按对象中的值搜索

Question

我有一个函数可以在节点参数中生成具有深度和子节点数量的对象。

问题是如何编写一个函数来按名称搜索树中的元素？

我试图让它像这样（下面的代码）但它没有进入子节点

const searchByName = (searchValue) => {
    for (let key in completeTree) {
        if(key.name === searchValue){
            return true
        }
    }
}

console.log(searchByName("namesurname"))

Answer 1

可以使用如下递归函数，因为您的代码每次都会生成名称，因此请多次运行它以查看找到和未找到的结果：

请注意，我在您的代码中更改了名称生成字符串。

const getRandomInt = (max) => {
    return Math.floor(Math.random() * Math.floor(max));
}

function getRandomIdentifier() {
    return Math.floor(Math.random() * Math.floor(70));
}

const generateName = (names) => {
    let first = getRandomInt(names.length)
    let second = getRandomInt(names.length)

    return names[first] + names[second]
}

const generateTree = (depth, level, names) => {
    let dateOptions = {
        day: "numeric",
        month: "numeric",
        year: "numeric",
        hour: "numeric",
        minute: "numeric",
        millisecond: "numeric" //millisecond
    }

    let date = new Date()

    let node = {
        name: generateName(names),
        id: getRandomIdentifier(), //unique
        modificationDate: date.toLocaleString('ru', dateOptions),
        type: "folder",
        children: []
    }

    if (depth > 0) {
        for (let i = 0; i < level; i++) {
            node.children.push(generateTree(depth - 1, level, names));
        }
    } else {
        node.type = "file"
    }

    return node
}

let completeTree = generateTree(1, 5, ['aaa', 'bbb', 'ccc'])


//recursive function like below can search in a tree of given structure
function searchByName(searchFor,currentNode){
  let result = "not found";
  if(currentNode.name.indexOf(searchFor)!=-1){
  //found
   result = currentNode;
  }
  if(currentNode.children.length>0){
    for(let i=0; i<currentNode.children.length; i++){
      result = searchByName(searchFor,currentNode.children[i]);
      if(result!="not found"){
        break;
      }
    }
  }
  return result;
}
console.log("Generated Tree:");
console.log(completeTree);
console.log("Search Result for bbbbbb:");
console.log(searchByName('bbbbbb',completeTree));
console.log("Search Result for aaabbb:");
console.log(searchByName('aaabbb',completeTree));
console.log("Search Result for aaaccc:");
console.log(searchByName('aaaccc',completeTree));

Answer 2

因为你不知道先验孩子的深度，你可以使用递归的方法：

const getRandomInt = (max) => {
    return Math.floor(Math.random() * Math.floor(max));
}

function getRandomIdentifier() {
    return Math.floor(Math.random() * Math.floor(70));
}

const generateName = (names) => {
    let first = getRandomInt(names.length)
    let second = getRandomInt(names.length)

    return names[first] + names[second]
}

const generateTree = (depth, level, names) => {
    let dateOptions = {
        day: "numeric",
        month: "numeric",
        year: "numeric",
        hour: "numeric",
        minute: "numeric",
        millisecond: "numeric" //millisecond
    }

    let date = new Date()

    let node = {
        name: generateName(names),
        id: getRandomIdentifier(), //unique
        modificationDate: date.toLocaleString('ru', dateOptions),
        type: "folder",
        children: []
    }

    if (depth > 0) {
        for (let i = 0; i < level; i++) {
            node.children.push(generateTree(depth - 1, level, names));
        }
    } else {
        node.type = "file"
    }

    return node
}

let completeTree = generateTree(1, 5, ['test', 'name', 'surname'])
console.log(JSON.stringify(completeTree, null, 2));

function searchName(current, search) {
  if(current.name === search) {
    return true;
  }
  for(var i=0; i<current.children.length; i++) {
    if(searchName(current.children[i], search)){
      return true;
    }
  }
  return false;
}

console.log("contains xxxx: " + searchName(completeTree, "xxx"))
console.log("contains " + completeTree.name + ": " + searchName(completeTree, completeTree.name))
console.log("contains " + completeTree.children[0].name + ": " + searchName(completeTree, completeTree.children[0].name))
console.log("contains " + completeTree.children[1].name + ": " + searchName(completeTree, completeTree.children[1].name))