Metadata-Version: 2.5
Name: stackure
Version: 1.20261013.19
Summary: Stackure for Python: passwordless auth for B2B apps as ASGI and WSGI middleware. Zero dependencies.
Project-URL: Homepage, https://stackure.com
Project-URL: Repository, https://github.com/syi-stackure/sdk-py
Project-URL: Documentation, https://stackure.com/docs/python
Project-URL: Bug Tracker, https://github.com/syi-stackure/sdk-py/issues
Author-email: Stackure <support@stackure.com>
License: MIT License
        
        Copyright (c) 2025 ysashank
        
        Permission is hereby granted, free of charge, to any person obtaining a copy
        of this software and associated documentation files (the "Software"), to deal
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        SOFTWARE.
License-File: LICENSE
Keywords: asgi,auth,authentication,b2b,magic-link,mcp,middleware,passwordless,rbac,stackure,wsgi
Classifier: Development Status :: 5 - Production/Stable
Classifier: Framework :: Django
Classifier: Framework :: FastAPI
Classifier: Framework :: Flask
Classifier: Intended Audience :: Developers
Classifier: License :: OSI Approved :: MIT License
Classifier: Programming Language :: Python :: 3
Classifier: Programming Language :: Python :: 3.14
Classifier: Topic :: Internet :: WWW/HTTP
Classifier: Topic :: Security
Classifier: Topic :: Software Development :: Libraries :: Python Modules
Classifier: Typing :: Typed
Requires-Python: >=3.14
Description-Content-Type: text/markdown

# Stackure Python SDK

[![Check build](https://github.com/syi-stackure/sdk-py/actions/workflows/check-build.yml/badge.svg)](https://github.com/syi-stackure/sdk-py/actions/workflows/check-build.yml)
[![PyPI version](https://img.shields.io/pypi/v/stackure.svg)](https://pypi.org/project/stackure/)
[![License: MIT](https://img.shields.io/badge/license-MIT-blue.svg)](./LICENSE)

[Stackure](https://stackure.com) is passwordless auth for B2B apps. This SDK adds it to an ASGI or WSGI app, such as FastAPI, Flask or Django, with zero dependencies.

## Install

```bash
pip install stackure
```

Needs Python 3.14 or later.

## Quickstart

Set `STACKURE_APP_ID` and `STACKURE_APP_SECRET` from the app's page in Stackure, then wrap your ASGI or WSGI app:

```python
import stackure

app = stackure.auth()(app)

# in a view
user = stackure.user_from_request(request)
```

## Docs

- Guide and reference: [stackure.com/docs/python](https://stackure.com/docs/python)
- For AI agents: [stackure.com/llms.txt](https://stackure.com/llms.txt)

## License

MIT
