nuxt中间件路由排除

直接上中间件代码

export default ({ store, route, redirect }) => {

  const token = store.state.token

  // 不需要登录能访问的的页面
  const publicPages = [
    '/area',
    '/home',
    '/home/edu_customize',
    '/home/edu_staging',
    '/home/free_price',
    '/home/free_study',
    '/home/whole_overseer',
    '/big_forum',
    '/big_forum/details',
    '/educational_classroom',
    '/organization_login',
    '/organization_register',
    '/school_lists',
    '/school_lists/local_consultation',
    '/school_lists/details',
    '/school_lists/contrast',
    '/security_services',
    '/security_services/contrast',
    '/student_login',
    '/student_register',
    '/study_release',
    '/training_agency',
    '/training_agency/service_instructions',
    '/training_agency/details',
  ]

  const authRequired = !publicPages.includes(route.path) // 访问其他页面需要登录才能访问

  // 如果用户未登录，则重定向到登录页
  if (!token && authRequired) {
    return redirect('/home')
  }

}

因为在有的项目里面,需要做路由排除,也就是没有token也能访问到的页面,就这样写了,有更好的方法可以在评论区留言