Allowed Subscription resolver to return *QueryError directly

Previously, any error returned by the Subscription resolver was immediately wrapped inside
its own `*QueryError` value even if the returned error was already a `*QueryError`.

Now, when receiving such types, we use it as-is without wrapping again.

Author: Matthieu Vachon

internal/exec/subscribe.go

@@ -49,9 +49,15 @@ func (r *Request) Subscribe(ctx context.Context, s *resolvable.Schema, op *query
 		result = callOut[0]
 
 		if f.field.HasError && !callOut[1].IsNil() {
-			resolverErr := callOut[1].Interface().(error)
-			err = errors.Errorf("%s", resolverErr)
-			err.ResolverError = resolverErr
+			switch resolverErr := callOut[1].Interface().(type) {
+			case *errors.QueryError:
+				err = resolverErr
+			case error:
+				err = errors.Errorf("%s", resolverErr)
+				err.ResolverError = resolverErr
+			default:
+				panic(fmt.Errorf("can only deal with *QueryError and error types, got %T", resolverErr))
+			}
 		}
 	}()
 

subscription_test.go

@@ -24,6 +24,7 @@ func (r *helloResolver) Hello() string {
 }
 
 var resolverErr = errors.New("resolver error")
+var resolverQueryErr = &qerrors.QueryError{Message: "query", ResolverError: resolverErr}
 
 type helloSaidResolver struct {
 	err      error
@@ -265,6 +266,27 @@ func TestSchemaSubscribe(t *testing.T) {
 				},
 			},
 		},
+		{
+			Name: "subscription_resolver_can_query_error",
+			Schema: graphql.MustParseSchema(schema, &rootResolver{
+				helloSaidResolver: &helloSaidResolver{err: resolverQueryErr},
+			}),
+			Query: `
+				subscription onHelloSaid {
+					helloSaid {
+						msg
+					}
+				}
+			`,
+			ExpectedResults: []gqltesting.TestResponse{
+				{
+					Data: json.RawMessage(`
+						null
+					`),
+					Errors: []*qerrors.QueryError{resolverQueryErr},
+				},
+			},
+		},
 		{
 			Name:   "schema_without_resolver_errors",
 			Schema: graphql.MustParseSchema(schema, nil),