I am uploading files to a WebDav server with pyCurl and the following function:
def fileupload(url, filename, filedata, upload_user, upload_pw):
# Initialize pycurl
c = pycurl.Curl()
url = url + filename
c.setopt(pycurl.URL, url)
c.setopt(pycurl.UPLOAD, 1)
c.setopt(pycurl.READFUNCTION, open(filename, 'rb').read)
c.setopt(pycurl.SSL_VERIFYPEER, 0)
c.setopt(pycurl.SSL_VERIFYHOST, 0)
c.setopt(pycurl.USERPWD, upload_user + ":" + upload_pw)
# Set size of file to be uploaded.
filesize = os.path.getsize(filename)
c.setopt(pycurl.INFILESIZE, filesize)
# Start transfer
c.perform()
c.close()
Now, I want to modify the function to upload a file coded in a base64 bytestring. How I have to modify the readfunction c.setopt(pycurl.READFUNCTION, open(filename, 'rb').read)
and the filsize filesize = os.path.getsize(filename)
to make this work? The base64 bytestring is provided in the variable filedata
.
As an bytestream noob, my try and error approach was unfortunately not successful...
Thanks in advance!
I think I found a solution with this code:
def fileupload(url, filename, filedata, upload_user, upload_pw):
filedata = base64.b64decode(filedata).decode('ascii', 'ignore')
# Initialize pycurl
c = pycurl.Curl()
url = url + filename
c.setopt(pycurl.URL, url)
c.setopt(pycurl.UPLOAD, 1)
c.setopt(pycurl.READFUNCTION, StringIO(filedata).read)
c.setopt(pycurl.SSL_VERIFYPEER, 0)
c.setopt(pycurl.SSL_VERIFYHOST, 0)
c.setopt(pycurl.USERPWD, upload_user + ":" + upload_pw)
# Set size of file to be uploaded.
filesize = len(filedata)
c.setopt(pycurl.INFILESIZE, filesize)
# Start transfer
c.perform()
c.close()