Unable to retrieve errors occured in Graphql mutation in flutter project - flutter

Question

I am using the package graphql_flutter for GraphQL operations in my flutter app. The queries and mutations are going well but I cannot retrieve the errors by following the ways mentioned in their doc. Every time I receive a generic error message which is,
ClientException: Failed to connect to http://127.0.0.1:3006/graphql:
That I get by doing,
print(result.exception.toString());
My mutation looks like,
final MutationOptions mutationOptions = MutationOptions(
documentNode: gql(mutationString),
variables: vars
);
final QueryResult result = await _instance._client.mutate(mutationOptions);
if (result.hasException) {
// none of the following prints the expected error.
print(result.exception.clientException.message);
print(result.exception.graphqlErrors);
print(result.exception.toString());
}
print(result.data);
return result.data;
Whereas in the apollo client, My error is :
{
"errors": [
{
"message": "Invalid Phone number provided",
"locations": [
{
"line": 2,
"column": 3
}
],
"path": [
"otp"
],
"extensions": {
"code": "INTERNAL_SERVER_ERROR",
....
But I get none of that.
Note: The success response is coming as expected. I would like to know how can I get the graphql errors.

Answer 1

I found the problem. It is because android cannot connect to 127.0.0.1 or localhost from the emulator. I replaced the host with my local IP address and it is working fine now.

Answer 2

Put this in a try/ catch block and see if it can catch any exceptions
final QueryResult result = await _instance._client.mutate(mutationOptions);

Answer 3

the original question was how to get the graphql errors.
I'm using graphql: ^5.0.0
I checked the docs and found this example:
if (result.hasException) {
if (result.exception.linkException is NetworkException) {
// handle network issues, maybe
}
return Text(result.exception.toString())
}
but that just gave me the exception, not the error, I casted the result exception to the type of error and was able to get the message:
if (result.hasException) {
if (result.exception!.linkException is ServerException) {
ServerException exception =
result.exception!.linkException as ServerException;
var errorMessage = exception.parsedResponse!.errors![0].message;
print(errorMessage);
throw Exception(errorMessage);
}
}
this seems like a lot of work for a simple message, I wonder if there is another simpler built-in way to do it