Reputation: 5943
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?
Upvotes: 39
Views: 79563
Reputation: 1
If you want to do it without an exception flow, reserving exceptions only for exception flow. You could do the following:
def ensure_sftp_directories_exist(sftp: SFTPClient, remote_dir: str) -> None:
cleaned_remote_dir = remote_dir.strip('/') # Remove leading and trailing slashes
sub_directories = cleaned_remote_dir.split("/") # Get all directories
current_directory = "/" # Root is start
for directory in sub_directories:
# Skip empty directory names for accidental double '/'
if not directory:
continue
if directory not in sftp.listdir(current_directory):
sftp.mkdir(f"{current_directory}/{directory}")
current_directory = f"{current_directory}/{directory}"
Notes:
Upvotes: 0
Reputation: 39536
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).
Upvotes: 3
Reputation: 187
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
Upvotes: 5
Reputation: 11
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))```
Upvotes: 1
Reputation: 3585
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_)
Upvotes: 12
Reputation: 19601
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.
Upvotes: 69
Reputation: 81
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)
Upvotes: 8