如何像设置子节点一样设置vue-路由器嵌套路由

Question

我正在尝试将一个作为常量导入的路由器放在主路由器的子程序中

路由器/组件/滑块/index.js

const sliderRouter = {
    path: '/slider',
    name: 'Slider',
    meta: {
        title: 'Slider'
    },
    component: () => import('@/views/components/slider'),
}

export default sliderRouter

路由器/index.js

...

import authRouter from './modules/auth'
import sliderRouter from './modules/components/slider'

...

export default new Router({
    mode: 'history',
    linkActiveClass: "active selected",
    routes: [
        authRouter, // this work
        {
            path: '/admin',
            name: 'primary',
            meta: { requiresAuth: true },
            // remastered version
            component: loadView('MainLayout'),
            children: [
                sliderRouter,  /* make something like this */
                //recent routes
                {
                    path: '/home',
                    name: 'home',
                    // route level code-splitting
                    // this generates a separate chunk (about.[hash].js) for this route
                    // which is lazy-loaded when the route is visited.
                    // classic version:
                    component: () => import(/* webpackChunkName: "about" */ '../views/Home.vue')
                    // component: About
                }
            ]
        },

真的有可能让滑块路由器以这样的方式工作并从MainLayout继承(就像一个典型的子节点)，并且URL也是‘/admin/滑块’吗？

Answer 1

我通过改变滑块的路径来解决这个问题。

path: '/slider',

至

path: 'slider',

因为它使用的是like子路由，它在名字之前不需要斜杠