如果任何目录没有读取权限，pysftp.Connection.walktree()就会失败
如果任何目录没有读取权限，pysftp.Connection.walktree()就会失败
提问于 2021-07-15 19:30:16
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)这部分运行得很好。但是，当任何文件夹没有读取权限时，walktree方法将引发权限异常。如何忽略拒绝访问文件夹并继续使用可访问文件夹的walktree？
Stack Overflow用户
回答已采纳
发布于 2021-07-15 20:05:19
你不能。Connection.walktree不允许这样的自定义。
但是这个方法非常简单，请查看its source code。只需复制代码并根据需要进行自定义即可。类似于以下内容(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)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/68392997
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