This example shows how to deserialize a graph while losing type information.
const orig: Graph = ...
const serialized = JSON.stringify(orig);
const builder = graphJson();
const deserialized = builder(JSON.parse(serialized));
Note, that because no custom hydrators were given, deserialized will have
type MutGraph<unknown, unknown> which probably isn't desired.
This example demonstrates using a simple data hydrator.
const orig: Graph<number, string> = ...
const serialized = JSON.stringify(orig);
const builder = graphJson()
.nodeDatum(data => data as number)
.linkDatum(data => data as string);
const deserialized = builder(JSON.parse(serialized));
Now deserialized has the same type as the original grah. In real use,
you'll probably want to use type guards to guarantee the data was
deserialized correctly.
create a Json graph constructor with default settings
If you serialize a graph or node using
JSON.stringify, this operator can be used to hydrate them back into a MutGraph. This is expecially useful for serializing the graph to do layouts in a separate worker.If you serialize a GraphNode the deserialized graph will be the same node that was serialized, but the rest of that node's connected component will be deserialized as well, and still reachable via Graph#nodes. The type information will inherently be lost, and will need to be cast.
Since serialization doesn't inherently imply deserialization, Json#nodeDatum and Json#linkDatum can be used to provide custom hydration for node and linke data.