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Mark L.45's avatar
Mark L.45
Collaborator | Level 9
10 years ago

Shared Links in SwiftyDropbox broken?

Running this function using SwiftyDropbox! Was working well til this morning, now it returns nothing useful, just the reply that the link already exists, and when I try and get the link... it returns...
API
Mark L.45's avatar
Mark L.45
Collaborator | Level 9
10 years ago

Hi Greg,

 

Printed the errors, so I try to create a shared link. It fails with ...

 

sharing_create_linkV2 /acme201/acmeSINGLE/acmeX101.001.png

error Optional([request-id abc63bdac24b4afe306dae1d] API route error - {

    ".tag" = "shared_link_already_exists";

})

 

I ask for the shared link and I get...

 

error Optional([request-id 18b2cd24390e59cf83cedcef] API route error - {

    ".tag" = "shared_link_not_found";

}) source /acme201/acmeSINGLE/acmeX101.001.png response.0 nil response.1 nil 0

,/acme201/acmeSINGLE/acmeX101.001.png,0

 

Maybe I am asking the wrong question. I used the same string in both create link and return link API calls; but I am unclear if the API really works that way. The string I supply is the one you see here in bold/italics.  Create shared link wants a PATH, and yet get shared link suggests it needs a URL?

 

What do I need to add an HTTP://something? to the PATH when I call client.sharing.getSharedLinkFile API

 

 

  • Mark L.45's avatar
    Mark L.45
    Collaborator | Level 9
    10 years ago
    *** STOP PRESS *** Managed to find a solution, used this code that works.

    _ = client.sharing.listSharedLinks(path:sourcePath,cursor:nil,directOnly:false).response(queue: DispatchQueue(label: "MyCustomSerialQueue")) { response, error2 in


    print("data error \(error2) source \(sourcePath) response.0 \(response?.links[0].url) \(posIndex)")
    sharedDataAccess.fnLink(index2seek: posIndex, fnLink: (response?.links[0].url)!)
    sharedDataAccess.fnPos(index2seek: posIndex, fnPos: posIndex)
    }
    • tim's avatar
      tim
      Explorer | Level 4
      10 years ago

      Just a note: if your code deals with a file that's in a folder which is also shared, then links[0].url will exist and be the shared link of the folder, not the file. If you want it to only return direct shared links, rather than use an ancestor's link, then specify directOnly as true.

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