MongoDB $graphLookup 获取所有级别的子项 - 嵌套结果

MongoDB $graphLookup get children all levels deep - nested result

https://www.slideshare.net/mongodb/webinar-working-with-graph-data-in-mongodb幻灯片 50 中所述,可以在视图上使用 $graphLookup 以获得 2 层深的树结构嵌套格式.

我有一个 MongoDB 集合,其中树节点作为文档,格式如下:

{ "_id" : { "$oid" : "5b1a952361c6fa3418a15660" }, 
"nodeId" : 23978995, 
"name" : "settings", 
"type" : "Node",
"parentId" : [ 23978893, 23979072, 23979081 ] }

我创建了一个像这样的视图:

db.createView("treeView", "node", [
{
 $graphLookup: {
    from: "node",
    startWith: "$nodeId",
    connectFromField: "nodeId",
    connectToField: "parentId",
    maxDepth: 0,
    as: "children"
 }
}
]);

然后我执行图形查找,例如:

db.node.aggregate([ 
{ $match: {"nodeId": 23978786 } },
{
 $graphLookup: {
    from: "treeView",
    startWith: "$nodeId",
    connectFromField: "nodeId",
    connectToField: "parentId",
    maxDepth: 0,
    as: "children"
 }
}
]);

我的问题是如何获得整个层次结构,所有级别的深度?

遗憾的是,您无法获得嵌套格式的完整深度。使用视图是一种解决方法,可让您执行该操作,但您需要为所需的每个嵌入级别创建一个新视图。 相反,我会考虑在不提供深度的情况下执行 graphLookup,从根级别开始,在单个查询中获取所有层次结构,然后在应用程序级别计算树。

这看起来像这样:

db.node.aggregate([
    { $match: {
        parentId: null
    }},
    { $graphLookup: {
        from: "node",
        startWith: "$nodeId",
        connectFromField: "nodeId",
        connectToField: "parentId",
        depthField: "depth",
        as: "children"
    }}
]);

这应该可以让您一次性获取整个层次结构,因此接下来,您需要根据 "children" 数组中的信息计算应用程序中的树。

@kmandalas 我最近两天遇到了这种问题, 我的 collection 有点不同,但概念与你的相同我希望我写的内容能帮助你得到结果,(我使用 SO 答案的参考)

我的 Collection 架构如下:

const Category = new Schema({
    sTitle: { type: String, trim: true },
    iParentId: { type: mongoose.Schema.Types.ObjectId, ref: 'Category' },
    bStatus: { type: Boolean, default: true },
    dUpdatedAt: { type: Date },
    dCreatedAt: { type: Date, default: Date.now }
});

首先,我使用 $graphLookup,然后我将所有 children 归为一个合适的 parent,例如:

{
  "_id": "5c6fa228c30bbf02cf12fe6c",
   "sTitle": "firstParent",
   "childrens":[{obj},{obj},{obj},{obj}]    // Childrens as well as grandChild
},
{
  "_id": "5c80d644ab57dd06d48cc474",
   "sTitle": "secondParent",
   "childrens":[]     //No Child
},
.....

得到这种结果后,我在节点 js 中创建了一棵树,没有使用任何 third-party 库。 树逻辑代码是:(!note:docs is $graphlooup output nothing else)

function list_to_tree(list) {
            var map = {}, node, roots = [], i;
            for (i = 0; i < list.length; i += 1) {
                map[list[i]._id] = i;
                list[i].children = [];
            }
            for (i = 0; i < list.length; i += 1) {
                node = list[i];
                if (node.iParentId !== null && map[node.iParentId] !== undefined) {
                    var node2 = {         //Because i need only _id,Title & childrens
                        _id: node._id,
                        Title: node.sTitle,
                        children: node.children
                    }
                    list[map[node.iParentId]].children.push(node2); //You can push direct "node"
                } else {
                    var node2 = {
                        _id: node._id,
                        Title: node.sTitle,
                        children: node.children
                    }
                    roots.push(node2);
                }
            }
            return roots;
        }
        let final_result = []     //For Storing all parent with childs
        if (docs.length >= 0) {   
            docs.map(single_doc => {  //For getting all parent Tree
                var single_child = list_to_tree(single_doc.children)
                var obj = {
                    _id: single_doc._id,
                    Title: single_doc.sTitle,
                    children: single_child
                }
                final_result.push(obj)
            })
        }
        console.log("Final Tree is : ",final_result)

希望对您有所帮助