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vue动态加载路由后台获取菜单数据拼装成路由格式无法加载模块 #101

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yangzilong-java opened this issue Aug 13, 2020 · 0 comments

yangzilong-java commented Aug 13, 2020

vue动态加载路由，后台获取菜单数据在component中路径拼接就找不模块，写死成一个字符串就可以。

`/**

后台查询的菜单数据拼装成路由格式的数据
@param routes
@param data
*/
export function generaMenu(routes, data) {
data.forEach(item => {
const url = '@/views/system/User'
const menu = {
path: item.path === '#' ? item.name : item.path,
// 这样不行
// component: item.path === '#' ? Layout : () => import(url),
// 而下面这样就可以
component: item.path === '#' ? Layout : () => import('@/views/system/User'),
name: item.name,
meta: item.meta,
children: []
}
if (item.children !== null) {
generaMenu(menu.children, item.children)
}
routes.push(menu)
})
}`

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