(containsBody, containsLambdaParameter);

    //Create a constant with the -> IEnumerable<T> target
    var targetAsConstant = Expression.Constant(target, target.GetType());

    //Where(p => searchValues.Contains(p.Id))
    var whereBody = Expression.Call(typeof(Enumerable), "Where", new[] { targetType }, targetAsConstant, containsLambda);

    //target.Where(p => searchValues.Contains(p.Id))
    var whereLambda = Expression.Lambda<Func<IEnumerable>>(whereBody).Compile();

    return whereLambda.Invoke();
}
于 2013-07-04T02:24:43.667 回答
5

为了避免使用泛型(因为在设计时类型未知),您可以使用一些反射并“手动”构建表达式

您需要通过在一个 Where 子句中定义“包含”表达式来做到这一点:

public IQueryable GetItemsFromContainsClause(Type type, IEnumerable<string> items)
    {
        IUnitOfWork session = new SandstoneDbContext();
        var method = this.GetType().GetMethod("ContainsExpression");
        method = method.MakeGenericMethod(new[] { type });

        var lambda = method.Invoke(null, new object[] { "Codigo", items });
        var dbset = (session as DbContext).Set(type);
        var originalExpression = dbset.AsQueryable().Expression;

        var parameter = Expression.Parameter(type, "");
        var callWhere = Expression.Call(typeof(Queryable), "Where", new[] { type }, originalExpression, (Expression)lambda);
        return dbset.AsQueryable().Provider.CreateQuery(callWhere);

    }

    public static Expression<Func<T, bool>> ContainsExpression<T>(string propertyName, IEnumerable<string> values)
    {
        var parameterExp = Expression.Parameter(typeof(T), "");
        var propertyExp = Expression.Property(parameterExp, propertyName);
        var someValue = Expression.Constant(values, typeof(IEnumerable<string>));
        var containsMethodExp = Expression.Call(typeof(Enumerable), "Contains", new[] { typeof(string) }, someValue, propertyExp);
        return Expression.Lambda<Func<T, bool>>(containsMethodExp, parameterExp);
    }

在这种情况下,“Codigo”是硬编码的,但它可以是获取您定义的类型的任何属性的参数。

您可以使用以下方法对其进行测试:

public void LambdaConversionBasicWithEmissor()
    {
        var cust= new Customer();
        var items = new List<string>() { "PETR", "VALE" };
        var type = cust.GetType();
        // Here you have your results from the database
        var result = GetItemsFromContainsClause(type, items);
    }
于 2013-07-04T20:29:52.923 回答
0

您可以使用以下一组类来解决您的问题。

首先,我们需要创建一个 Contains 类,该类将决定从源数组中选择哪些项目。

class Contains
{
    public bool Value { get; set; }

    public Contains(object[] items, object item)
    {
        Value = (bool)(typeof(Enumerable).GetMethods()
                                         .Where(x => x.Name.Contains("Contains"))
                                         .First()
                                         .MakeGenericMethod(typeof(object))
                                         .Invoke(items, new object[] { items, item }));
    }
}

然后我们需要创建一个 Where 类,该类将用于根据将选择的项目形成谓词。应该清楚的是,在我们的例子中,我们将使用 Contains 类作为我们的谓词方法。

class Where
{
    public object Value { get; set; }

    public Where(object[] items, object[] items2)
    {
        Value = typeof(Enumerable).GetMethods()
                                  .Where(x => x.Name.Contains("Where"))
                                  .First()
                                  .MakeGenericMethod(typeof(object))
                                  .Invoke(items2, new object[] { items2, new Func<object, bool>(i => new Contains(items, i).Value) });
    }
}

最后一步是简单地调用我们从 Where 类中得到的结果,它实际上是 Enumerable.WhereArrayIterator 类型而不是 List 类型,因为 Where Extension 方法的结果是延迟执行的产物。

因此我们需要创建一个非延迟对象,通过调用它的 ToList 扩展方法,并得到我们的结果。

class ToList
{
    public List<object> Value { get; set; }

    public ToList(object[] items, object[] items2)
    {
        var where = new Where(items, items2).Value;

        Value = (typeof(Enumerable).GetMethods()
                                  .Where(x => x.Name.Contains("ToList"))
                                  .First()
                                  .MakeGenericMethod(typeof(object))
                                  .Invoke(where, new object[] { where })) as List<object>;
    }
}

最后,您可以使用以下类简单地测试整个过程。

class Program
{
    static void Main()
    {
        var items = new object[] { 1, 2, 3, 4 };
        var items2 = new object[] { 2, 3, 4, 5 };

        new ToList(items, items2).Value.ForEach(x => Console.WriteLine(x));

        Console.Read();
    }
}
于 2013-07-02T01:26:28.543 回答