Comments (2)
@gergelyke It sounds like you should just return a GraphQLList
. You'd query with:
{
users(olderThan: 40) {
name
}
}
from graphql-spec.
Correct, returning a List type is exactly how you should handle a field like that.
from graphql-spec.
Related Issues (20)
- Conditional resolving / optional expanding
- The directive that is ``@oneOf`` could not implement some ``filter query``. HOT 1
- Field error from list arg with nullable variable entry (nullable=optional clash?) HOT 6
- Using variables in scalars HOT 8
- GraphQL Mutation how to make fields optional to update fields only by selection and generate schema HOT 1
- [RFC] Non-existent request field operator HOT 8
- (Graphql-schema) Cannot query field "closeDiscussion" on type "Mutation" HOT 1
- GraphQL spec is contradictory: can selection sets be empty or not? HOT 1
- Allow to return Input Object types HOT 1
- Formalize Global Object Identification.
- Grammar should have single root node HOT 2
- [Feature request]: Support non-list variables for list arguments HOT 9
- Redundant field aliases identical to field names
- request for clarification/examples of nested cursor pagination HOT 1
- RFC: Standardize `application/graphql` with `.graphql` extension HOT 6
- Repository names with dashes cannot be requested HOT 7
- Typed Scalars HOT 10
- Clarify validation rule on non-null type wrapping another non-null type HOT 3
- Confused about type definitions's optional fields definitions HOT 6
- Clarification of how to handle errors that occur in a subscription event stream HOT 2
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from graphql-spec.