Can't get data when 400 error. in Flutter - ajax

Question

I'm trying to implement login with the Dio package in my app. When I send the correct email and password I get a 200 status code and user data. But when I send the email or password incorrect backend sends 400 error code and data like this {"message": "User Not Exist","data": [],"status": false} the problem is I'm unable to get the data when I have 400 error because in dio catchError method I can get just error and stacktrace.
Future login(String username, String password) async {
try {
String url = "$baseUrl/admin/user/login";
print(url);
var res = await dio.post(
url,
data: {"email": username, "password": password},
);
if (res.statusCode == 400) {
print(res.data); <----- This dont print anything.
return false;
} else {
print(res.data);
return true;
}
// await Future.delayed(Duration(seconds: 4));
} catch (e, s) {<----- here I have just error and stacktrace not the data
print("stacktrace $s");
print("error $e");
}
}

Answer 1

I solved this issue using on DioError catch instead of the catch method.
Future login(String username, String password) async {
try {
String url = "$baseUrl/admin/user/login";
print(url);
var res = await dio.post(
url,
data: {"email": username, "password": password},
);
if (res.statusCode == 400) {
print(res.data); <----- This dont print anything.
return false;
} else {
print(res.data);
return true;
}
// await Future.delayed(Duration(seconds: 4));
} on DioError catch (e) {
print(e.response.data);<-----------here I can get the data
return e.response.data;
}
}

Answer 2

Take try..catch out and use this form:
dio.post(...).then((response) {...}).catch(...)
Then you can use response as you wish.
You can read this article