33

我想用 Python 在远程服务器上上传文件。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)

我正在考虑在 Paramiko 中执行命令来创建路径（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()

但是这个解决方案对我来说听起来并不好，因为我关闭了连接然后再次重新打开它。有更好的方法吗？

4

7 回答 7

58

SFTP 支持常用的 FTP 命令（chdir、mkdir 等），因此请使用以下命令：

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()

要完全模拟mkdir -p，您可以递归地使用 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.

于 2013-02-11T19:57:29.460 回答
8

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_)
于 2013-12-06T11:25:49.647 回答
6

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)
于 2015-01-28T18:47:32.283 回答
3

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

于 2019-09-24T10:31:00.240 回答
2

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).

于 2020-05-14T08:31:00.473 回答
1

Paramiko 包含一个 mkdir 函数：

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

于 2013-02-11T19:55:44.073 回答
0

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))```
于 2018-12-20T23:59:57.240 回答