当使用GetX Flutter从一个页面导航到另一个页面时,从第一个页面传递的参数不会在第二个页面上接收。
第一控制器:
class LoginController extends GetxController {
var _loading = false.obs;
get loading => _loading.value;
set loading(value) {
_loading.value = value;
}
var _loginResponse = LoginResponse().obs;
late TextEditingController emailController, passwordController;
@override
void onInit() {
super.onInit();
emailController = TextEditingController();
passwordController = TextEditingController();
}
@override
void onReady() {
super.onReady();
}
@override
void onClose() {
super.onClose();
emailController.dispose();
passwordController.dispose();
}
Future<void> loginCall(
{required String email, required String password}) async {
if (!GetUtils.isEmail(email)) {
Get.snackbar('Invalid email', 'Please enter valid email',
snackPosition: SnackPosition.BOTTOM);
return;
}
if (password.length < 8) {
Get.snackbar('Invalid password', 'Minimum 8 chars',
snackPosition: SnackPosition.BOTTOM);
return;
}
_loading.value = true;
var result =
await NetworkRequest.loginCall(email: email, password: password);
result != null
? _loginResponse.value = result
: Get.snackbar(
'Something went wrong', 'Hold back and try again after sometime.',
snackPosition: SnackPosition.BOTTOM);
if(_loginResponse.value.responseCode == 1) {
**Get.offAll(GoogleAuthenticationScreen(), arguments: _loginResponse.value.responseData!.session.toString());**
}
_loading.value = false;
}
}
第二控制器:
class GoogleAuthenticationController extends GetxController {
var _loading = false.obs;
late TextEditingController pinController;
var _authResponse = GoogleTwoFactorResponse().obs;
@override
void onInit() {
super.onInit();
pinController = TextEditingController();
var session = Get.arguments.toString();
twoFactorCall(token: session);
}
@override
void onClose() {
super.onClose();
pinController.dispose();
}
Future<void> twoFactorCall({required String token}) async {
_loading.value = true;
var result = await NetworkRequest.googleAuthentication(token: token);
result != null
? _authResponse.value = result
: Get.snackbar(
'Something went wrong', 'Hold back and try again after sometime.',
snackPosition: SnackPosition.BOTTOM);
Get.snackbar(result!.responseMessage!, "",
snackPosition: SnackPosition.BOTTOM);
_loading.value = false;
}
}
Getx版本获取:^4.3.8
Doctor摘要(要查看所有详细信息,请运行flutter doctor -v):[✓] Flutter(Channel stable,2.2.2,on macOS 11.4 20F71 darwin-x64,locale en-IN)[✓] Android toolchain - develop for Android devices(Android SDK version 30.0.2)[✓] Xcode - develop for iOS and macOS [✓] Chrome - develop for web [✓] Android Studio(version 4.1)[✓]连接的设备(2可用)
·未发现问题!
2条答案
按热度按时间f3temu5u1#
你可以试试
onInIt方法的外部。希望它能工作
unhi4e5o2#
是的,我有答案。我已经使用了从一个类到另一个类的传递参数,下面显示代码。
这里LoginController用于LoginScreen类
这里你的参数传入GoogleAuthenticationScreen。现在你必须将参数从GoogleAuthenticationScreen传递到GoogleAuthenticationController。
这边请。