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Forum Discussion
mbm123
4 years agoExplorer | Level 4
Access without an "app"
Hey, I was wondering if it is possible to access a dropbox account without the token generated in the app console. What I mean is, can I use an api on my python script that will query an input drop...
- 4 years ago
Yes, that's correct.
vimalapalanisamy22
3 years agoNew member | Level 2
Currently we are using few Dropbox api method. We are using Authorization flow in which response code ,scope ,client id passed.this will give me Authorization code and i need to pass them to token api which will return access token. Is there any way to automate this process- exchanging Authorization code for access token. Or Dropbox can use api for Authorization flow for instead of this website.
- Greg-DB3 years agoDropbox Staff
vimalapalanisamy22 No, it is not possible to automate the app authorization flow. That will need to be processed manually once per user. The resulting access token and optional refresh token can be stored and re-used programmatically. Refer to the OAuth Guide and authorization documentation for more information.
- vimala_palanisamy223 years agoExplorer | Level 3Let's say ,I am team admin in my team there are 3 members-not admin role are there ..all of them are creating full Dropbox apps. I have added them in the settings->app permission ->other apps..now as a team admin how can I generate a token for the apps that was created by other members without using Authorization flow.
- Greg-DB3 years agoDropbox Staff
vimala_palanisamy22 You cannot get an access token for someone else's app without using the OAuth app authorization flow. If you need an access token for someone else's app to connect to your account, you will need to use the OAuth app authorization flow as documented. (The owner of an app can use the "Generate" button on the app's page on the App Console to create an access token for their own account for that app without using the actual OAuth app authorization flow, but that is only available to the owner of the app in particular.)
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