在 Python 中使用 SFTP 上传文件，但如果路径不存在则创建目录
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Upload files using SFTP in Python, but create directories if path doesn't exist
提问by franzlorenzon
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:
我想用 Python 在远程服务器上上传一个文件。我想事先检查远程路径是否真的存在，如果不存在，则创建它。在伪代码中：
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:
我正在考虑在 Paramiko 中执行一个命令来创建路径（例如mkdir -p remote_path）。我想出了这个：
# 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?
但是这个解决方案对我来说听起来不太好，因为我关闭了连接然后再次重新打开它。有没有更好的方法来做到这一点？
采纳答案by isedev
SFTP supports the usual FTP commands (chdir, mkdir, etc...), so use those:
SFTP 支持常用的 FTP 命令（chdir、mkdir 等...），因此请使用这些命令：
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:
要完全模拟mkdir -p，您可以递归地处理 remote_path：
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.
当然，如果remote_path 还包含远程文件名，则需要将其拆分，将目录传递给mkdir_p 并使用文件名代替'.'。在 sftp.put 中。
回答by Benjamin
Paramiko contains a mkdir function:
Paramiko 包含一个 mkdir 函数：
回答by gabhijit
Something simpler and slightly more readable too
一些更简单，更易读的东西
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_)
回答by Mickey Afaneh
Had to do this today. Here is how I did it.
今天不得不这样做。这是我如何做到的。
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)
回答by Daniil Burmashev
Assuming sftp operations are expensive, I would go with:
假设 sftp 操作很昂贵，我会选择：
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))```
回答by ritesh
you can use pysftp package:
您可以使用 pysftp 包：
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
您可以查看此链接以获取更多详细信息：https: //pysftp.readthedocs.io/en/release_0.2.9/cookbook.html#pysftp-connection-mkdir
回答by Mikhail Gerasimov
My version:
我的版本：
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 mkdirwithout trying listdir/statfirst due to EAFP principle(it's also more performant to make one network request than several).
由于EAFP 原则，我们mkdir不尝试listdir/stat首先尝试（发出一个网络请求比发出多个网络请求的性能更高）。
