int expectedCount = 4;
Expression<Func<Thing,bool>> expression = ...;

//Expression looks like (LocaleID = 1 && GenderID ==1 && (TimeFrameID == 2007 || TimeFrameID == 2008))

using (XYZDataContext context = new XYZDataContext())
{
    int count = context.Things.Where(expression).Count();
    //...
}

这是表达式的DebugView：

.Lambda #Lambda1<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>(XYZ.DataAccess.Thing $o)
{
    .Invoke (.Lambda #Lambda2<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>)($o) & .Invoke (.Lambda #Lambda3<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>)($o)
}

.Lambda #Lambda2<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>(XYZ.DataAccess.Thing $o)
{
    .Invoke (.Lambda #Lambda4<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>)($o) & .Invoke (.Lambda #Lambda5<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>)($o)
}

.Lambda #Lambda3<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>(XYZ.DataAccess.Thing $o)
{
    .Invoke (.Lambda #Lambda6<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>)($o) | .Invoke (.Lambda #Lambda7<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>)($o)
}

.Lambda #Lambda4<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>(XYZ.DataAccess.Thing $o)
{
    $o.LocaleID == .Constant<System.Nullable`1[System.Int32]>(1)
}

.Lambda #Lambda5<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>(XYZ.DataAccess.Thing $o)
{
    $o.GenderID == .Constant<System.Nullable`1[System.Int32]>(1)
}

.Lambda #Lambda6<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>(XYZ.DataAccess.Thing $o)
{
    $o.TimeframeID == .Constant<System.Nullable`1[System.Int32]>(2007)
}

.Lambda #Lambda7<System.Func`2[XYZ.DataAccess.Thing,System.Boolean]>(XYZ.DataAccess.Thing $o)
{
    $o.TimeframeID == .Constant<System.Nullable`1[System.Int32]>(2008)
}

这个表达对我来说是正确的,这是相当微不足道的.当我读取调试视图时,我看到：

((LocaleID == 1 && GenderID == 1) && (TimeFrameID == 2007 || TimeFrameID == 2008))

…哪个是对的.

更新1

删除其中一个内部或子句,它工作正常.因此,无论如何,同时使用内部或子句都会破坏从LINQ到sql的转换.

更新2

我无法让调试器进入.NET Framework代码 – 我尝试使用Reflector和Visual Studio一样.我进去了一次,但总的来说,踩踏不起作用.我在StackOverflowException中遇到的一次是在：

ExecutionContext.Run(ExecutionContext executionContext,ContextCallback callback,object state,bool ignoreSyncCtx)

更新3

以下是用于创建Expression的代码.有太多的代码要发布,但它的核心在下面.我有一些类,允许我构建一个复杂的多级查询并将其序列化为JSON和XML.核心是,每个查询都是使用以下方法构建的,然后是Or’d和And’d：

public class LinqSearchField<T,V> : ISearchField
{
    public string Name { get; private set; }
    public Expression<Func<T,V>> Selector { get; private set; }

    public Expression<Func<T,bool>> LessThan(V value)
    {
        return Expression.Lambda<Func<T,bool>>(Expression.LessThan(this.Selector.Body,GetConstant(value)),this.Selector.Parameters);
    }

    public Expression<Func<T,bool>> LessThanOrEqual(V value)
    {
        return Expression.Lambda<Func<T,bool>>(Expression.LessThanOrEqual(this.Selector.Body,bool>> Equal(V value)
    {
        return Expression.Lambda<Func<T,bool>>(Expression.Equal(this.Selector.Body,bool>> NotEqual(V value)
    {
        return Expression.Lambda<Func<T,bool>>(Expression.NotEqual(this.Selector.Body,bool>> GreaterThan(V value)
    {
        return Expression.Lambda<Func<T,bool>>(Expression.GreaterThan(this.Selector.Body,bool>> GreaterThanOrEqual(V value)
    {
        return Expression.Lambda<Func<T,bool>>(Expression.GreaterThanOrEqual(this.Selector.Body,this.Selector.Parameters);
    }

    private ConstantExpression GetConstant(V value)
    {
        return Expression.Constant(value,typeof(V));
    }

    public Expression<Func<T,bool>> Null()
    {
        return Expression.Lambda<Func<T,Expression.Constant(null)),bool>> NotNull()
    {
        return Expression.Lambda<Func<T,this.Selector.Parameters);
    }
}

这是和代码(OR代码是相同的,但使用Expression.And代替)：

public static Expression<Func<T,bool>> And<T>(this Expression<Func<T,bool>> expression1,Expression<Func<T,bool>> expression2)
{
    ParameterExpression[] parameters = expression1.Parameters.Union(expression2.Parameters).Distinct(new ParameterExpressionComparer()).ToArray();
    InvocationExpression invocationExpression1 = Expression.Invoke(expression1,parameters);
    InvocationExpression invocationExpression2 = Expression.Invoke(expression2,parameters);
    Expression binaryExpression = null;

    //And the current expression to the prevIoUs one.
    binaryExpression = Expression.AndAlso(invocationExpression1,invocationExpression2); //Or OrElse.

    //Wrap the expression in a lambda.
    return Expression.Lambda<Func<T,bool>>(binaryExpression,parameters);
}

更新4

它可能会不受欢迎,但这是一个sample which reproduces this issue.我真的需要弄清楚这里发生了什么.