SideMenu的getMenuItem方法处理，会导致目录下只有一个子菜单的时候，只显示子菜单，不显示父目录

[cage-taofei]

function getMenuItem(route, basePath = '') {
let menuItem = {
label: (route.meta && route.meta.title) || route.name,
key: route.name,
path: resolvePath(basePath, route.path),
icon: getIcon(route.meta),
order: route.meta?.order || 0,
}

const visibleChildren = route.children
? route.children.filter((item) => item.name && !item.isHidden)
: []

if (!visibleChildren.length) return menuItem

if (visibleChildren.length === 1) {
// 单个子路由处理
const singleRoute = visibleChildren[0]
menuItem = {
...menuItem,
label: singleRoute.meta?.title || singleRoute.name,
key: singleRoute.name,
path: resolvePath(menuItem.path, singleRoute.path),
icon: getIcon(singleRoute.meta),
}
const visibleItems = singleRoute.children
? singleRoute.children.filter((item) => item.name && !item.isHidden)
: []

if (visibleItems.length === 1) {
 // 这里建议改一下
  **menuItem = getMenuItem(visibleItems[0], menuItem.path)**
} else if (visibleItems.length > 1) {
  menuItem.children = visibleItems
    .map((item) => getMenuItem(item, menuItem.path))
    .sort((a, b) => a.order - b.order)
}

} else {
menuItem.children = visibleChildren
.map((item) => getMenuItem(item, menuItem.path))
.sort((a, b) => a.order - b.order)
}
return menuItem
}

[mizhexiaoxiao]

有兴趣提交PR吗