10

如何将几个相似的 SELECT 表达式组合成一个表达式？

   private static Expression<Func<Agency, AgencyDTO>> CombineSelectors(params Expression<Func<Agency, AgencyDTO>>[] selectors)
    {

        // ???

        return null;
    }

    private void Query()
    {
        Expression<Func<Agency, AgencyDTO>> selector1 = x => new AgencyDTO { Name = x.Name };
        Expression<Func<Agency, AgencyDTO>> selector2 = x => new AgencyDTO { Phone = x.PhoneNumber };
        Expression<Func<Agency, AgencyDTO>> selector3 = x => new AgencyDTO { Location = x.Locality.Name };
        Expression<Func<Agency, AgencyDTO>> selector4 = x => new AgencyDTO { EmployeeCount = x.Employees.Count() };

        using (RealtyContext context = Session.CreateContext())
        {
            IQueryable<AgencyDTO> agencies = context.Agencies.Select(CombineSelectors(selector3, selector4));

            foreach (AgencyDTO agencyDTO in agencies)
            {
                // do something..;
            }
        }
    }
4

3 回答 3

19

不简单; 您需要重写所有表达式 - 好吧，严格来说，您可以回收其中的大部分，但问题是x每个表达式都有不同（即使看起来相同），因此您需要使用访问者来替换所有表达式最后 的参数x。幸运的是，这在 4.0 中并不算太糟糕：

static void Main() {
    Expression<Func<Agency, AgencyDTO>> selector1 = x => new AgencyDTO { Name = x.Name };
    Expression<Func<Agency, AgencyDTO>> selector2 = x => new AgencyDTO { Phone = x.PhoneNumber };
    Expression<Func<Agency, AgencyDTO>> selector3 = x => new AgencyDTO { Location = x.Locality.Name };
    Expression<Func<Agency, AgencyDTO>> selector4 = x => new AgencyDTO { EmployeeCount = x.Employees.Count() };

    // combine the assignments from the 4 selectors
    var convert = Combine(selector1, selector2, selector3, selector4);

    // sample data
    var orig = new Agency
    {
        Name = "a",
        PhoneNumber = "b",
        Locality = new Location { Name = "c" },
        Employees = new List<Employee> { new Employee(), new Employee() }
    };

    // check it
    var dto = new[] { orig }.AsQueryable().Select(convert).Single();
    Console.WriteLine(dto.Name); // a
    Console.WriteLine(dto.Phone); // b
    Console.WriteLine(dto.Location); // c
    Console.WriteLine(dto.EmployeeCount); // 2
}
static Expression<Func<TSource, TDestination>> Combine<TSource, TDestination>(
    params Expression<Func<TSource, TDestination>>[] selectors)
{
    var zeroth = ((MemberInitExpression)selectors[0].Body);
    var param = selectors[0].Parameters[0];
    List<MemberBinding> bindings = new List<MemberBinding>(zeroth.Bindings.OfType<MemberAssignment>());
    for (int i = 1; i < selectors.Length; i++)
    {
        var memberInit = (MemberInitExpression)selectors[i].Body;
        var replace = new ParameterReplaceVisitor(selectors[i].Parameters[0], param);
        foreach (var binding in memberInit.Bindings.OfType<MemberAssignment>())
        {
            bindings.Add(Expression.Bind(binding.Member,
                replace.VisitAndConvert(binding.Expression, "Combine")));
        }
    }

    return Expression.Lambda<Func<TSource, TDestination>>(
        Expression.MemberInit(zeroth.NewExpression, bindings), param);
}
class ParameterReplaceVisitor : ExpressionVisitor
{
    private readonly ParameterExpression from, to;
    public ParameterReplaceVisitor(ParameterExpression from, ParameterExpression to)
    {
        this.from = from;
        this.to = to;
    }
    protected override Expression VisitParameter(ParameterExpression node)
    {
        return node == from ? to : base.VisitParameter(node);
    }
}

这使用找到的第一个表达式中的构造函数，因此您可能需要检查所有其他人在各自NewExpression的 s 中是否使用平凡的构造函数。不过，我把它留给了读者。

编辑：在评论中，@Slaks 指出更多的 LINQ 可以缩短这个时间。他当然是对的——不过为了便于阅读，有点密集：

static Expression<Func<TSource, TDestination>> Combine<TSource, TDestination>(
    params Expression<Func<TSource, TDestination>>[] selectors)
{
    var param = Expression.Parameter(typeof(TSource), "x");
    return Expression.Lambda<Func<TSource, TDestination>>(
        Expression.MemberInit(
            Expression.New(typeof(TDestination).GetConstructor(Type.EmptyTypes)),
            from selector in selectors
            let replace = new ParameterReplaceVisitor(
                  selector.Parameters[0], param)
            from binding in ((MemberInitExpression)selector.Body).Bindings
                  .OfType<MemberAssignment>()
            select Expression.Bind(binding.Member,
                  replace.VisitAndConvert(binding.Expression, "Combine")))
        , param);        
}
于 2011-05-30T21:35:07.707 回答
1

如果所有选择器都只会初始化AgencyDTO对象（如您的示例），您可以将表达式转换为NewExpression实例，然后Expression.New使用Members表达式调用。

您还需要将原始表达式中ExpressionVisitor的 s 替换为您正在创建的表达式ParameterExpression的单个ParameterExpression。

于 2011-05-30T21:13:18.317 回答
1

万一其他人偶然发现了与我类似的用例（我的选择根据所需的详细程度针对不同的类）：

简化场景：

public class BlogSummaryViewModel
{
    public string Name { get; set; }

    public static Expression<Func<Data.Blog, BlogSummaryViewModel>> Map()
    {
        return (i => new BlogSummaryViewModel
        {
            Name = i.Name
        });
    }
}

public class BlogViewModel : BlogSummaryViewModel
{
    public int PostCount { get; set; }

    public static Expression<Func<Data.Blog, BlogViewModel>> Map()
    {
        return (i => new BlogViewModel
        {
            Name = i.Name,
            PostCount = i.Posts.Count()
        });
    }
}

我调整了@Marc Gravell 提供的解决方案，如下所示：

public static class ExpressionMapExtensions
{
    public static Expression<Func<TSource, TTargetB>> Concat<TSource, TTargetA, TTargetB>(
        this Expression<Func<TSource, TTargetA>> mapA, Expression<Func<TSource, TTargetB>> mapB)
        where TTargetB : TTargetA
    {
        var param = Expression.Parameter(typeof(TSource), "i");

        return Expression.Lambda<Func<TSource, TTargetB>>(
            Expression.MemberInit(
                ((MemberInitExpression)mapB.Body).NewExpression,
                (new LambdaExpression[] { mapA, mapB }).SelectMany(e =>
                {
                    var bindings = ((MemberInitExpression)e.Body).Bindings.OfType<MemberAssignment>();
                    return bindings.Select(b =>
                    {
                        var paramReplacedExp = new ParameterReplaceVisitor(e.Parameters[0], param).VisitAndConvert(b.Expression, "Combine");
                        return Expression.Bind(b.Member, paramReplacedExp);
                    });
                })),
            param);
    }

    private class ParameterReplaceVisitor : ExpressionVisitor
    {
        private readonly ParameterExpression original;
        private readonly ParameterExpression updated;

        public ParameterReplaceVisitor(ParameterExpression original, ParameterExpression updated)
        {
            this.original = original;
            this.updated = updated;
        }

        protected override Expression VisitParameter(ParameterExpression node) => node == original ? updated : base.VisitParameter(node);
    }
}

扩展类的Map方法则变为：

    public static Expression<Func<Data.Blog, BlogViewModel>> Map()
    {
        return BlogSummaryViewModel.Map().Concat(i => new BlogViewModel
        {
            PostCount = i.Posts.Count()
        });
    }
于 2019-01-10T15:06:15.540 回答