使用require.context根据views文件夹下（多级文件夹）所有.vue文件路径实现路由自动匹配生成如下格式routes，routes: [{ path: '/', redirect: '/login' }, { path: '/home', name: 'home', component: () => import('@/views/Home.vue'), redirect: '/main', children: [{ path: '/main', name: 'main', component: () => import('@/views/Main/index.vue'), children: [{ path: '/xxx', name: 'xxx', component: () => import('@/views/Main/Xxx.vue'), }, { path: '/yyy', name: 'yyy', component: () => import('@/views/Main/Yyy.vue'), } ] }, { path: '/zz', name: 'zz', component: () => import('@/views/Zz/index.vue'), children: [{ path: '/mmm', name: 'mmm', component: () => import('@/views/Zz/Mmm.vue'), }] }, ] }, { path: '/login', name: 'login', component: () => import('@/views/Login.vue') } ]

你可以使用以下代码实现：

const files = require.context('@/views', true, /\.vue$/)
const routes = []

files.keys().forEach(key => {
  const name = key.match(/\/(\w+)\.vue$/)[1]
  const path = key.replace(/^\.\//, '').replace(/\.vue$/, '').replace(/\/index$/, '')
  const route = {
    path: `/${path    name: name,
    component: files(key).default
  }
  if (path === 'Home') {
    route.redirect = '/main'
    route.children = [{
      path: 'main',
      name: 'main',
      component: () => import('@/views/Main/index.vue'),
      children: [{
        path: 'xxx',
        name: 'xxx',
        component: () => import('@/views/Main/Xxx.vue')
      }, {
        path: 'yyy',
        name: 'yyy',
        component: () => import('@/views/Main/Yyy.vue')
      }]
    }, {
      path: 'zz',
      name: 'zz',
      component: () => import('@/views/Zz/index.vue'),
      children: [{
        path: 'mmm',
        name: 'mmm',
        component: () => import('@/views/Zz/Mmm.vue')
      }]
    }]
  }
  routes.push(route)
})

routes.push({
  path: '/',
  redirect: '/login'
}, {
  path: '/login',
  name: 'login',
  component: () => import('@/views/Login.vue')
})

export default routes

这里使用了 `require.context` 获取所有 `.vue` 文件，然后通过正则表达式获取文件名和路径，最后根据路径和组件生成路由对象，如果是 `Home` 组件，则添加子路由。最后还要添加默认路由和登录路由。