我有以下带有n级孩子的JavaScript，并且想要搜索id，如果任何项目来自匹配id，则需要将对象从root返回到匹配项目。
我希望将找到的项目的整个层次结构从root返回到包含它的孩子的对象。
我尝试使用lodash和下划线，但找不到简单的解决方案。
input: {
"children": [{
"name": "Home",
"title": "Home",
"id": "home1",
"children": []
},
{
"name": "BUSINESS AND ROLE SPECIFIC",
"title": "BUSINESS AND ROLE SPECIFIC",
"id": "BAR1",
"children": [{
"name": "Global Businesses",
"title": "Global Businesses",
"id": "GB1",
"children": [{
"name": "Commercial Banking",
"title": "Commercial Banking",
"id": "CB1",
"children": [{
"name": "FLAGSHIP PROGRAMMES",
"title": "FLAGSHIP PROGRAMMES",
"id": "FG1",
"children": []
}]
}]
}]
},
{
"name": "RISK MANAGEMENT",
"title": "RISK MANAGEMENT",
"id": "RM1",
"children": []
}
]
}
Search: {
id: 'FG1'
}
return :{
"name": "BUSINESS AND ROLE SPECIFIC",
"title": "BUSINESS AND ROLE SPECIFIC",
"id": "BAR1",
"children": [{
"name": "Global Businesses",
"title": "Global Businesses",
"id": "GB1",
"children": [{
"name": "Commercial Banking",
"title": "Commercial Banking",
"id": "CB1",
"children": [{
"name": "FLAGSHIP PROGRAMMES",
"title": "FLAGSHIP PROGRAMMES",
"id": "FG1",
"children": [{}]
}]
}]
}]
}
答案 0 :(得分：2)
您可以使用此功能：
function findChild(obj, condition) {
if (Object.entries(condition).every( ([k,v]) => obj[k] === v )) {
return obj;
}
for (const child of obj.children || []) {
const found = findChild(child, condition);
// If found, then add this node to the ancestors of the result
if (found) return Object.assign({}, obj, { children: [found] });
}
}
// Sample data
var input = { "children": [{ "name": "Home", "title": "Home", "id": "home1", "children": [] }, { "name": "BUSINESS AND ROLE SPECIFIC", "title": "BUSINESS AND ROLE SPECIFIC", "id": "BAR1", "children": [{ "name": "Global Businesses", "title": "Global Businesses", "id": "GB1", "children": [{ "name": "Commercial Banking", "title": "Commercial Banking", "id": "CB1", "children": [{ "name": "FLAGSHIP PROGRAMMES", "title": "FLAGSHIP PROGRAMMES", "id": "FG1", "children": [] }] }] }] }, { "name": "RISK MANAGEMENT", "title": "RISK MANAGEMENT", "id": "RM1", "children": [] } ]},
search = { id: 'FG1' };
console.log(findChild(input, search));.as-console-wrapper { max-height: 100% !important; top: 0; }
你也可以使用它来搜索多个条件，这些条件必须同时为真：
search = { "name": "Global Businesses", "title": "Global Businesses" };
...会为您提供具有指定名称和标题的对象。
你在评论中提问：
是否可以提供数字以便不在输入中删除给定节点的子节点。喜欢，
const donotRemoveChildNode = 2; console.log(findChild(input, search, donotRemoveChildNode ));...所以如果匹配条件，它不会删除特定节点的子节点吗？
在这里，如果我们搜索
{ id: 'FG1'}并提供donotRemoveChildNode = 2，则不会删除＆＃34;商业银行＆＃34;的第一级孩子。
我想说donotRemoveChildNode必须是3，因为＆＃34;商业银行＆＃34;的祖先层次结构中有三个级别的children数组。节点。值为0将显示最顶层children属性的第一级子项。
以下是额外参数的工作原理 - 我在数据中添加了一些记录来说明输出的差异：
function findChild(obj, condition, removeChildNodesBefore = Infinity) {
if (Object.entries(condition).every( ([k,v]) => obj[k] === v )) {
return obj;
}
for (const child of obj.children || []) {
let found = findChild(child, condition, removeChildNodesBefore - 1);
if (found) {
return Object.assign({}, obj, {
children: removeChildNodesBefore <= 0
? obj.children.map( sibling =>
sibling == child ? found
: Object.assign({}, sibling, {children: []})
)
: [found]
});
}
}
}
var input = { "children": [{ "name": "Home", "title": "Home", "id": "home1", "children": [] }, { "name": "BUSINESS AND ROLE SPECIFIC", "title": "BUSINESS AND ROLE SPECIFIC", "id": "BAR1", "children": [{ "name": "Global Businesses", "title": "Global Businesses", "id": "GB1", "children": [{ "name": "test", "title": "test", "id": "xxx", "children": [{ "name": "testDeep", "title": "test", "id": "deep", "children": []}]}, { "name": "Commercial Banking", "title": "Commercial Banking", "id": "CB1", "children": [{ "name": "test", "title": "test", "id": "yyy", "children": []}, { "name": "FLAGSHIP PROGRAMMES", "title": "FLAGSHIP PROGRAMMES", "id": "FG1", "children": [] }] }] }] }, { "name": "RISK MANAGEMENT", "title": "RISK MANAGEMENT", "id": "RM1", "children": [] } ]},
search = { id: 'FG1' }
console.log(findChild(input, search, 3));.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 1 :(得分：1)
function getBranch(branches, leaf_id)
{
var result_branch = null;
branches.some(function(branch, idx) {
if (branch.id == leaf_id) {
result_branch = Object.assign({}, branch);
result_branch.children.forEach(function(child, idx) {
delete result_branch.children[idx].children;
});
return true;
} else {
let target_branch = getBranch(branch.children, leaf_id);
if (target_branch) {
result_branch = Object.assign({}, branch);
delete result_branch.children
result_branch.children = [target_branch];
return true;
}
}
return false;
});
return result_branch;
}
console.log(getBranch(input.children, 'GB1'));
答案 2 :(得分：0)
一种方法是首先循环根子项，然后创建另一个函数以查看其中任何子项中是否存在Id。
var data = {
"children": [{
"name": "Home",
"title": "Home",
"id": "home1",
"children": []
},
{
"name": "BUSINESS AND ROLE SPECIFIC",
"title": "BUSINESS AND ROLE SPECIFIC",
"id": "BAR1",
"children": [{
"name": "Global Businesses",
"title": "Global Businesses",
"id": "GB1",
"children": [{
"name": "Commercial Banking",
"title": "Commercial Banking",
"id": "CB1",
"children": [{
"name": "FLAGSHIP PROGRAMMES",
"title": "FLAGSHIP PROGRAMMES",
"id": "FG1",
"children": []
}]
}]
}]
},
{
"name": "RISK MANAGEMENT",
"title": "RISK MANAGEMENT",
"id": "RM1",
"children": []
}
]
};
function hasId( id, data ) {
if (data.id === id) return true;
if (data.children) {
for (const child of data.children) {
if (hasId( id, child)) return true;
}
}
return false;
}
function search( id, data ) {
for (const child of data.children) {
if (hasId(id, child)) return child;
}
return null;
}
console.log(search( "FG1", data ));