GraphQL Entities
Last updated: 3 minutes read.
Defining the base entity
- Must be defined with the @key directive.
- The “fields” argument of the @key directive must reference a valid field that can uniquely identify the entity.
- Multiple primary keys are possible.
An example is provided below:
Subgraph 1 (base entity)
type MyEntity @key(fields: "id") @key(fields: "name") {
id: ID!
name: String!
}
Extending entities
Entities cannot be shared types (be defined in more than one single subgraph; see Entity stubs below).
The base entity remains unaware of fields added through extension; only the extension itself is aware of them.
Attempting to extend a non-entity with an extension that includes the @key directive or attempting to extend a base entity with an extension that does not include the @key directive will both result in errors.
The primary key reference should be listed as a field with the @external directive.
Below is an example extension for MyEntity (which was defined above in Subgraph 1):
Subgraph 2 (extension):
extend type MyEntity @key(fields: "id") {
id: ID! @external
newField: String!
}
Entity stubs
If one subgraph references a base entity (an entity defined in another subgraph) without adding new fields, that reference must be declared as a stub. In federation v1, stubs appear similar to extensions but do not add any new fields.
An entity stub contains the minimal amount of information necessary to identify the entity (referencing exactly one of the primary keys from the base entity regardless of whether there are multiple primary keys on the base entity).
The identifying primary key should feature the @external directive.
For example, a stub of MyEntity (which was defined above in Subgraph 1):
Subgraph 3 (stub):
extend type MyEntity @key(fields: "id") {
id: ID! @external
}
What is a shared type?
Types that are identical by name and structure and feature in more than one subgraph are shared types.
Can I extend a shared type?
Subgraphs are normalized before federation. This means you can extend a type if the resolution of the extension after normalization is exactly identical to the resolution of the type after normalization in other subgraphs.
Unless the resolution of the extension in a single subgraph is exactly identical to all other subgraphs, extension is not possible.
Here is a valid example where both subgraphs resolve to identical enums after normalization:
Subgraph 1:
enum Example {
A,
B
}
extend enum Example {
C
}
Subgraph 2:
enum Example {
A,
B,
C
}
Here, the enum named Example in Subgraph 1 resolves to be identical to the enum named Example in Subgraph 2.
However, if we were to include Subgraph 3, which does not feature the “C” value, the enum is no longer identical in all 3 subgraphs. Consequently, federation would fail.
Subgraph 3:
enum Example {
A,
B
}