MongoDB $ graphLookup让所有级别的孩子都深入 - 嵌套结果

时间:2018-06-08 20:24:24

标签: mongodb aggregation-framework tree-structure

https://www.slideshare.net/mongodb/webinar-working-with-graph-data-in-mongodb中所示,幻灯片50 ,可以在视图上使用$graphLookup,以便在嵌套中获得2级深度树结构格式

我有一个带树节点的MongoDB集合作为文档,格式如下:

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

我创建了一个类似的视图:

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

我执行图形查找,如:

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

我的问题是如何才能获得整个层次结构,所有层次都很深?

2 个答案:

答案 0 :(得分:3)

不幸的是,您无法以嵌套格式获得完整深度。使用视图是一种允许您执行该操作的变通方法,但您需要为所需的每个嵌入级别创建新视图。 相反,在考虑应用程序级别的树之前,我会考虑执行graphLookup而不提供深度,从根级别开始,在单个查询中获取所有层次结构。

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"
    }}
]);

这应该允许您一次性获取整个层次结构,接下来,您需要根据“children”数组中的信息计算应用程序中的树。

答案 1 :(得分:0)

@kmandalas从最近两天开始,我就面临这种问题, 我的收藏集有些不同,但概念与您的相同。我希望我写的内容能帮助您获得结果,(我使用SO答案的引用)

我的收藏模式如下:

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,然后将所有孩子归为一个合适的父母,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,我在不使用任何第三方库的情况下在节点js中创建了一棵树。 树逻辑代码是:(!note:docs是$ graphlooup,什么也没输出)

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)

希望对您有帮助