c# - 使用递归属性 ASP.NET/C#

Question

我正在寻找一种方法，通过 ASP.NET MVC 使用 jQuery Dyantree 获取通用结果集并将其转换为树结构。我有一个像这样的测试类:

public class GenericResults
{
    public string ClaimId { get; set; }
    public string DrugName { get; set; }
    public int PatientId { get; set; }

    //simulates database call, this could be any type of result set
    public static List<GenericResults> MockDatabaseCall()
    {
        return new List<GenericResults>()
        {
            new GenericResults { ClaimId = "abc", DrugName="Drug 1",  PatientId=1 },
            new GenericResults { ClaimId = "bcd", DrugName="Drug 2",  PatientId=1 },
            new GenericResults { ClaimId = "def", DrugName="Drug 3",  PatientId=1 },
            new GenericResults { ClaimId = "fgi", DrugName="Drug 4",  PatientId=1 },
            new GenericResults { ClaimId = "ijk", DrugName="Drug 5",  PatientId=2 },
            new GenericResults { ClaimId = "klm", DrugName="Drug 6",  PatientId=2 },
            new GenericResults { ClaimId = "mno", DrugName="Drug 7",  PatientId=2 },
            new GenericResults { ClaimId = "pqr", DrugName="Drug 8",  PatientId=2 },
            new GenericResults { ClaimId = "qrs", DrugName="Drug 9",  PatientId=2 },
            new GenericResults { ClaimId = "stu", DrugName="Drug 10", PatientId=2 },
        };
    }
}

Dynatree 的数据结构如下:

public string Title { get; set; }
public List<TreeView> Children { get; set; }
public bool IsFolder
{
    get
    {
        return this.Children.Count > 0;
    }
}

在这个例子中，我们确实不需要 ClaimId 属性，但我想重用我已经编写的存储过程来提供给树，所以我想包含它。我需要分组的内容成为它自己的 TreeView 对象，而 Children 属性是所有具有分组内容值的记录。到目前为止，MockDatabaseCallMethod，如果我在 PatientId 上分组，我将需要两个 TreeView 对象:一个对应于每个唯一的 PatientId。我希望能够递归地执行此操作，但不确定如何做，因为有些项目将成为叶子(没有 child )，这很好。这是我的尝试:

public class TreeView
{
    public string Title
    {
        get;
        set;
    }

    public List<TreeView> Children { get; set; }

    public bool IsFolder
    {
        get
        {
            return this.Children.Count > 0;
        }
    }

    //made void for the time being to limit compilation errors
    public static void ListToTree(string displayName,string groupedOn)
    {
        //seed data
        List<GenericResults> results = GenericResults.MockDatabaseCall();

        //"group" on the property passed by the user
        var query = results.Select(x => x.GetType().GetProperty(groupedOn).GetValue(x,null)).Distinct();

        foreach (var i in query)
        {
            var treeView = new TreeView();
            treeView.Title = displayName;

            //iterate over results, if object has the property value of what's currently being iterated over
            //create a new TreeView object for it
            treeView.Children = results.Where(x => x.GetType().GetProperty(groupedOn).GetValue(x, null) == i)
                                       .Select(n => new TreeView
                                                    {
                                                        Title = displayName,
                                                        Children = x <--no idea what to do here
                                                    });
        }
    }
}

理想情况下，我想要一个 ToTree 扩展方法，但首先我想获得一些关于此的指示。谢谢 :)

Answer 1

我不确定我是否理解正确，但如果我理解正确，我会建议以下解决方案:

-) 在 TreeView更改属性 public List<TreeView> Children { get; set; }至 public IEnumerable<GenericResults> Children { get; set; } .

-) 替换你的 void ListToTree(string displayName使用以下方法:

public static List<TreeView> ListToTree(string displayName, string groupedOn)
{
    //seed data
    List<GenericResults> results = GenericResults.MockDatabaseCall();

    List<TreeView> trees = new List<TreeView>();

    //"group" on the property passed by the user
    var query = results.Select(x => x.GetType().GetProperty(groupedOn).GetValue(x, null)).Distinct();

    foreach (int i in query)
    {
        var treeView = new TreeView();
        treeView.Title = displayName;
        treeView.Children = results.Where(x => ((int)x.GetType().GetProperty(groupedOn).GetValue(x, null)) == i);

        trees.Add(treeView);
    }

    return trees;
}

不过我不确定我是否理解正确;-)