Unhandled Exception: type 'Null' is not a subtype of type 'Future'

Question

i have tried to turn Future to nullable and also the string but something goes wrong any help please?

floatingActionButton: FloatingActionButton(
    onPressed: () {
      getName().then((value) {
        print(value);
        throw("error!!!!!");
      }).catchError((error) => print(error.toString()));
    },
    child: Icon(Icons.add),
),

out side the build method somewhere in my class

Future<String> getName() async =>'Some String';

i'm just trying to handle the error in the then () in .catch why it tells me that — Ahmed Abd ElGhany, Aug 26 '21 at 01:31
It would help if you tell us in which line the problem is. Copy the whole error message in your question. — stacktrace2234, Aug 26 '21 at 02:36

Miftakhul Arzak · Answer 1 · 2021-08-26T04:11:36.850

0

You can try this

getName().then((value) {
                print(value);
                throw ("error!!!!");
              }).catchError((error) {
                print("catch error :: " + error.toString());
              });

that code print catch error :: error!!!!. Is that you want?

edited Aug 26 '21 at 04:11

answered Aug 26 '21 at 03:43

Miftakhul Arzak

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i wanted to catch the then() exception in the .catchError() that's why i throw an exception in the then callback – Ahmed Abd ElGhany Aug 26 '21 at 03:56
didn't work for me unless i made a condition to check if the value FutureOr – Ahmed Abd ElGhany Aug 26 '21 at 23:56

score 0 · Accepted Answer · answered Aug 26 '21 at 04:10

so this fixed my code

floatingActionButton: FloatingActionButton(
        onPressed: () => getName().then((value){
          //  the return type of an error handler is not acceptable
          // as the value of future.
          // after lambda expression is executed it gone so
          // we wouldn't know where the error came from
          // because also the return type is dynamic so we 
          //can't get to a datatype interface so it may be null since it's dynamic 
          if(value is FutureOr) {
            throw("error");
          }
        }).catchError((error) =>  print(error.toString()) ) ,

Unhandled Exception: type 'Null' is not a subtype of type 'Future'

2 Answers2