I'm attempting to simply push a local file to an FTP destination.
However my script always seems to hang during the "put" operation.
filepath = '/home/foo/local.csv'
conn = sftp.Connection(host=HOST, username=USERNAME,
port=PORT, password=PASSWORD)
remote_path = 'foo/local.csv'
conn.put(filepath, remote_path)
conn.close()
The file successfully gets uploaded to FTP but the control flow gets stuck on the put command. Any ideas how to resolve this?
Thank you.
Related
I am uploading files to a remote sftp server using the pysftp module in Python. About 25% of the time a file won't upload and I receive the error: '[Errno 2] No such file.'
I am connecting to the remote sftp server with simply a username and password. No SSH key is being used. Upon establishing the connection I pass an instance of cnopts() with hostkeys set to None since no SSH keys are being used. I then loop through each file and and perform a put() to upload each file to the sftp server. The first couple files typically upload successfully, but then by the thrid or fourth file I typically receive the [Errno 2] error. If I rerun the script on the same file that just failed, it will upload just fine. Therefore, it doesn't seem to be an issue with the local or remote path since everything is the same the first run to the second.
Connection Code
elif self.ctype == 'sftp':
if self.pkpath == None:
#set pysftp to not check for ssh key, only use password
cnopts = pysftp.CnOpts()
cnopts.hostkeys = None
#connecting to sftp server
self.conn = pysftp.Connection(self.address, self.user, password=self.password,
cnopts=cnopts)
else:
self.conn = pysftp.Connection(self.address, self.user, password=self.password,
private_key=self.pkpath)
Put Code (uploadFile function)
def uploadFile(self,fileName,fdestname=None):
...
if fdestname is None:
fattr = self.conn.put(fileName, confirm=True, preserve_mtime=True)
else:
fattr = self.conn.put(fileName, fdestname, confirm=True, preserve_mtime=True)
Looping Code
elif inputdict['source'] == 'unprocessed':
Sftp.uploadFile(os.path.join(inputdict['unprocessedfolder'],vimsf), vimsfm)
print Sftp.lst()
Since the local and remote file paths are correct and subsequent loops upload the previously failed files just fine, I'd expect that all files would upload successfully on the first try.
In this particular case, the '[Errno 2] No such file.'error was showing up because the input on the put method was set to True. The files were being removed from the destination folder so quickly after the upload, the confirmation process could not complete. It would then return the '[Errno 2] No such file.' error. The issue was intermittent since about 50% of the time, the files would stick around long enough in the destination directory that the confirmation could complete successfully. Setting the confirm bool to False solved the issue.
Very good catch here. Was pulling my hair out, but turned out that on the post side, they were processing the file quickly and removing it. This caused the error.
head, tail = os.path.split(filename)
with pysftp.Connection('hotname', username='uname', password='pass',
cnopts=cnopts) as sftp:
with sftp.cd('/mydir/subdir/'):
sftp.put(filename,tail,confirm=False)
I use Paramiko to put a file to an SFTP server:
import paramiko
transport = paramiko.Transport((host, port))
transport.connect(username=username, password=password)
sftp = paramiko.SFTPClient.from_transport(transport)
sftp.put(local_path, remote_path)
Now, I would like to check if it worked. The idea is that I compare the checksum of the local file and the remote one (that is located on the SFTP server).
Does Paramiko functionality allows to do that? If it is the case, how exactly it works?
With the SFTP, running over an encrypted SSH session, there's no chance the file contents could get corrupted while transferring. So unless it gets corrupted when reading the local file or writing the remote file, you can be pretty sure that the file was uploaded correctly, if the .put does not throw any error.
try:
sftp.put(local_path, remote_path)
except:
# Something went wrong
If you want to test explicitly anyway:
While there's the check-file extension to the SFTP protocol to calculate a remote file checksum, it's not widely supported. Particularly it's not supported by the most widespread SFTP server implementation, the OpenSSH. See What SFTP server implementations support check-file extension.
If you are lucky to connect to another SFTP server that supports the extension, you can use the Paramiko's SFTPFile.check method.
If not, your only option is to download the file back and compare locally.
If you have a shell access to the server, you can of course try to run some shell checksum command (sha256sum) over a separate shell/SSH connection (or channel) and parse the results. But that's not an SFTP solution anymore. See Comparing MD5 of downloaded files against files on an SFTP server in Python.
if the file doesn't upload then the method will throw an error, so u can check for error
import paramiko
transport = paramiko.Transport((host, port))
transport.connect(username=username, password=password)
sftp = paramiko.SFTPClient.from_transport(transport)
try:
sftp.put(local_path, remote_path)
except IOError:
#'Failure'
except OSError:
#'Failure'
I am using the below Python code to upload a file via SFTP using Paramiko. The connection "seems" to be fine, the code executes to the end, just the file isn't reaching the destination when I check in FileZilla.
I have checked and set permissions on the file to 777 (just to be sure). I have also checked my file path string in a separate terminal and the path is valid.
import paramiko
.
.
transport = paramiko.Transport((host, port))
transport.connect(username = username, password = password)
sftp = paramiko.SFTPClient.from_transport(transport)
sftp.put(filePath, "/") # Upload file to root FTP folder
sftp.close()
transport.close()
What can I do to debug this? Anything I can print out, check connection succeeded etc?
The second argument of SFTPClient.put (remotepath) is path to a file, not a folder:
the destination path on the SFTP server. Note that the filename should be included. Only specifying a directory may result in an error.
Try this:
sftp.put(filePath, "/filename")
I wrote a simple code to upload a file to a SFTP server in Python. I am using Python 2.7.
import pysftp
srv = pysftp.Connection(host="www.destination.com", username="root",
password="password",log="./temp/pysftp.log")
srv.cd('public') #chdir to public
srv.put('C:\Users\XXX\Dropbox\test.txt') #upload file to nodejs/
# Closes the connection
srv.close()
The file did not appear on the server. However, no error message appeared. What is wrong with the code?
I have enabled logging. I discovered that the file is uploaded to the root folder and not under public folder. Seems like srv.cd('public') did not work.
I found the answer to my own question.
import pysftp
srv = pysftp.Connection(host="www.destination.com", username="root",
password="password",log="./temp/pysftp.log")
with srv.cd('public'): #chdir to public
srv.put('C:\Users\XXX\Dropbox\test.txt') #upload file to nodejs/
# Closes the connection
srv.close()
Put the srv.put inside with srv.cd
Do not use pysftp it's dead. Use Paramiko directly. See also pysftp vs. Paramiko.
The code with Paramiko will be pretty much the same, except for the initialization part.
import paramiko
with paramiko.SSHClient() as ssh:
ssh.load_system_host_keys()
ssh.connect(host, username=username, password=password)
sftp = ssh.open_sftp()
sftp.chdir('public')
sftp.put('C:\Users\XXX\Dropbox\test.txt', 'test.txt')
To answer the literal OP's question: the key point here is that pysftp Connection.cd works as a context manager (so its effect is discarded without with statement), while Paramiko SFTPClient.chdir does not.
import pysftp
with pysftp.Connection(host="www.destination.com", username="root",
password="password",log="./temp/pysftp.log") as sftp:
sftp.cwd('/root/public') # The full path
sftp.put('C:\Users\XXX\Dropbox\test.txt') # Upload the file
No sftp.close() is needed, because the connection is closed automatically at the end of the with-block
I did a minor change with cd to cwd
Syntax -
# sftp.put('/my/local/filename') # upload file to public/ on remote
# sftp.get('remote_file') # get a remote file
I'm trying to use paramiko to get a file via SFTP.
It connects, I can list directories and it even downloads the first megabyte or so of the file but then it just hangs. No exception, no error, nothing. It just hangs there indefinitely.
Here's the code I'm working with:
import paramiko
t = paramiko.Transport( host )
t.connect( username=uname, password=passwd )
f = paramiko.SFTPClient.from_transport( t )
print f.listdir()
f.get( fname, fname ) #it hangs on this line :\
I have sftp access to the host in question but no shell access.
The host contains a single file that I need to fetch regularly and process in a python script.
Any help with this problem or alternate solutions to doing SFTP in Python are greatly appreciated :)
I was experiencing the same problem as Ulfur. He posted his own fix/workaround as a comment to another answer, so I decided to add it as a proper answer to make it more visible.
The basic idea is to not use the .get() method, but to loop over the lines. The following is a Python 3 implementation.
transport = None
sftp = None
sftp_path = 'some/sftp/path'
dst_path = '/some/local/path'
try:
transport = paramiko.Transport((SFTP_HOST, SFTP_PORT))
transport.set_log_channel('delapi')
transport.connect(username=SFTP_USER, password=SFTP_PASS)
sftp = paramiko.SFTPClient.from_transport(transport)
with sftp.open(sftp_path, 'r') as remote:
with open(dst_path, 'w') as local:
for line in remote:
local.write(line)
except SSHException:
print("SSH error")
finally:
if sftp:
sftp.close()
if transport:
transport.close()
I suggest you fire up Wireshark on the client and see what's happening at the protocol level. You won't be able to read the data in the packets as it will be encrypted, but you will see what's going on at the TCP/IP level and that might provide a clue.