Question

我有List<Scenario>这样：

ScenarioNumber      Key     Value
1                   foo     bar
1                   fez     ben
2                   far     baz
2                   fem     bit

我有List<Mapping>这样：

Key     MappingValue
foo     abc
fez     xyz
far     123
fem     bob

问题：我希望通过Key加入这两项，然后将ScenarioNumber分组到List<ScenarioNode>。

public class ScenarioNode
{
    public int ScenarioNumber { get; set; }

    public List<ScenarioArgument> Arguments { get; set; }
}

ScenarioArgument是一个包含属性的类：Key，Value和MappingValue。

我有以下内容，我被困在Arguments =部分：

var scenarioFormats = from s in this.investment.Scenarios
                      join m in this.keyMappings 
                      on new { Key = s.Key, Level = "DEAL" } 
                      equals new { Key = m.Key, Level = m.Level } into sm
                      from scen in sm.DefaultIfEmpty()
                      group s by s.ScenarioNumber into sg
                      select new ScenarioNode
                      {
                          ScenarioNumber = sg.Key,
                          Arguments = new List<ScenarioArgument> { } // <<-- now what?                 
                      };

Answer 1

你可以试试这个：

var scenarioFormats = from s in this.investment.Scenarios
                      join m in this.keyMappings on new { Key = s.Key, Level = "DEAL" } equals new { Key = m.Key, Level = m.Level } 
                      group new{Scenario=s,KeyMapping=m} by s.ScenarioNumber into sg
                      select new ScenarioNode
                      {
                        ScenarioNumber = sg.Key,
                        Arguments = sg.Select(e=>new ScenarioArgument
                                                 {
                                                   Key=e.Scenario.Key,
                                                   Value=e.Scenario.Key,
                                                   MappingValue=e.KeyMapping.MappingValue,
                                                 } 
                                             )                          
                      };

首先应用加入，然后按Scenario对Mapping和Scenario.ScenarioNumber进行分组，这样，最后您可以预测{{1}中您期望的结果使用结果组

的属性

Answer 2

您可以投射包含Scenario和MappingValue的匿名类型：

var scenarioFormats = from s in this.investment.Scenarios
                      join m in this.keyMappings 
                      on new { Key = s.Key, Level = "DEAL" }
                      equals new { Key = m.Key, Level = m.Level } into sm
                      from scen in sm.DefaultIfEmpty()
                      group new { Scenario = s, scen?.MappingValue } by s.ScenarioNumber into sg
                      select new ScenarioNode
                      {
                          ScenarioNumber = sg.Key,
                          Arguments = sg.Select(s => new ScenarioArgument
                          {
                              Key = s.Scenario.Key,
                              Value = s.Scenario.Value,
                              MappingValue = s.MappingValue
                           }).ToList()
                        };

LINQ连接，然后分组，进入强类型的类列表

2 个答案: