1. 简化 ruoyi 获取菜单列表时,和前端耦合性比较重的情况

2. 删除 MenuDO 表非关键性字段,进一步简化模型
This commit is contained in:
YunaiV
2021-01-07 01:57:19 +08:00
parent c9372b0a5c
commit 3bf173d744
15 changed files with 131 additions and 329 deletions

View File

@ -18,7 +18,7 @@ export function login(username, password, code, uuid) {
// 获取用户详细信息
export function getInfo() {
return request({
url: '/get-info',
url: '/get-permission-info',
method: 'get'
})
}

View File

@ -3,7 +3,7 @@ import request from '@/utils/request'
// 获取路由
export const getRouters = () => {
return request({
url: '/get-routers',
url: '/list-menus',
method: 'get'
})
}

View File

@ -43,19 +43,28 @@ const permission = {
// 遍历后台传来的路由字符串,转换为组件对象
function filterAsyncRouter(asyncRouterMap, isRewrite = false) {
return asyncRouterMap.filter(route => {
// 将 ruoyi 后端原有耦合前端的逻辑,迁移到此处
// 处理 meta 属性
route.meta = {
title: route.menuName,
icon: route.icon
}
// 处理 component 属性
if (route.children) { // 父节点
// debugger
if (route.parentId === 0) {
route.component = Layout
} else {
route.component = ParentView
}
} else { // 根节点
route.component = loadView(route.component)
}
// filterChildren
if (isRewrite && route.children) {
route.children = filterChildren(route.children)
}
if (route.component) {
// Layout ParentView 组件特殊处理
if (route.component === 'Layout') {
route.component = Layout
} else if (route.component === 'ParentView') {
route.component = ParentView
} else {
route.component = loadView(route.component)
}
}
if (route.children != null && route.children && route.children.length) {
route.children = filterAsyncRouter(route.children, route, isRewrite)
}