从结构化数据构建JSON层次结构

C# | .NET 4.5 | Entity Framework 5

C＃| .NET 4.5 |实体框架5

I have data coming back from a SQL Query in the form of ID,ParentID,Name. I'd like to take that data and parse it into a Hierarchical JSON string. So far it seems to be much more of a daunting task than it should be. Since I'm using Entity the data comes back nicely to me as an IEnumerable. Now I believe I just need some form of recursion, but I'm not quite sure where to start. Any help is appreciated.

我有一个以ID，ParentID，Name形式从SQL查询返回的数据。我想获取该数据并将其解析为Hierarchical JSON字符串。到目前为止，这似乎是一项艰巨的任务，而不是应有的任务。因为我正在使用Entity，所以数据作为IEnumerable很好地回馈给我。现在我相信我只需要某种形式的递归，但我不确定从哪里开始。任何帮助表示赞赏。

Data Returns as

数据返回为

id   parentId    name
1    1           TopLoc
2    1           Loc1
3    1           Loc2
4    2           Loc1A

Code is

代码是

public static string GetJsonLocationHierarchy(long locationID)
{
    using (EntitiesSettings context = new EntitiesSettings())
    {
        // IEnumerable of ID,ParentID,Name
        context.GetLocationHierarchy(locationID);
    }
}

The end result I'd hope would be something like this:

最终的结果我希望是这样的：

{
    "id": "1",
    "parentId": "1",
    "name": "TopLoc",
    "children": [
        {
            "id": "2",
            "parentId": "1",
            "name": "Loc1",
            "children": [
                {
                    "id": "4",
                    "parentId": "2",
                    "name": "Loc1A",
                    "children": [
                        {}
                    ]
                }
            ]
        },
        {
            "id": "3",
            "parentId": "1",
            "name": "Loc2",
            "children": [
                {}
            ]
        }
    ]
}

2 个解决方案

#1

One way to turn a flat table into a hierarchy is to put all your nodes into a dictionary. Then iterate over the dictionary, and for each node, look up its parent and add it to the parent's children. From there, you just need to find the root and serialize it.

将平面表转换为层次结构的一种方法是将所有节点放入字典中。然后迭代字典，并为每个节点查找其父节点并将其添加到父节点的子节点。从那里，你只需要找到根并序列化它。

Here is an example program to demonstrate the approach:

以下是演示该方法的示例程序：

class Program
{
    static void Main(string[] args)
    {
        IEnumerable<Location> locations = new List<Location>
        {
            new Location { Id = 1, ParentId = 1, Name = "TopLoc" },
            new Location { Id = 2, ParentId = 1, Name = "Loc1" },
            new Location { Id = 3, ParentId = 1, Name = "Loc2" },
            new Location { Id = 4, ParentId = 2, Name = "Loc1A" },
        };

        Dictionary<int, Location> dict = locations.ToDictionary(loc => loc.Id);

        foreach (Location loc in dict.Values)
        {
            if (loc.ParentId != loc.Id)
            {
                Location parent = dict[loc.ParentId];
                parent.Children.Add(loc);
            }
        }

        Location root = dict.Values.First(loc => loc.ParentId == loc.Id);

        JsonSerializerSettings settings = new JsonSerializerSettings
        {
            ContractResolver = new CamelCasePropertyNamesContractResolver(),
            Formatting = Formatting.Indented
        };
        string json = JsonConvert.SerializeObject(root, settings);

        Console.WriteLine(json);
    }
}

class Location
{
    public Location()
    {
        Children = new List<Location>();
    }

    public int Id { get; set; }
    public int ParentId { get; set; }
    public string Name { get; set; }
    public List<Location> Children { get; set; }
}

Here is the output:

这是输出：

{
  "id": 1,
  "parentId": 1,
  "name": "TopLoc",
  "children": [
    {
      "id": 2,
      "parentId": 1,
      "name": "Loc1",
      "children": [
        {
          "id": 4,
          "parentId": 2,
          "name": "Loc1A",
          "children": []
        }
      ]
    },
    {
      "id": 3,
      "parentId": 1,
      "name": "Loc2",
      "children": []
    }
  ]
}

#2

I tried something like this in javascript.

我在javascript中尝试过类似的东西。

list = [
  {
    id: 1,
    title: 'home',
    parent: null
  },
  {
    id: 2,
    title: 'about',
    parent: null
  },
  {
    id: 3,
    title: 'team',
    parent: 2
  },
  {
    id: 4,
    title: 'company',
    parent: 2
  }
]

function treeify(list, idAttr, parentAttr, childrenAttr) {
    if (!idAttr) idAttr = 'id';
    if (!parentAttr) parentAttr = 'parent';
    if (!childrenAttr) childrenAttr = 'children';
    var treeList = [];
    var lookup = {};
    list.forEach(function(obj) {
        lookup[obj[idAttr]] = obj;
        obj[childrenAttr] = [];
    });
    list.forEach(function(obj) {
        if (obj[parentAttr] != null) {
            lookup[obj[parentAttr]][childrenAttr].push(obj);
        } else {
            treeList.push(obj);
        }
    });
    return treeList;
};

console.log(JSON.stringify(treeify(list)));

#1