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feature request: from_dict() #88

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armijnhemel opened this issue May 11, 2022 · 1 comment

armijnhemel commented May 11, 2022

Currently there is a method to_dict() but there is no equivalent from_dict(). It might be useful to have this as well (use case: templating).

The text was updated successfully, but these errors were encountered:

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tdruez commented Jun 27, 2022

@armijnhemel you can already provide a dict to the PackageURL class using the ** unpacking operator:

>>> from packageurl import PackageURL
>>> purl = PackageURL.from_string("pkg:type/namespace/[email protected]")
>>> purl_as_dict = purl.to_dict()
>>> purl_as_dict
{'type': 'type', 'namespace': 'namespace', 'name': 'name', 'version': '1.0', 'qualifiers': None, 'subpath': None}

>>> purl_from_dict = PackageURL(**purl_as_dict)
>>> str(purl_from_dict)
'pkg:type/namespace/[email protected]'

Note that you can use this method with any Python class.

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