我有一个对象数组,看起来像这样:
[{ name: 'test',
size: 0,
type: 'directory',
path: '/storage/test' },
{ name: 'asdf',
size: 170,
type: 'directory',
path: '/storage/test/asdf' },
{ name: '2.txt',
size: 0,
type: 'file',
path: '/storage/test/asdf/2.txt' }]
可以有任意数量的路径,这是遍历目录中的文件和文件夹的结果。
我要做的就是确定它们的“根”节点。最终,这将被存储在mongodb中,并使用物化路径来确定它的关系。
在本例中,/storage/test
是一个没有父级的根。/storage/test/asdf
拥有/storage/test
的父类,而/storage/test/asdf/2.txt
的父类也是它的父类。
我的问题是,您将如何迭代此数组,以确定父数组和相关的子数组?任何正确方向的帮助都是很棒的!
谢谢
发布于 2013-10-23 12:06:40
你可以这样做:
var arr = [] //your array;
var tree = {};
function addnode(obj){
var splitpath = obj.path.replace(/^\/|\/$/g, "").split('/');
var ptr = tree;
for (i=0;i<splitpath.length;i++)
{
node = { name: splitpath[i],
type: 'directory'};
if(i == splitpath.length-1)
{node.size = obj.size;node.type = obj.type;}
ptr[splitpath[i]] = ptr[splitpath[i]]||node;
ptr[splitpath[i]].children=ptr[splitpath[i]].children||{};
ptr=ptr[splitpath[i]].children;
}
}
arr.map(addnode);
console.log(require('util').inspect(tree, {depth:null}));
输出
{ storage:
{ name: 'storage',
type: 'directory',
children:
{ test:
{ name: 'test',
type: 'directory',
size: 0,
children:
{ asdf:
{ name: 'asdf',
type: 'directory',
size: 170,
children: { '2.txt': { name: '2.txt', type: 'file', size: 0, children: {} } } } } } } } }
发布于 2013-10-23 11:50:19
假设/
永远不会出现在文件列表中,下面这样的代码应该是可行的:
function treeify(files) {
var path = require('path')
files = files.reduce(function(tree, f) {
var dir = path.dirname(f.path)
if (tree[dir]) {
tree[dir].children.push(f)
} else {
tree[dir] = { implied: true, children: [f] }
}
if (tree[f.path]) {
f.children = tree[f.path].children
} else {
f.children = []
}
return (tree[f.path] = f), tree
}, {})
return Object.keys(files).reduce(function(tree, f) {
if (files[f].implied) {
return tree.concat(files[f].children)
}
return tree
}, [])
}
它会将您在问题中提到的数组转换为如下所示:
[ { name: 'test',
size: 0,
type: 'directory',
path: '/storage/test',
children:
[ { name: 'asdf',
size: 170,
type: 'directory',
path: '/storage/test/asdf',
children:
[ { name: '2.txt',
size: 0,
type: 'file',
path: '/storage/test/asdf/2.txt',
children: [] } ] } ] } ]
我实际上没有用任何其他数据源测试过这一点,所以您的里程数可能会有所不同,但至少它应该会推动您朝着正确的方向前进。
发布于 2018-06-25 02:29:43
基于@user568109的解决方案,但在数组而不是对象中返回结果:
function filesToTreeNodes(arr) {
var tree = {}
function addnode(obj) {
var splitpath = obj.fileName.replace(/^\/|\/$/g, "").split('/');
var ptr = tree;
for (let i = 0; i < splitpath.length; i++) {
let node: any = {
fileName: splitpath[i],
isDirectory: true
};
if (i == splitpath.length - 1) {
node.isDirectory = false
}
ptr[splitpath[i]] = ptr[splitpath[i]] || node;
ptr[splitpath[i]].children = ptr[splitpath[i]].children || {};
ptr = ptr[splitpath[i]].children;
}
}
function objectToArr(node) {
Object.keys(node || {}).map((k) => {
if (node[k].children) {
objectToArr(node[k])
}
})
if (node.children) {
node.children = Object.values(node.children)
node.children.forEach(objectToArr)
}
}
arr.map(addnode);
objectToArr(tree)
return Object.values(tree)
}
这是更好地理解输入/输出格式的签名:
export interface TreeNode {
isDirectory: string
children: TreeNode[]
fileName: string
}
export interface File {
fileName: string
}
export type fileToTreeNodeType = (files: File[]) => TreeNode[]
https://stackoverflow.com/questions/19531453
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