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

Project description

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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Files for url-for-s3, version 0.1.2
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