在Python中使用SFTP上载文件,但如果路径不存在则创建目录

Question

我想用Python上传远程服务器上的文件.我想事先检查远程路径是否真的存在,如果不存在,则要创建它.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不太好,因为我关闭连接然后再次重新打开它.有没有更好的方法呢？

Answer 1

SFTP支持通常的FTP命令(chdir,mkdir等...),所以使用:

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,您可以递归地通过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()

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并使用文件名而不是'.' 在sftp.put中.

Answer 2

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_)

Answer 3

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)

Answer 4

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

您可以检查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