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如何在javascript中使用递归查找对象?

javascript

user944513 · 技术社区 · 6 年前

let tree = {
    name: "A",
    children: [
        {
            name: 'A-1',
            children: [
                {name: "A-1-A"},
                {name: "A-1-B"},
            ]
        },
        {
            name: 'B-1',
            children: [
                {
                    name: "B-1-A",
                    children: [
                        {name: "B-11-A"},
                        {name: "B-11-B"}
                    ]
                },
                {name: "B-1-B"},
            ]
        },

    ]


};

object 从 tree object 使用递归。

searchFn(tree,'A-1') 它应该会回来 { name: 'A-1', children: [ {name: "A-1-A"}, {name: "A-1-B"}, ] }

我这样叫它 searchFn(tree,'A-1-A') 它应该还这个

{name: "A-1-A"}

我试过了但没用

 function searchFn(obj ,searchText){
        if(obj.name === searchText) return obj
        if(obj.children.length > 0){
          return   searchFn(obj.children.pop(),searchText)
        }
      return null
    }

2 回复 | 直到 6 年前

Nina Scholz 6 年前

您需要迭代对象的子对象并为结果获取一个变量。

function searchFn(object, searchText) {
    var result;
    if (object.name === searchText) return object;
    (object.children || []).some(o => result = searchFn(o, searchText));
    return result || null;
}


let tree = { name: "A", children: [{ name: 'A-1', children: [{ name: "A-1-A" }, { name: "A-1-B" }] }, { name: 'B-1', children: [{ name: "B-1-A", children: [{ name: "B-11-A" }, { name: "B-11-B" }] }, { name: "B-1-B" }] }] };

console.log(searchFn(tree, 'foo'));
console.log(searchFn(tree, 'A-1'));

.as-console-wrapper { max-height: 100% !important; top: 0; }

Alexandre Senges 6 年前

    class Node {
        static fromStruct = (struct) => {
            const tree = new Node(struct.name);
            if(!struct.children) return tree;
            for(const child of struct.children) {
                tree.children.push(Node.fromStruct(child));
            }
            return tree;
        }

        constructor(name){
            this.name = name;
            this.children = [];
        }
    
        addChild = (parentName, childName) => {
            const parent = this.searchFn(parentName);
            if(!parent) throw new Error("Parent not found");
            parent.children.push(new Node(childName));
        }
    
        searchFn = (name) => {
            if(this.name === name) return this;
            for(const child of this.children) {
                const found = child.searchFn(name);
                if(found !== null) return found;
            }
            return null;
        }
    }


    const data = {
        name: "A",
        children: [
            {
                name: 'A-1',
                children: [
                    {name: "A-1-A"},
                    {name: "A-1-B"},
                ]
            },
            {
                name: 'B-1',
                children: [
                    {
                        name: "B-1-A",
                        children: [
                            {name: "B-11-A"},
                            {name: "B-11-B"}
                        ]
                    },
                    {name: "B-1-B"},
                ]
            },
    
        ]
    };

    const tree = Node.fromStruct(data);
    console.log(tree.searchFn("A-1-A"));

xdeepakv 6 年前

const tree = {"name":"A","children":[{"name":"A-1","children":[{"name":"A-1-A"},{"name":"A-1-B"}]},{"name":"B-1","children":[{"name":"B-1-A","children":[{"name":"B-11-A"},{"name":"B-11-B"}]},{"name":"B-1-B"}]}]}

const searchFn = (tree, name) => {
  let result = null;
  if (typeof tree !== "object") return result;
  if (tree.name === name) return tree;
  if (tree.children && tree.children.length) {
    tree.children.some(data => (result = searchFn(data, name)));
  }
  return result;
};
console.log(searchFn(tree, "A-1-A"));
console.log(searchFn(tree, "A-1"));
console.log(searchFn(tree, ""));

.as-console-row {color: blue!important}

Code Maniac 6 年前

你可以试试这样的东西

let tree = { name: "A", children: [{ name: 'A-1', children: [{ name: "A-1-A" }, { name: "A-1-B" }] }, { name: 'B-1', children: [{ name: "B-1-A", children: [{ name: "B-11-A" }, { name: "B-11-B" }] }, { name: "B-1-B" }] }] };

function searchFn(obj, text){
  if(obj.name === text) return obj
  else{
    if(obj && obj.children && Array.isArray(obj.children)){
      for(let value of obj.children){
        if(searchFn(value, text)){
          return searchFn(value, text)
        }
      }
    }
  }
  return null
}

console.log(searchFn(tree, 'A-1'))
console.log(searchFn(tree, 'A-1-A'))
console.log(searchFn(tree, 'A-A-A'))

Param Siddharth 6 年前

函数的问题是它只检查 children

function searchFn(obj ,searchText) { 
    if(obj.name === searchText) return obj; 
    if(obj.children) {
        for (let x of obj.children) {
            let y = searchFn(x,searchText);
            if (y)
                return y;
        }
    }
    return null; 
}