之前可能会被问到但是我找不到它,但是你如何将参数传递给命名路线？
这就是我构建路线的方式
Widget build(BuildContext context) {
return new Navigator(
initialRoute: 'home/chooseroom',onGenerateRoute: (RouteSettings settings) {
WidgetBuilder builder;
switch (settings.name) {
case 'home/chooseroom':
// navigates to 'signup/choose_credentials'.
builder = (BuildContext _) => new ChoosePage();
break;
case 'home/createpage':
builder = (BuildContext _) => new CreateRoomPage();
break;
case 'home/presentation':
builder = (BuildContext _) => new Presentation();
break;
default:
throw new Exception('Invalid route: ${settings.name}');
}
return new MaterialPageRoute(builder: builder,settings: settings);
},);
这就是你怎么称呼它
Navigator.of(上下文).pushNamed( ‘家/演示文稿’)
但是如果我的小部件是新的演示文稿(arg1,arg2,arg3)怎么办？
解决方法
基本上你有2个选择：
>使用一些第三方软件包进行路由 – 我认为最好的是Fluro.
>利用onGenerateRoute.此选项仅限于您可以传递的参数(字符串/数字)
要使用第二个选项,假设您要传递三个参数：Navigator.of(context).pushNamed(‘home / presentation：arg1：1337：hello’)
MaterialApp (
...,onGenerateRoute: handleRoute,routes:...,)
Route<dynamic> handleRoute(RouteSettings settings) {
WidgetBuilder builder;
final List<String> uri = settings.name.split('/');
if (uri[0].startsWith('home')) {
// handle all home routes:
if(uri[1].startsWith('presentation:'){
// cut slice by slice
final String allArgs =
uri[1].substring('presentation:'.length);
final List<String> args = allArgs.split(':');
// use your string args
print(args[0]); // prints "arg1"
int x = int.parse(args[1]); // becomes 1337
print(args[2]); // prints "hello"
builder = (ctx)=> Presentation(args[0],args[1],args[2]);
...
