我正在尝试找到一种使用路由（香草js）和子路由展平阵列的方法，以便获得具有相同级别的对象但它们的name和{ {1}}属性基于其“级别”。
例如，我希望path成为name之类的东西，dashboard.settings.user.......也会将“有效”路径的路径连接在一起。
我一直在尝试使用递归函数，但无法弄清楚如何跟踪级别。
更新：
path更新2：
基于Nina解决方案，我要求一种保留原始对象/值的方法，但是将var routesConfig = [
{
name: 'dashboard',
path: '/dashboard',
children: [
{
name: 'settings',
path: '/settings',
children: [
{
name: 'user',
path: '/user'
},
{
name: 'email',
path: '/email',
children: [
{
name: 'template',
path: '/template',
children: [
{
name: 'error',
path: '/error'
},
{
name: 'success',
path: '/success'
}
],
name: 'lists',
path: '/lists',
children: [
{
name: 'signup',
path: '/signup'
},
{
name: 'newsletter',
path: '/newsletter'
}
]
},
{
name: 'sender',
path: '/sender'
}
]
}
]
},
{
name: 'contact',
path: '/contact',
children: [
{
name: 'person',
path: '/person',
children: [
{
name: 'john',
path: '/john'
},
{
name: 'jane',
path: '/jane'
}
]
}
]
}
]
},
]
flattenRoutes();
function flattenRoutes(){
let routes = [];
routesConfig.map(route => {
const {name, path, children} = route;
routes.push({name, path});
if(children && children.length){
iterateRouteChildren(route, routes, 1, []);
}
});
console.log("routes", routes);
}
function iterateRouteChildren(parent, routes, depth, tree){
for(let i = 0; i < parent.children.length; i++){
let childRoute = parent.children[i];
let name = childRoute.name;
let path = childRoute.path;
if(depth === 1){
tree = [];
}else{
tree.push(childRoute.name)
}
routes.push({name, path, tree: tree.filter(Boolean).join('.')});
if(childRoute.children && childRoute.children.length){
iterateRouteChildren(childRoute, routes, depth + 1, tree);
}
}
}
和name重写为它们的“新”值，例如应该路由“ email”将其path读为：name，并将其dashboard.settings.email重写为：path
因此，请对每个对象进行突变，并保留其不变的属性（但仍将其保留为一个平面数组，因此不要嵌套子数组）。
更新3：
尝试澄清。我嵌套的routesConfig将包含具有“名称”，“路径”，“其他属性”和也许“子代”之类的属性的对象。
我想弄平整棵树，所以一切都在同一水平上，但是所有子项都应根据其深浅程度来重写其“名称”和“路径”。示例：
/dashboard/settings/email谢谢
答案 0 :(得分：0)
您可以通过将最后的name和path移交给嵌套对象来采取迭代和递归的方法。
function getFlat(array, n = '', p = '') {
return array.reduce((r, { children = [], ...o }) => {
var name = n + (n && '.') + o.name,
path = p + o.path;
return r.concat(Object.assign({}, o, { name, path }), getFlat(children, name, path));
}, []);
}
var routes = [{ name: 'dashboard', path: '/dashboard', children: [{ name: 'settings', path: '/settings', children: [{ name: 'user', path: '/user' }, { name: 'email', path: '/email', children: [{ name: 'template', path: '/template', children: [{ name: 'error', path: '/error' }, { name: 'success', path: '/success' }] }, { name: 'lists', path: '/lists', children: [{ name: 'signup', path: '/signup' }, { name: 'newsletter', path: '/newsletter' }] }, { name: 'sender', path: '/sender' }] }] }, { name: 'contact', path: '/contact', children: [{ name: 'person', path: '/person', children: [{ name: 'john', path: '/john' }, { name: 'jane', path: '/jane' }] }] }] }],
flat = getFlat(routes);
console.log(flat);.as-console-wrapper { max-height: 100% !important; top: 0; }