CODEWITHSUNDEEP
google-drive-api
Jon Saalbach
Jon Saalbach

Reputation: 385

Get shared link through Google Drive API

I am working on an app using Google Drive. I want the user to be able to share files by link, setting the permissions to anyone and withLink as described in the Google Developers documentation.

However, I cannot figure out what link to share. When I share a file in the Google Drive browser interface, I see the Link to share in the format:

https://docs.google.com/presentation/d/[...]/edit?usp=sharing

A link in this format is not part of the file resource object, nor it is returned from the http call setting the permissions. I hope someone can explain to me how to get this link through the REST api?

Upvotes: 28

Views: 51131

Answers (5)

rubbles
rubbles

Reputation: 11

For python, I only needed to get the file "id". Then "created" the link like this:

def create_folder(folder_name, folder_id):
    """Create a folder and prints the folder ID and Folder link
    Returns : Folder Id

    """

    try:
        # create drive api client
        service = build("drive", "v3", credentials=creds)
        file_metadata = {
            "name": folder_name,
            "mimeType": "application/vnd.google-apps.folder",
            "parents": [folder_id],
        }

        file = (
            service.files().create(body=file_metadata, fields="id").execute()
        )
        id = file.get("id")
        print(
            f'Folder ID: "{id}".',
            f'https://drive.google.com/drive/u/0/folders/{id}',
        )
        return id

    except HttpError as error:
        print(f"An error occurred: {error}")
        return None

Upvotes: 1

Heitor Althmann
Heitor Althmann

Reputation: 367

Here's a practical example on how to get the WebViewLink file property (A.K.A. File edit link):

$file = $service->files->get($file_id, array('fields' => 'webViewLink'));
$web_link_view = $file->getWebViewLink();

OR

$sheetsList = $drive_service->files->listFiles([
  'fields' => 'files(id, name, webViewLink, webContentLink)',
]);

$web_link_view = $sheetsList->current()->getWebViewLink();

Pay attention that you should load the file specifying which fields you wanna bring with it (In this case, webViewLink). If you don't do that, only id and name will be available.

In case you need to adjust the file sharing settings, this is how you do it:

$permissions = new \Google_Service_Drive_Permission();
$permissions->setRole('writer');
$permissions->setType('anyone');

$drive_service->permissions->create($file_id, $permissions);

Possible values for setRole() and setType() can be found here: https://developers.google.com/drive/api/v3/reference/permissions/create

Upvotes: 3

Claudio Cherubino
Claudio Cherubino

Reputation: 15004

You can use the alternateLink property in the File resource to get a link that can be shared for opening the file in the relevant Google editor or viewer:

https://developers.google.com/drive/v2/reference/files

Update

[With API V3](https://developers.google.com/drive/api/v3/reference/files it is suggested to use the webViewLink property.

Upvotes: 28

rzymek
rzymek

Reputation: 9281

To actually enable link sharing using Google Drive API:

drive.permissions.create({
  fileId: '......',
  requestBody: {
    role: 'reader',
    type: 'anyone',
  }
});

Get the webLinkView value using:

const webViewLink = await drive.files.get({
    fileId: file.id,
    fields: 'webViewLink'
}).then(response => 
    response.data.webViewLink
);

Upvotes: 19

Miguel Q
Miguel Q

Reputation: 3618

In my case using the PHP Api v3, for the link to be non-empty you must define that you request this field... and if you have the right permissions:

so something like this:

$file =self::$service->files->get("1ogXyGxcJdMXt7nJddTpVqwd6_G8Hd5sUfq4b4cxvotest",array("fields"=>"webViewLink"));

Upvotes: 4

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