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Share a file, then return the URL ONLY

I am using Python to share an image. Then, I want to be able to only print to the console the short url. How do I accomplish this?

With this:

print(client.share(path))

I get:

{'visibility': 'PUBLIC', 'expires': 'Tue, 01 Jan 2030 00:00:00 +0000', 'url': 'https://....'}

What I want:

https://....

Any help is greatly appreciated!

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Accepted answers

Greg-DB

The share method you're using returns a dict, so you can access the fields like:

print(client.share(path)['url'])

That uses API v1 though, which is deprecated. You should switch to using API v2. Using API v2, creating a shared linked is accomplished using the sharing_create_shared_link_with_settings method, which returns a SharedLinkMetadata object. Using it would look like this (also including the construction of the client itself):

dbx = dropbox.Dropbox(ACCESS_TOKEN)
print(dbx.sharing_create_shared_link_with_settings(path).url)

All comments

JohnnyNPC

I'm not too familiar with de-serializing JSON in python, but you are getting a JSON object if you convert that JSON into an object you can just retrieve the URL member of that object

Greg-DB

The share method you're using returns a dict, so you can access the fields like:

print(client.share(path)['url'])

dbx = dropbox.Dropbox(ACCESS_TOKEN)
print(dbx.sharing_create_shared_link_with_settings(path).url)

mnoah66

That worked! Thank you