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Python function that generates a URL to a given S3 resource.

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Python function that generates a URL to a given S3 resource.


url_for_s3('static', bucket_name='my-cool-foobar-bucket',
           scheme='https', filename='pics/logo.png')

Will return:

Note: this function assumes that the given resource exists on S3 and is publicly accessible.

Based loosely on Flask-S3’s url_for().

For a complete, working Flask app that demonstrates url-for-s3 in action, have a look at flask-s3-save-example.

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Filename, size & hash SHA256 hash help File type Python version Upload date
url_for_s3-0.1.2-py2.py3-none-any.whl (3.7 kB) Copy SHA256 hash SHA256 Wheel py2.py3 Aug 28, 2015

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