如果任何目录没有读取权限，pysftp.Connection.walktree() 将失败

Question

我正在使用 pysftp 连接，使用 walktree() 方法列出所有文件，如下所示

with pysftp.Connection(domain, username=USER,
                       password=PWD, port=port,
                       cnopts=cnopts) as sftp:

    # call back method to list the files for the directory found in walktree
    def dir_found_list_files(path):
        if path:
            files = sftp.listdir_attr(path)
            for file in files:
                if 'r' in file.longname[:4]:
                    filelist.append(path+'/'+file.filename)

    # call back method when file found in walktree
    def file_found(path):
        pass

    # call back method when unknown found in walktree
    def unknown_file(path):
        pass

    # sftp walk tree to list files
    sftp.walktree("/", file_found, dir_found_list_files, unknown_file, recurse=True)

这部分工作正常。但是当任何文件夹没有读取权限时，walktree 方法会引发权限异常。如何忽略访问被拒绝的文件夹并继续使用 walktree 获取可访问的文件夹？

Answer 1

你不能。 Connection.walktree 不允许这样的自定义。

但是方法很简单，检查its source code。只需复制代码并根据需要对其进行自定义即可。像这样的东西（未经测试）：

def robust_walktree(sftp, remotepath, fcallback, dcallback, ucallback, recurse=True)
    try:
        entries = sftp.listdir(remotepath)
    except IOError as e:
        if e.errno != errno.EACCES:
            raise
        else
            entries = []

    for entry in entries:
        pathname = posixpath.join(remotepath, entry)
        mode = sftp.stat(pathname).st_mode
        if S_ISDIR(mode):
            # It's a directory, call the dcallback function
            dcallback(pathname)
            if recurse:
                # now, recurse into it
                robust_walktree(sftp, pathname, fcallback, dcallback, ucallback)
        elif S_ISREG(mode):
            # It's a file, call the fcallback function
            fcallback(pathname)
        else:
            # Unknown file type
            ucallback(pathname)

如果任何目录没有读取权限，pysftp.Connection.walktree() 将失败

pysftp.Connection.walktree() fails if any directory don't have read permission

python

sftp

paramiko

pysftp