python ftplib 检查文件是否是文件夹？
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ftplib checking if a file is a folder?
提问by UberJumper
How can i check if a file on a remote ftp is a folder or not using ftplib?
如何使用 ftplib 检查远程 ftp 上的文件是否是文件夹？
Best way i have right now is to do a nlst, and iterate through calling size on each of the files, if the file errors out then it is a folder?
我现在最好的方法是做一个 nlst，并迭代调用每个文件的大小，如果文件出错，那么它是一个文件夹？
Is there a better way? I cannot parse the output of list, since there is about a dozen different ftp servers(many extremely old.)
有没有更好的办法？我无法解析列表的输出，因为大约有十几个不同的 ftp 服务器（许多非常老旧。）
What should i do?
我该怎么办？
回答by Antoine P.
FTP is quite a rudimentary protocol and there's no built-in protocol query allowing you to get the type (file, dir) of each node, so a heuristic like the one you found is the only solution.
FTP 是一个相当基本的协议，没有内置的协议查询允许您获取每个节点的类型（文件、目录），因此像您发现的那样的启发式方法是唯一的解决方案。
If getting the size of each node doesn't work, perhaps you should consider calling FTP.nlst()on each of those nodes: those that error out will be files rather than dirs.
如果获取每个节点的大小不起作用，也许您应该考虑调用FTP.nlst()每个节点：那些出错的将是文件而不是目录。
回答by Ceylan B.
There are not "isdir" and "isfile" definitions in ftplib. If you don't have to use ftplib, i recommend you to use ftputil.
ftplib 中没有“isdir”和“isfile”定义。如果您不必使用 ftplib，我建议您使用ftputil。
First of all you have to install ftputil package. To achive this, use this command: python -m pip install ftputil. After the installation, you can import the library to your code. I think it's enough explanation. So, let's go to the implementation:
首先，您必须安装 ftputil 包。为了实现这一目标，使用这个命令：python -m pip install ftputil。安装后，您可以将库导入到您的代码中。我觉得解释就够了。那么，让我们进入实现：
import ftputil
with ftputil.FTPHost("host", "username", "password") as ftp_host:
ftp_host.chdir("/directory/")
list = ftp_host.listdir(ftp_host.curdir)
for fname in list:
if ftp_host.path.isdir(fname):
print(fname + " is a directory")
else:
print(fname + " is not a directory")
回答by SilentGhost
FTP.dirreturns a directory listing, that you can parse with a callback function to find out whether it's a directory. For example, like this:
FTP.dir返回一个目录列表，您可以使用回调函数对其进行解析，以确定它是否是一个目录。例如，像这样：
def parse(line):
if line[0] == 'd':
print(line.rpartition(' ')[2]) # gives you the name of a directory
ftp.dir(parse)
回答by Giampaolo Rodolà
You can use MLST command:
您可以使用 MLST 命令：
import ftplib
f = ftplib.FTP()
f.connect("localhost")
f.login()
print f.sendcmd('MLST /')
Against pyftpdlib serverthe code above prints:
针对pyftpdlib 服务器，上面的代码打印：
250-Listing "/":
modify=20111212124834;perm=el;size=12288;type=dir;unique=807g100001; /
250 End MLST.
What you have to do is parse that string and look for "type=dir" or "type=cdir" (current dir, as in ".") or "type=pdir" (parent dir as in "..") via a regular expression. If you get a match, it means that the provided path refers to a directory.
您需要做的是解析该字符串并通过以下方式查找“type=dir”或“type=cdir”（当前目录，如“.”）或“type=pdir”（父目录，如“..”）一个正则表达式。如果匹配，则表示提供的路径是指目录。
回答by Evan Fosmark
def is_file(filename):
return ftp.size(filename) is not None
This works because ftp.size returns None if it is a directory.
这是有效的，因为 ftp.size 返回 None 如果它是一个目录。
回答by Evan Fosmark
def is_file(filename):
current = ftp.pwd()
try:
ftp.cwd(filename)
except:
ftp.cwd(current)
return True
ftp.cwd(current)
return False
Here is another solution. Half way writing it out I realized it has a problem. If you don't have permission to change to a folder, it'll read it as a file instead. It'll work if you have access to any folder.
这是另一种解决方案。写到一半才发现有问题。如果您无权更改文件夹，它会将其作为文件读取。如果您有权访问任何文件夹，它将起作用。
I still posted it because maybe it'll give some ideas.
我仍然发布它，因为它可能会提供一些想法。
回答by jameh
I have used this when dealing with ftplib:
我在处理 ftplib 时使用过这个：
import ftplib
from ftplib import FTP
ftp=FTP("ftp address")
ftp.login("user","password")
path = "/is/this/a/file/or/a/folder/"
try:
ftp.cwd(path)
print "it's a folder!"
except ftplib.error_perm:
print "it's not a folder!"
回答by Daniel Lee
This is what I use for finding the directories in a given directory. I return the directory names as a set because I use sets later on in the program, but you could leave off the set creation at the end and return the directory names as a list, tuple, etc.
这是我用于在给定目录中查找目录的方法。我将目录名称作为集合返回，因为我稍后会在程序中使用集合，但是您可以在最后不创建集合，并将目录名称作为列表、元组等返回。
def get_directories(ftp_server):
"""
Returns a set of directories in the current directory on the FTP server
Stdout output of ftp_server.dir() is redirected to an IO object and then
reset, because ftp_server.dir() only prints its results to stdout.
@param ftp_server: Open connection to FTP server
@return: Set of directory names
"""
# Redirect stdout
old_stdout = sys.stdout
sys.stdout = new_stdout = StringIO()
# Get directory listing
ftp_server.dir()
# Reset stdout
sys.stdout = old_stdout
directory_listing = new_stdout.getvalue().split("\n")
# Directories are listed starting with "d" for "directory"
only_directories = (x for x in directory_listing if x.startswith("d"))
# This only deals with directory names without spaces.
directory_names = set(dirname.split()[-1] for dirname in only_directories)
return directory_names
回答by Diego Vélez
I have found a solution, but it may not be the best and I think it can be helpful for you.
我找到了一个解决方案，但它可能不是最好的，我认为它可能对您有所帮助。
> def isfile(remote_name):
> try:
> ftp.sendcmd('MDTM ' + remote_name)
> except:
> return False
> else:
> return True
This function will return TRUE if 'remote_name' is a regular file, otherwise will return False.
如果“remote_name”是常规文件，则此函数将返回 TRUE，否则将返回 False。
回答by Nick
My entire solution looks like this:
我的整个解决方案如下所示：
def is_file(filename):
try:
ftp.cwd(filename)
except Exception as e:
if "550" in e:
return True
return False
