You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
How can I extract the fragment from the schema without knowing the $id but I know the path? For example getErrors() returns the path, not the ref or $id. For all schemas the path will exist if defined but the $id is optional.
The reason for this requirement is that I am looking for a way to get access to user defined property keywords of a schema (which are allowed as per spec and correctly ignored by this implementation - see http://json-schema.org/latest/json-schema-core.html#rfc.section.6.4 )
Thanks.
The text was updated successfully, but these errors were encountered:
In an attempt to cleanup this repo we are trying to filter the issues and see which ones might be closed. Is it safe to assume this is a rather old issue, which now has a answer, and the issue can be closed? Feel free to close it yourself with some comments if helpful.
I have the following example schema:
Which I then load as per the following example:
I can then grab a fragment from the schema if I know the $id of a property:
print_r($schemaStorage->resolveRef('file://mySchema#/properties/test/properties/blah'));
How can I extract the fragment from the schema without knowing the $id but I know the path? For example getErrors() returns the path, not the ref or $id. For all schemas the path will exist if defined but the $id is optional.
See https://json-schema.org/understanding-json-schema/structuring.html#structuring
The reason for this requirement is that I am looking for a way to get access to user defined property keywords of a schema (which are allowed as per spec and correctly ignored by this implementation - see http://json-schema.org/latest/json-schema-core.html#rfc.section.6.4 )
Thanks.
The text was updated successfully, but these errors were encountered: