我想改变这个:
<Router history={browserHistory}>
<Route path='/' component={Layout}>
<Route path='signup' component={SignupPage} />
<Route path='assemblies/' component={AssemblyPages}>
<Route path='categories/' component={Categories}>
</Route>
</Route>
</Router>
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至
import AssembliesRoute from './AssembliesRoute';
<Router history={browserHistory}>
<Route path='/' component={Layout}>
<Route path='signup' component={SignupPage} />
<AssembliesRoute />
</Route>
</Router>
//AssembliesRoute.js
export default class extends compondent {
render(){
return <Route path='assemblies/' component={AssemblyPages}>
<Route path='categories/' component={Categories}>
</Route>
}
}
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基本上,我想将所有嵌套在程序集路由中的路由,并在程序集文件夹中的特定文件中处理它们,然后将这些路由返回到路由器.但是当我尝试这样做时,我找不到路线.这可能吗?