c# - 在 Linq 中展平和分组复杂对象并保留空子项

Question

我有一个名为 RouteExport 的复杂对象的列表，我试图根据 CustomerNumber 值将其展平和分组，以便它返回一个匿名对象，如下所示

{ CustomerNumber = "1235", Route = route1, Section = section2, Sequence = sequence2 } 

或

{ CustomerNumber = "1234", Route = route1, Section = null, Sequence = null }

模型看起来像这样:

public class RouteExport
{
    public string Name { get; set; }
    public string Term { get; set; }
    public List<string> CustomerNumbers { get; set; }
    public List<SectionExport> Sections { get; set; }
}

public class SectionExport
{
    public string Name { get; set; }
    public List<string> CustomerNumbers { get; set; }
    public List<SequenceExport> Sequences { get; set; }
}

public class SequenceExport
{
    public string Name { get; set; }
    public List<string> CustomerNumbers { get; set; }
}

每个对象都有一个 CustomerNumber 列表，如果客户在该路线/路段/序列中，则该列表包含客户的号码。我想根据该客户编号对每个对象进行分组。到目前为止，我一直在使用这个:

var flattendExport = exportViewModel.ExportContainer.Routes
            .SelectMany(rt => rt.Sections
                .SelectMany(sec => sec.Sequences
                    .SelectMany(seq => seq.CustomerNumbers
                        .Select(custNum => new { rt, sec, seq, custNum })))).ToList();

展平对象但不按 CustomerNumber 分组，也不为部分或序列返回 null。

我如何进行查询以返回按每个对象的 CustomerNumbers 分组并返回 null(如果它们不在部分或序列中)的扁平化列表？

用测试用例更新

我没有充分表述原始问题，因此进行了一些修改。我想对客户的号码进行分组，以便此数据:

var data = new List<RouteExport>
{
    new RouteExport
    {
        CustomerNumbers = new List<string> { "1", "2" },
        Sections = new List<SectionExport>
        {
            new SectionExport()
            {
                CustomerNumbers = new List<string> { "1" },
                Sequences = new List<SequenceExport> {
                    new SequenceExport()
                    {
                        CustomerNumbers = new List<string> { "1" }
                    }
                }
            }
        }
    }
};

返回以下结果:

{ CustomerNumber = "1", Route = route1, Section = section1, Sequence = sequence1},
{ CustomerNumber = "2", Route = route1, Section = null, Sequence = null}

Answer 1

当 Sections 集合为空时，您可以使用 LINQ 的 DefaultIfEmpty 插入您需要的空值。参见 https://msdn.microsoft.com/en-us/library/bb355419(v=vs.110).aspx

在第二个.SelectMany(...) 的末尾添加:

.DefaultIfEmpty(new { rt, null, null, cystNum })

“对于没有路段的每条路线，创建一个虚拟结果”。

编辑:鉴于您更改的要求，它看起来像这样:

 var res = data.SelectMany(r => r.CustomerNumbers
            .SelectMany(c => r.Sections.Where(s => s.CustomerNumbers.Contains(c))
                .SelectMany(s => s.Sequences
                  .Select(seq => new { CustomerNumber = c, Route = r, Section = s, Sequence = seq }))
                .DefaultIfEmpty(new { CustomerNumber = c, Route = r, Section = (SectionExport)null, Sequence = (SequenceExport)null })))
                .ToList();