Davide Brunato

If this solution satisfy you please close the issue. Changing the default decode behavior may not have a very wide consensus, and can be a harsh decision to take even...

Hi, the *XMLSchema.xsd* document is already loaded in `XMLSchema` class (into _meta_schema_ instance) and it's used to validate every XSD that is passed to initialize `XMLSchema()` instance (using the `XMLSchema.check_schema()`...

> I appreciate. Will be validating xml schemas generated via automation so kinda need it. But in this case you don't need to replace the XSD 1.0/1.1 meta-schema, you can...

> Actually, I like the JsonML format of the xml from your converter. Can you verify the following assumptions are correct? > > * If I generate my schema in...

> I do not have a need to serialize to python objects. > For python object i intend both low level data (basic data types into nested dictionaries/lists) or high...

Hi @MartyStache, you can try new minor release v2.1.0, that has an automatic usage of pre-build meta-schema. In you case also this should works: ```pycon >>> import xmlschema >>> >>>...

Added a [section on meta-schemas and XSD file validation](https://xmlschema.readthedocs.io/en/latest/usage.html#meta-schemas-and-xsd-sources) in docs. So in your case you can alternatively provide the XSD 1.1 meta-schema as _schema_ argument: ```pycon >>> import xmlschema...

Hi, it's better to let xmlschema to fetch namespaces, e.g.: ```pycon >>> from xmlschema import XMLSchema >>> schema = XMLSchema('camt.054.001.02.xsd') >>> schema.validate('document.xml') ``` About paths: usually _schema_path_ is not needed,...

Found a bug in `XMLSchema.iter_decode()` (called by `validate()`) if a *schema_path* is provided without a *path*. It will be fixed in the next bugfix release. After that your script should...

Bugfix release v2.1.1 is available. For your case the script can be changed as follows: ```python from xml.etree import ElementTree from xmlschema import XMLSchema def main() -> None: schema =...