我正在尝试采用平坦的路径数组,并创建嵌套的对象数组。我遇到的麻烦是生成子节点的递归部分...
起始数组:
const paths = [
'/',
'/blog',
'/blog/filename',
'/blog/slug',
'/blog/title',
'/website',
'/website/deploy',
'/website/infrastructure',
'/website/infrastructure/aws-notes',
];
具有所需的输出结构:
[
{
path: '/',
},
{
path: '/blog',
children: [
{
path: '/blog/filename',
},
{
path: '/blog/slug',
},
{
path: '/blog/title',
}
]
},
{
path: '/website',
children: [
{
path: '/website/deploy',
},
{
path: '/website/infrastructure',
children: [
{
path: '/website/infrastructure/aws-notes',
}
],
},
],
},
]
到目前为止,我在这里已经尝试过一些方法,但最终以无限循环或不良结构结束:
const getPathParts = (path) => path.substring(1).split('/');
const getPathLevel = (path) => getPathParts(path).length - 1;
const getTree = (paths) => paths.reduce((tree, path, i, paths) => {
const pathParts = getPathParts(path);
const pathDepth = getPathLevel(path);
const current = pathParts[pathDepth];
const parent = pathParts[pathDepth - 1] || null;
const item = {
path,
children: [],
};
if (pathDepth > 0 || parent !== null) {
// recursive check for parent, push this as a child to that parent?
return [...tree];
}
return [
...tree,
item,
];
}, []);
我已经尝试过array.find|some|filter
来检索父节点,但是我不知如何将节点作为子节点推入正确的嵌套节点中。注意:我已经为示例提取了一些代码,请原谅任何语法/拼写问题。
答案 0 :(得分:1)
您可以采用嵌套方法,将路径分离并检查路径是否存在。如果没有,推一个新的物体。
稍后返回实际对象的子代。继续进行,直到没有更多路径项目可用为止。
const
paths = ['/', '/blog', '/blog/filename', '/blog/slug', '/blog/title', '/website', '/website/deploy', '/website/infrastructure', '/website/infrastructure/aws-notes'],
result = paths.reduce((r, path) => {
path.split(/(?=\/)/).reduce((a, _, i, p) => {
var temp = a.find(o => o.path === p.slice(0, i + 1).join(''));
if (!temp) {
a.push(temp = { path, children: [] });
}
return temp.children;
}, r);
return r;
}, []);
console.log(result);
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对于仅使用最终目录的路径进行创建的情况,可以采用部分创建路径。
这种方法可以防止空子数组。
const
paths = [
'/',
'/blog/filename',
'/blog/slug',
'/blog/title',
'/website/deploy',
'/website/infrastructure/aws-notes'
],
result = [];
paths.reduce((r, string) => {
string.split(/(?=\/)/).reduce((o, _, i, p) => {
o.children = o.children || [];
var path = p.slice(0, i + 1).join(''),
temp = o.children.find(o => o.path === path);
if (!temp) {
o.children.push(temp = { path });
}
return temp;
}, r);
return r;
}, { children: result });
console.log(result);
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