在过去的几天里,我一直在互联网上搜索这个问题的解决方案,却没有找到我想要的东西。基本上,我的问题是:
因此,经过一些搜索之后,我想我已经决定了解决这个问题的一个很好的方法,就是编写我自己的ConcatenatingQueryable IQueryable实现,它需要两个IQueryable的实现,并在每个独立的IQueryable上执行表达式树,然后连接结果。但是,当它返回堆栈溢出时,我似乎遇到了问题。基于http://blogs.msdn.com/b/mattwar/archive/2007/07/30/linq-building-an-iqueryable-provider-part-i.aspx,到目前为止我已经实现了以下内容:
class Program
{
static void Main(string[] args)
{
var source1 = new[] { 1, 2 }.AsQueryable();
var source2 = new[] { -1, 1 }.AsQueryable();
var matches = new ConcatenatingQueryable<int>(source1, source2).Where(x => x <= 1).ToArray();
Console.WriteLine(string.Join(",", matches));
Console.ReadKey();
}
public class ConcatenatingQueryable<T> : IQueryable<T>
{
private readonly ConcatenatingQueryableProvider<T> provider;
private readonly Expression expression;
public ConcatenatingQueryable(IQueryable<T> source1, IQueryable<T> source2)
: this(new ConcatenatingQueryableProvider<T>(source1, source2))
{}
public ConcatenatingQueryable(ConcatenatingQueryableProvider<T> provider)
{
this.provider = provider;
this.expression = Expression.Constant(this);
}
public ConcatenatingQueryable(ConcatenatingQueryableProvider<T> provider, Expression expression)
{
this.provider = provider;
this.expression = expression;
}
Expression IQueryable.Expression
{
get { return expression; }
}
Type IQueryable.ElementType
{
get { return typeof(T); }
}
IQueryProvider IQueryable.Provider
{
get { return provider; }
}
public IEnumerator<T> GetEnumerator()
{
// This line is calling Execute below
return ((IEnumerable<T>)provider.Execute(expression)).GetEnumerator();
}
IEnumerator IEnumerable.GetEnumerator()
{
return ((IEnumerable)provider.Execute(expression)).GetEnumerator();
}
}
public class ConcatenatingQueryableProvider<T> : IQueryProvider
{
private readonly IQueryable<T> source1;
private readonly IQueryable<T> source2;
public ConcatenatingQueryableProvider(IQueryable<T> source1, IQueryable<T> source2)
{
this.source1 = source1;
this.source2 = source2;
}
IQueryable<TS> IQueryProvider.CreateQuery<TS>(Expression expression)
{
var elementType = TypeSystem.GetElementType(expression.Type);
try
{
return (IQueryable<TS>)Activator.CreateInstance(typeof(ConcatenatingQueryable<>).MakeGenericType(elementType), new object[] { this, expression });
}
catch (TargetInvocationException tie)
{
throw tie.InnerException;
}
}
IQueryable IQueryProvider.CreateQuery(Expression expression)
{
var elementType = TypeSystem.GetElementType(expression.Type);
try
{
return (IQueryable)Activator.CreateInstance(typeof(ConcatenatingQueryable<>).MakeGenericType(elementType), new object[] { this, expression });
}
catch (TargetInvocationException tie)
{
throw tie.InnerException;
}
}
TS IQueryProvider.Execute<TS>(Expression expression)
{
return (TS)Execute(expression);
}
object IQueryProvider.Execute(Expression expression)
{
return Execute(expression);
}
public object Execute(Expression expression)
{
// This is where I suspect the problem lies, as executing the
// Expression.Constant from above here will call Enumerate again,
// which then calls this, and... you get the point
dynamic results1 = source1.Provider.Execute(expression);
dynamic results2 = source2.Provider.Execute(expression);
return results1.Concat(results2);
}
}
internal static class TypeSystem
{
internal static Type GetElementType(Type seqType)
{
var ienum = FindIEnumerable(seqType);
if (ienum == null)
return seqType;
return ienum.GetGenericArguments()[0];
}
private static Type FindIEnumerable(Type seqType)
{
if (seqType == null || seqType == typeof(string))
return null;
if (seqType.IsArray)
return typeof(IEnumerable<>).MakeGenericType(seqType.GetElementType());
if (seqType.IsGenericType)
{
foreach (var arg in seqType.GetGenericArguments())
{
var ienum = typeof(IEnumerable<>).MakeGenericType(arg);
if (ienum.IsAssignableFrom(seqType))
{
return ienum;
}
}
}
var ifaces = seqType.GetInterfaces();
if (ifaces.Length > 0)
{
foreach (var iface in ifaces)
{
var ienum = FindIEnumerable(iface);
if (ienum != null)
return ienum;
}
}
if (seqType.BaseType != null && seqType.BaseType != typeof(object))
{
return FindIEnumerable(seqType.BaseType);
}
return null;
}
}
}我对这个界面没有太多的经验,也有点不知所措。有人对如何做这件事有什么建议吗?如果需要的话,我也愿意完全放弃这种方法。
重申一下,我得到了一个StackOverflowException,堆栈跟踪只是上面两行注释行之间的一串调用,每对调用之间都有“外部代码”。我添加了一个示例Main方法,它使用了两个微小的枚举,但是您可以想象这些是更大的数据源,需要很长时间来枚举。
提前感谢您的帮助!
发布于 2014-09-15 20:34:45
当您分解传递给IQueryProvider的表达式树时,您将看到LINQ方法的调用链。请记住,通常LINQ通过链接扩展方法工作,其中前一个方法的返回值作为第一个参数传递给下一个方法。
如果从逻辑上讲,这意味着链中的第一个LINQ方法必须有一个源参数,从代码中可以很清楚地看出,它的源代码实际上是最初启动整个过程的同一个IQueryable (您的ConcatenatingQueryable)。
当你建造这个的时候,你就有了正确的想法--你只需要走一小步就行了。我们需要做的是重新指出,首先LINQ方法使用实际的源,然后允许执行遵循它的自然路径。
下面是一些执行此操作的示例代码:
public object Execute(Expression expression)
{
var query1 = ChangeQuerySource(expression, Expression.Constant(source1));
var query2 = ChangeQuerySource(expression, Expression.Constant(source2));
dynamic results1 = source1.Provider.Execute(query1);
dynamic results2 = source2.Provider.Execute(query2);
return Enumerable.Concat(results1, results2);
}
private static Expression ChangeQuerySource(Expression query, Expression newSource)
{
// step 1: cast the Expression as a MethodCallExpression.
// This will usually work, since a chain of LINQ statements
// is generally a chain of method calls, but I would not
// make such a blind assumption in production code.
var methodCallExpression = (MethodCallExpression)query;
// step 2: Create a new MethodCallExpression, passing in
// the existing one's MethodInfo so we're calling the same
// method, but just changing the parameters. Remember LINQ
// methods are extension methods, so the first argument is
// always the source. We carry over any additional arguments.
query = Expression.Call(
methodCallExpression.Method,
new Expression[] { newSource }.Concat(methodCallExpression.Arguments.Skip(1)));
// step 3: We call .AsEnumerable() at the end, to get an
// ultimate return type of IEnumerable<T> instead of
// IQueryable<T>, so we can safely use this new expression
// tree in any IEnumerable statement.
query = Expression.Call(
typeof(Enumerable).GetMethod("AsEnumerable", BindingFlags.Static | BindingFlags.Public)
.MakeGenericMethod(
TypeSystem.GetElementType(methodCallExpression.Arguments[0].Type)
),
query);
return query;
}https://stackoverflow.com/questions/25855152
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