Question

如果我提出一个愚蠢的问题，我感到很抱歉，但这令我不安，而且我无法找到最佳解决方案。

我有一个JSON数据，如下所示：

{
    "my_data": [
        {
            "name": "bugs_db",
            "type": "database",
            "children": [
                {
                    "name": "oss",
                    "type": "ui"
                },
                {
                    "name": "dashboard",
                    "type": "ui"
                },
                {
                    "name": "dev-dash",
                    "type": "ui"
                }
            ]
        },
        {
            "name": "oss",
            "type": "ui",
            "children": [
                {
                    "name": "active-directory",
                    "type": "nfs"
                },
                {
                    "name": "passive-directory",
                    "type": "FAT32"
                }
            ]
        },
        {
            "name": "jira_db",
            "type": "database",
            "children": [

            ]
        },
        {
            "name": "active_directory",
            "type": "nfs",
            "children": []
        }
    ]
}

我正在尝试处理此数据，以便对于选定的根（名称），将存在数据的层次关系。对于exa。结果数据应如下所示（如果我选择＆＃34; bugs_db＆＃34;）。

{
    "name": "bugs_db",
    "kind": "root",
    "children": [
        {
            "name": "oss",
            "type": "ui",
            "children": [
                {
                   "name": "active-directory",
                   "type": "nfs",
                   "children": []
                },
                {
                    "name": "passive-directory",
                    "type": "FAT32"
                }
            ]
        },
        {
            "name" : "dashboard",
            "type": "ui"
        },
        {
            "name": "dev-dash",
            "type": "ui"
        }
    ]
}

我曾尝试编写直到第1级的代码..

var selectedApp = "bugs_db";
var all_data = {
  name: selectedApp,
  type: "root",
  children: []
}
for(var i = 0; i < data.my_data.length; i++){
  var currentObj = data.my_data[i];
  if(currentObj.name == selectedApp && currentObj.children.length){
      for(var j = 0; j < currentObj.children.length; j++){
           let childObj = {
              name: currentObj.children[j].name,
              type: currentObj.children[j].type,
              children: []
           }
           allData.children.push(childObj);
      }
  }
}

但是上面的代码没有进一步的层次结构。我知道这可以使用某种递归函数来完成..但我不确定如何......可能就是为什么会陷入无限循环。

任何人都可以帮助我。如果您需要更多信息，请与我们联系。

Answer 1

您可以通过构建新对象来获取地图和递归方法。

function getTree(name) {
    var object = map.get(name);
    return object && Object.assign({}, object, { children: object.children.map(o => Object.assign({}, o, getTree(o.name))) });
}

var object = { my_data: [{ name: "bugs_db", type: "database", children: [{ name: "oss", type: "ui" }, { name: "dashboard", type: "ui" }, { name: "dev-dash", type: "ui" }] }, { name: "oss", type: "ui", children: [{ name: "active-directory", type: "nfs" }, { name: "passive-directory", type: "FAT32" }] }, { name: "jira_db", type: "database", children: [] }, { name: "active_directory", type: "nfs", children: [] }] },
    map = new Map(object.my_data.map(o => [o.name, o]));

console.log(getTree("bugs_db"));

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通过循环检查。

function getTree(name, visited = new Set) {
    var object = map.get(name);

    if (!object) {
        return object;
    }

    if (visited.has(name)) {
        return object && Object.assign({}, object, { circular: true, children: [] });
    }

    visited.add(name);
    return Object.assign({}, object, { children: object.children.map(o => Object.assign({}, o, getTree(o.name, visited))) });
}

var object = { my_data: [{ name: "bugs_db", type: "database", children: [{ name: "oss", type: "ui" }, { name: "dashboard", type: "ui" }, { name: "dev-dash", type: "ui" }, { name: "bugs_db", type: "exception" }] }, { name: "oss", type: "ui", children: [{ name: "active-directory", type: "nfs" }, { name: "passive-directory", type: "FAT32" }] }, { name: "jira_db", type: "database", children: [] }, { name: "active_directory", type: "nfs", children: [] }] },
    map = new Map(object.my_data.map(o => [o.name, o]));

console.log(getTree("bugs_db"));

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Answer 2

你想要这样的东西吗？

var data = {
    "my_data": [
        {
            "name": "bugs_db",
            "type": "database",
            "children": [
                {
                    "name": "oss",
                    "type": "ui"
                },
                {
                    "name": "dashboard",
                    "type": "ui"
                },
                {
                    "name": "dev-dash",
                    "type": "ui"
                }
            ]
        },
        {
            "name": "oss",
            "type": "ui",
            "children": [
                {
                    "name": "active-directory",
                    "type": "nfs"
                },
                {
                    "name": "passive-directory",
                    "type": "FAT32"
                }
            ]
        },
        {
            "name": "jira_db",
            "type": "database",
            "children": [

            ]
        },
        {
            "name": "active-directory",
            "type": "nfs",
            "children": [
            {
            	"name": "ubuntu",
            	"type": "os"
            }
            ]
        },
        {
            "name": "ubuntu",
            "type": "os",
            "children": []
        }
    ]
};
var selectedApp = "bugs_db";
var all_data = {
  name: selectedApp,
  type: "root",
  children: []
}
for(var i = 0; i < data.my_data.length; i++){
  var currentObj = data.my_data[i];
  if(currentObj.name == selectedApp && currentObj.children.length){
      for(var j = 0; j < currentObj.children.length; j++){
           let childObj = {
              name: currentObj.children[j].name,
              type: currentObj.children[j].type,
           }
           findChildren(childObj)
           all_data.children.push(childObj);
      }
  }
}

function findChildren(obj){
	obj.children = data.my_data.filter(o => o.name == obj.name && o.type == obj.type ).map(o => o.children)[0];
  if (obj.children)
  obj.children.forEach(child => {
  	findChildren(child);
  });
}




console.log(all_data)

Answer 3

你可以这样做：

const selectedApp = "bugs_db";
const data = {"my_data": [{"name": "bugs_db","type": "database","children": [{"name": "oss","type": "ui"},{"name": "dashboard","type": "ui"},{"name": "dev-dash","type": "ui"}]},{"name": "oss","type": "ui","children": [{"name": "active-directory","type": "nfs"},{"name": "passive-directory","type": "FAT32"}]},{"name": "jira_db","type": "database","children": []},{"name": "active_directory","type": "nfs","children": []}]};
const all_data = {
  name: selectedApp,
  type: "root",
  children: []
};

all_data.children = data.my_data
  .find(el => el.name == selectedApp).children
  .map(obj => {
    const found = data.my_data.find(el => el.name == obj.name);
    if (found && found.children) {
      obj.children = found.children;
    }
    return obj;
  });

console.log(all_data);

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处理JSON以创建层次关系

3 个答案: