CODEWITHSUNDEEP
pythonpython-3.xgoogle-apigoogle-drive-api
OxxoCode
OxxoCode

Reputation: 31

Copying a file with Drive API v3 doesn't change title or location

I'm trying to use v3 of the Drive API to copy a file. Although I'm successful in getting a copy of the file to be created, neither the copied file's title nor location is changed as I had hoped. Looking at code snippets available online, I'm not sure what's wrong with my code, so any help would be greatly appreciated.

def copy(service, file_id, dest_id): #Drive service, id of file to be copied, id of destination folder
    service.files().copy(
        fileId=file_id,
        supportsAllDrives=True,
        body = {
                'title':'copiedFile',
                'parents' : [ {'kind':"drive#fileLink",
                                'id': dest_id}],
                }
        ).execute()

Upvotes: 1

Views: 2723

Answers (3)

Ed Nutting
Ed Nutting

Reputation: 271

For anyone else coming to this in 2024 using the Google Drive v3 API, the answer is to use the 'name' parameter instead of 'title':

function copyFile(fileId, parentId, newFileName) {
    const copiedFile = Drive.Files.copy({
        name: newFileName,
        parents: [{
            id: parentId,
        }]
    }, fileId, {
        supportsAllDrives: true
    });
    return copiedFile.id;
}

Upvotes: 0

OxxoCode
OxxoCode

Reputation: 31

Managed to figure this out own by own, it turned out to be a fairly simple approach - here's the code for a function that handles copying a single file.

#Handles copying of a file.
#If no new destination (parent) is provided, it is copied
#into the same location the original file was located.

def copy(service, fileID, parents=None):
    mimetype = service.files().get(fileId=fileID, fields='mimeType', supportsAllDrives=True).execute()['mimeType']
    if mimetype == 'application/vnd.google-apps.folder': #If the provided file id points to a folder
        #Code to handle copying a folder
    else:
        file_metadata = {}
        if parents is not None: #If custom destination is specified, add it to the file metadata
            file_metadata['parents'] = [parents]

        #Copy the file
        file = service.files().copy(
            fileId=fileID,
            body=file_metadata,
            fields='id, name', #Can be whatever fields you want, I just like having access to the resulting file id & name
            supportsAllDrives=True).execute()

Upvotes: 2

glotchimo
glotchimo

Reputation: 685

You need to copy the file first, then update it by first removing the current parents and then adding the ID of the new parent folder. Here's a snippet that should do the trick:

copy = (
    service.files()
    .copy(
        fileId=file_id,
        body={"title": "copiedFile"},
    )
    .execute()
)

service.files().update(
    fileId=copy.get("id"),
    addParents=dest_id,
    removeParents=copy.get("parents"),
    fields="id, parents",
).execute()

Upvotes: 1

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