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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:
...
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)
Greg-DB
Dropbox Community Moderator
Dropbox Community ModeratorThe 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)
That worked! Thank you
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