Question

我无法让这个工作，如果有人可以提供帮助，我将不胜感激。

所以我从数据库中获取XML结果，如下所示：

<matches>
  <issuer client_name="MTR" score="6" match_list="MTR CORPORATION LIMITED"/>
  <issuer client_name="PEOPLE''S REPUBLIC OF CHINA" score="4"
          match_list="DEMOCRATIC PEOPLE'S REPUBLIC OF KOREA;GOVERNMENT OF THE
          HONG KONG SPECIAL ADMINISTRATIVE REGION OF THE PEOPLE'S REPUBLIC OF
          CHINA;MONGOLIAN PEOPLE'S REPUBLIC;PEOPLE'S DEMOCRATIC REPUBLIC OF
          ALGERIA;PEOPLE'S REPUBLIC OF CHINA"/>
</matches>

从这个XML开始，我需要在做一些逻辑之后填充一个对象，比如匹配client_name我在XML结果中回到我发送到数据库以获得匹配的那个。

       XDocument.Load(new StringReader(
                              row.GetString(row.GetOrdinal("xml_ret"))))))
                 .Single().Descendants("matches")
                 .Select(x =>
                     new Pool() {
                         Constituents = (IEnumerable<Constituent>)
                            //(... cannot work this how can IsReference populate)
                         //ClientName = x.Attribute("client_name").Value,
                         //Score = x.Attribute("score").Value,
                         //MatchList = x.Attribute("match_list").Value,
                     });

以非LINQ的方式，我可以填充像这样的对象：

foreach (Constituent constituent in pool.Constituents)
        {
            if (!string.IsNullOrEmpty(constituent.Name)
                && string.IsNullOrEmpty(constituent.Curve))
            {
                i++;

                ConstituentMatch match = new ConstituentMatch();
                ConstituentMatch.Group group =new ConstituentMatch.Group("High");

                //High Match group
                ICollection<string> curves = new List<string>();
                curves.Add("EUR" + i);
                curves.Add("USD" + i);

                ICollection<string> names = new List<string>();

                ConstituentMatch.Group.Entry entry =
                    new ConstituentMatch.Group.Entry(constituent.Name + " Ltd.",
                                                     curves);
                group.Add(entry);
                entry =
                    new ConstituentMatch.Group.Entry(constituent.Name + " Inc.",
                                                     curves);
                group.Add(entry);

                match.AddGroup(group);

            }
        }

但是我如何使用LINQ来做到这一点，因为我相信你可以做到这一点，我无法解决这个问题。

构成类看起来像：

public sealed class Constituent
{
    private string name;

    public string Name
    {
        get { return name; }
        set { name = value; }
    }

    public ConstituentMatch Match {get;set;}

    public Constituent(string name)
    {
        this.name = name;
    }

    public Constituent() : this(string.Empty) { }
}

成分匹配类看起来像这样：

public sealed class ConstituentMatch
{
    private readonly Dictionary<string, Group> matches = new Dictionary<string, Group>();

    public IEnumerable<string> GroupNames
    {
        get { return matches.Keys; }
    }

    public Group this[string name]
    {
        get { return matches[name]; }
    }

    public IEnumerable<Group> Groups
    {
        get { return matches.Values; }
    }

    public void AddGroup(Group group)
    {
        matches[group.Name] = group;
    }

    /// <summary>
    /// Match group e.g. Poor, High, All, Begins With
    /// </summary>
    public sealed class Group
    {
        private readonly string name;
        private readonly ICollection<Entry> matches = new List<Entry>();

        public string Name
        {
            get { return name; }
        }

        public Group(string name)
        {
            this.name = name;
        }

        public void Add(Entry entry)
        {
            matches.Add(entry);
        }

        public override bool Equals(object obj)
        {
            bool result = false;
            if (obj is Group)
            {
                Group other = obj as Group;
                result = name == other.name;
            }
            return result;
        }

        public override int GetHashCode()
        {
            return name.GetHashCode();
        }

        public sealed class Entry
        {
            private string legalName;
            private IEnumerable<string> curves;

            private double notional     = double.NaN;
            private char seniriority    = 'N';

            public string LegalName
            {
                get { return legalName; }
            }

            public IEnumerable<string> Curves
            {
                get { return curves; }
            }

            public Entry(string legalName, IEnumerable<string> curves)
                       : this(legalName, curves, double.NaN, 'N') { }

            public Entry(string legalName,
                         IEnumerable<string> curves,
                         double notional,
                         char seniriority)
            {
                this.legalName = legalName;
                this.curves = curves;
                this.notional = notional;
                this.seniriority = seniriority;
            }
        }
    }
}

Answer 1

语言集成查询正如其名称所示，是查询对象和数据的技术，而不是用于修改它们。

Answer 2

类似的东西应该有用

var haystack = new Pool().Constituents;
var indexedhaystack = haystack.Select((item, index)=> new {
    item, index
});
var pool = new Pool()
{
    Constituents = from l in indexedhaystack
                   select new Constituent()
                   {
                        //your stuff here
                   }
};

...延长......

var constituents = new Pool().Constituents.Select((c, i) =>
    new
    {
        Constituent = c,
        Index = i
    });
var items = from c in constituents
            where !string.IsNullOrEmpty(c.Constituent.Name)
                  && string.IsNullOrEmpty(c.Constituent.Curve)
            let curves = new[]{
                            "EUR" + c.Index.ToString(),
                            "USD" + c.Index.ToString()
                               }
            let match = new ConstituentMatch(){
                new Group("High") {
                    new Entry(
                                c.Constituent.Name + " Ltd.",
                                curves),
                    new Entry(
                                c.Constituent.Name + " Inc.",
                                curves)
                }
            }
            select new
            {
                Name = c.Constituent.Name,
                Curves = curves,
                Match = match
            };

...

public class Constituent
{
    public string Name { get; set; }
    public string Curve { get; set; }
}
public class Pool
{
    public List<Constituent> Constituents { get; set; }
}
public class Entry
{
    public Entry(string entry, IEnumerable<string> curves)
    {
    }
}
public class Group : List<Entry>
{
    public Group(string group) { }
}
public class ConstituentMatch : List<Group>
{

}

如何在Lambda表达式中填充IEnumerable？

2 个答案: