5

这是我第一次听说LINQ，我对此一无所知。请对我温柔一点。

我有这组数据。

+---------+--------+---------------+
| RadioID | NodeID | SourceRadioID |
+---------+--------+---------------+
| R0      |      1 |               |
| R1      |      1 |               |
| R2      |      1 |               |
| R3      |      1 |               |
| R4      |      1 |               |
| R5      |      2 |               |
| R6      |      2 |               |
| R7      |      2 | R0            |
| R8      |      2 |               |
| R9      |      2 |               |
| R10     |     11 |               |
| R11     |     11 | R9            |
| R12     |     11 |               |
| R13     |     11 |               |
+---------+--------+---------------+

我需要做的是编写一个返回列表的方法NodeID。例子，

List<int> dependentNode = GetChildNode(1); // int ParentNode

我的预期结果是 NodeIDs: 2 and 11。

NodeID = 2被包括在内，因为有一个RadioID = R7连接到RadioID = R0的属于NodeID = 1。

NodeID = 11也包括在内，因为RadioID = R11连接到Radio = R9哪个属于NodeID = 2（也连接到NodeID = 1）。

我查找这篇文章，但我总是得到StackOverFlowException

这是完整的代码：

public class RadioEntity
{
    public string RadioID { get; set; }
    public int NodeID { get; set; }
    public string SourceRadioID { get; set; }
}

public class SampleDemo
{
    public void SampleMethod()
    {

        Func<int, int,List<int>> GetChildNode = null;
        GetChildNode = (x, y) =>
            {
                return (from _x in GetRadio()
                        where (GetRadio().Where(i => i.NodeID == x).Select(i => i.RadioID)).Contains(_x.RadioID)
                        from _y in new[] { _x.NodeID }.Union(GetChildNode(_x.NodeID, y + 1))
                        select _y).ToList<int>();
            };

        var _res = GetChildNode(1, 0);

    }

    public List<RadioEntity> GetRadio()
    {
        List<RadioEntity> _returnVal = new List<RadioEntity>();
        _returnVal.Add(new RadioEntity() { RadioID = "R0", NodeID = 1, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R1", NodeID = 1, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R2", NodeID = 1, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R3", NodeID = 1, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R4", NodeID = 1, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R5", NodeID = 2, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R6", NodeID = 2, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R7", NodeID = 2, SourceRadioID = "R0" });
        _returnVal.Add(new RadioEntity() { RadioID = "R8", NodeID = 2, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R9", NodeID = 2, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R10", NodeID = 11, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R11", NodeID = 11, SourceRadioID = "R9" });
        _returnVal.Add(new RadioEntity() { RadioID = "R12", NodeID = 11, SourceRadioID = "" });
        _returnVal.Add(new RadioEntity() { RadioID = "R13", NodeID = 11, SourceRadioID = "" });

        return _returnVal;
    }

}

您可以建议是否有更好的方法来做到这一点。在这里对不起新手。

4

1 回答 1

2

如果您是新手，请不要对递归和 lambdas 太聪明。

    public List<int> GetChildren(int id)
    {
        var nodes = GetRadio();
        var parent = nodes.Single(n => n.NodeID == id);
        var children = nodes.Where(n => n.SourceRadioID == parent.RadioID).Select(n => n.NodeID);

        return children.Union(children.SelectMany(GetChildren)).ToList();
    }

更新 1

public List<int> GetChildren(int id)
{
    IEnumerable<RadioEntity> parent = GetRadio().Where(x => x.NodeID == id);
    IEnumerable<int> children = (
                                    from r in GetRadio()
                                    where parent.Select(x=>x.RadioID)
                                                .Contains(r.SourceRadioID)
                                    select r
                                ).Select(n => n.NodeID);

    return children.Union(children.SelectMany(GetChildren)).ToList();
}
于 2013-05-01T03:22:55.003 回答