为 return 自定义流创建函数
Make function to return custom stream
是否可以创建一个 returns 自定义流并像这样处理它的函数?
user.logIn('owner', '1234')
.listen(
success (Object user) {
print(user);
},
error: (Object user, Object error) {
print(error);
}
);
类似于:
class LoginResult {
bool success = false;
String username;
}
Stream<LoginResult> onLogin() async* {
while(...) {
yield new LoginResult()
..success = isSuccess
..userName = 'someUser';
}
}
或
StreamController<LoginResult> onLoginController = new StreamController<LoginResult>();
// might not be necessary if you only need one listener at most
Stream<LoginResult> _onLogin = onLoginController.stream.asBroadcastStream();
Stream<LoginResult> get onLogin => _onLogin
...
onLoginController.add(new LoginResult()
..success = isSuccess
..userName = 'someUser');
然后就可以像
一样使用了
是否可以创建一个 returns 自定义流并像这样处理它的函数?
user.logIn('owner', '1234')
.listen(
success (Object user) {
print(user);
},
error: (Object user, Object error) {
print(error);
}
);
类似于:
class LoginResult {
bool success = false;
String username;
}
Stream<LoginResult> onLogin() async* {
while(...) {
yield new LoginResult()
..success = isSuccess
..userName = 'someUser';
}
}
或
StreamController<LoginResult> onLoginController = new StreamController<LoginResult>();
// might not be necessary if you only need one listener at most
Stream<LoginResult> _onLogin = onLoginController.stream.asBroadcastStream();
Stream<LoginResult> get onLogin => _onLogin
...
onLoginController.add(new LoginResult()
..success = isSuccess
..userName = 'someUser');
然后就可以像
一样使用了