Upload files using SFTP in Python, but create directories if path doesn't exist

Question

I want to upload a file on a remote server with Python. I’d like to check beforehand if the remote path is really existing, and if it isn’t, to create it. In pseudocode:

if(remote_path not exist):
    create_path(remote_path)
upload_file(local_file, remote_path)

I was thinking about executing a command in Paramiko to create the path (e.g. mkdir -p remote_path). I came up with this:

# I didn't test this code

import paramiko, sys

ssh = paramiko.SSHClient()
ssh.connect(myhost, 22, myusername, mypassword)
ssh.exec_command('mkdir -p ' + remote_path)
ssh.close

transport = paramiko.Transport((myhost, 22))
transport.connect(username = myusername, password = mypassword)

sftp = paramiko.SFTPClient.from_transport(transport)
sftp.put(local_path, remote_path)
sftp.close()

transport.close()

But this solution doesn’t sound good to me, because I close the connection and then reopen it again. Is there a better way to do it?

Asked By: franzlorenzon

||

Source

Answer 1

Paramiko contains a mkdir function:

http://paramiko-docs.readthedocs.org/en/latest/api/sftp.html#paramiko.sftp_si.SFTPServerInterface.mkdir

Answered By: Benjamin

Answer 2

SFTP supports the usual FTP commands (chdir, mkdir, etc…), so use those:

sftp = paramiko.SFTPClient.from_transport(transport)
try:
    sftp.chdir(remote_path)  # Test if remote_path exists
except IOError:
    sftp.mkdir(remote_path)  # Create remote_path
    sftp.chdir(remote_path)
sftp.put(local_path, '.')    # At this point, you are in remote_path in either case
sftp.close()

To fully emulate mkdir -p, you can work through remote_path recursively:

import os.path

def mkdir_p(sftp, remote_directory):
    """Change to this directory, recursively making new folders if needed.
    Returns True if any folders were created."""
    if remote_directory == '/':
        # absolute path so change directory to root
        sftp.chdir('/')
        return
    if remote_directory == '':
        # top-level relative directory must exist
        return
    try:
        sftp.chdir(remote_directory) # sub-directory exists
    except IOError:
        dirname, basename = os.path.split(remote_directory.rstrip('/'))
        mkdir_p(sftp, dirname) # make parent directories
        sftp.mkdir(basename) # sub-directory missing, so created it
        sftp.chdir(basename)
        return True

sftp = paramiko.SFTPClient.from_transport(transport)
mkdir_p(sftp, remote_path) 
sftp.put(local_path, '.')    # At this point, you are in remote_path
sftp.close()

Of course, if remote_path also contains a remote file name, then it needs to be split off, the directory being passed to mkdir_p and the filename used instead of ‘.’ in sftp.put.

Answered By: isedev

Answer 3

Something simpler and slightly more readable too

def mkdir_p(sftp, remote, is_dir=False):
    """
    emulates mkdir_p if required. 
    sftp - is a valid sftp object
    remote - remote path to create. 
    """
    dirs_ = []
    if is_dir:
        dir_ = remote
    else:
        dir_, basename = os.path.split(remote)
    while len(dir_) > 1:
        dirs_.append(dir_)
        dir_, _  = os.path.split(dir_)

    if len(dir_) == 1 and not dir_.startswith("/"): 
        dirs_.append(dir_) # For a remote path like y/x.txt 

    while len(dirs_):
        dir_ = dirs_.pop()
        try:
            sftp.stat(dir_)
        except:
            print "making ... dir",  dir_
            sftp.mkdir(dir_)

Answered By: gabhijit

Answer 4

Had to do this today. Here is how I did it.

def mkdir_p(sftp, remote_directory):
    dir_path = str()
    for dir_folder in remote_directory.split("/"):
        if dir_folder == "":
            continue
        dir_path += r"/{0}".format(dir_folder)
        try:
            sftp.listdir(dir_path)
        except IOError:
            sftp.mkdir(dir_path)

Answered By: Mickey Afaneh

Answer 5

Assuming sftp operations are expensive,
I would go with:

def sftp_mkdir_p(sftp, remote_directory):
    dirs_exist = remote_directory.split('/')
    dirs_make = []
    # find level where dir doesn't exist
    while len(dirs_exist) > 0:
        try:
            sftp.listdir('/'.join(dirs_exist))
            break
        except IOError:
            value = dirs_exist.pop()
            if value == '':
                continue
            dirs_make.append(value)
        else:
            return False
    # ...and create dirs starting from that level
    for mdir in dirs_make[::-1]:
        dirs_exist.append(mdir)
        sftp.mkdir('/'.join(dirs_exist))```

Answered By: Daniil Burmashev

Answer 6

you can use pysftp package:

import pysftp as sftp

#used to pypass key login
cnopts = sftp.CnOpts()
cnopts.hostkeys = None

srv = sftp.Connection(host="10.2.2.2",username="ritesh",password="ritesh",cnopts=cnopts)
srv.makedirs("a3/a2/a1", mode=777)  # will happily make all non-existing directories

you can check this link for more details:
https://pysftp.readthedocs.io/en/release_0.2.9/cookbook.html#pysftp-connection-mkdir

Answered By: ritesh

Answer 7

My version:

def is_sftp_dir_exists(sftp, path):
    try:
        sftp.stat(path)
        return True
    except Exception:
        return False


def create_sftp_dir(sftp, path):
    try:
        sftp.mkdir(path)
    except IOError as exc:
        if not is_sftp_dir_exists(sftp, path):
            raise exc


def create_sftp_dir_recursive(sftp, path):
    parts = deque(Path(path).parts)

    to_create = Path()
    while parts:
        to_create /= parts.popleft()
        create_sftp_dir(sftp, str(to_create))

We try mkdir without trying listdir/stat first due to EAFP principle (it’s also more performant to make one network request than several).

Answered By: Mikhail Gerasimov

Upload files using SFTP in Python, but create directories if path doesn't exist

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