[issue41029] parallel sftp from local to remote file creates empty directory in home path

Christian Heimes added the comment: ParallelSSH is a 3rd party library and not part of Python's standard library. Please report the issue at https://github.com/ParallelSSH/parallel-ssh/ -- nosy: +christian.heimes resolution: -> third party stage: -> resolved status: open -> closed

[issue41029] parallel sftp from local to remote file creates empty directory in home path

reated in home. -- components: Library (Lib) messages: 371845 nosy: sugan19 priority: normal severity: normal status: open title: parallel sftp from local to remote file creates empty directory in home path type: behavior versions: Python 3.8 ___ Python t

[issue39729] stat.S_ISXXX can raise OverflowError for remote file modes

Change by Terry J. Reedy : -- nosy: +christian.heimes ___ Python tracker ___ ___ Python-bugs-list mailing list Unsubscribe:

[issue39729] stat.S_ISXXX can raise OverflowError for remote file modes

Arnon Yaari added the comment: Maybe so, but IMHO it is a lesser concern - the operating systems that use "unsigned long" add bits on the right, and don't change the less significant bits that the S_ISXXX macros check (if I am not mistaken, those bits are POSIX standards). This issue is a

[issue39729] stat.S_ISXXX can raise OverflowError for remote file modes

Ammar Askar added the comment: Size of the mode aside, isn't this approach problematic anyway? One platform could have an entirely different way of signalling ISDIR vs another. In this case using the host's S_ISDIR for the remote's mode would result in possibly incorrect values. --

[issue39729] stat.S_ISXXX can raise OverflowError for remote file modes

;. -- components: Library (Lib) messages: 362499 nosy: wiggin15 priority: normal severity: normal status: open title: stat.S_ISXXX can raise OverflowError for remote file modes type: behavior versions: Python 3.8, Python 3.9 ___ Python tracker <h

I get an error when I used urllib2.urlopen() to open a remote file in a ftp server

Hi everybody: I get an error when I used urllib2.urlopen() to open a remote file in a ftp server, My code is the following: file = 'ftp:/192.168.250.14:2180/RTVE/VIDEOS/Thisisit.wmv' mydata = urllib2.urlopen(file) Traceback (most recent call last): File console, line 1, in module File /usr

Re: I get an error when I used urllib2.urlopen() to open a remote file in a ftp server

On Thu, Jan 6, 2011 at 10:26 AM, Ariel isaacr...@gmail.com wrote: Hi everybody: I get an error when I used urllib2.urlopen() to open a remote file in a ftp server, My code is the following: file = 'ftp:/192.168.250.14:2180/RTVE/VIDEOS/Thisisit.wmv' Looks to me like you're missing a slash

Re: I get an error when I used urllib2.urlopen() to open a remote file in a ftp server

You are right, Thanks. On Thu, Jan 6, 2011 at 12:55 PM, Ian Kelly ian.g.ke...@gmail.com wrote: On Thu, Jan 6, 2011 at 10:26 AM, Ariel isaacr...@gmail.com wrote: Hi everybody: I get an error when I used urllib2.urlopen() to open a remote file in a ftp server, My code is the following

Including a remote file -- permission denied?

I have a Python script running on the default OSX webserver, stored in /Library/WebServer/CGI-Executables. That script spits out a list of files on a network drive, a la os.listdir('/Volumes/code/ directory/'). If I just execute this from the terminal, it works as expected, but when I try to

Re: How to open a remote file using python.

On Feb 22, 9:02 pm, venutaurus...@gmail.com venutaurus...@gmail.com wrote: On Feb 23, 9:25 am, MRAB goo...@mrabarnett.plus.com wrote: venutaurus...@gmail.com wrote: Hello all,            I am writing an application where I need to open a shared file on a remote machine using python

Re: How to open a remote file using python.

On Sun, Feb 22, 2009 at 9:02 PM, venutaurus...@gmail.com venutaurus...@gmail.com wrote: On Feb 23, 9:25 am, MRAB goo...@mrabarnett.plus.com wrote: venutaurus...@gmail.com wrote: Hello all, I am writing an application where I need to open a shared file on a remote machine using

How to open a remote file using python.

Hello all, I am writing an application where I need to open a shared file on a remote machine using python script. I tried using the following function: f = urllib.open(\\remote_machine\\folder1\\file1.doc) I also tried using class urllib.FancyURLopener(...) but

Re: How to open a remote file using python.

On Sun, Feb 22, 2009 at 8:13 PM, venutaurus...@gmail.com venutaurus...@gmail.com wrote: Hello all, I am writing an application where I need to open a shared file on a remote machine using python script. I tried using the following function: f =

Re: How to open a remote file using python.

venutaurus...@gmail.com wrote: Hello all, I am writing an application where I need to open a shared file on a remote machine using python script. I tried using the following function: f = urllib.open(\\remote_machine\\folder1\\file1.doc) I also tried using class

Re: How to open a remote file using python.

On Feb 23, 9:25 am, MRAB goo...@mrabarnett.plus.com wrote: venutaurus...@gmail.com wrote: Hello all,            I am writing an application where I need to open a shared file on a remote machine using python script. I tried using the following function: f =

FTPLIB fails to retreive the remote file

Hi All, I'm trying to download a remote file through FTP. Here's the script: ### #!/usr/bin/env python import ftplib import os import time ddir=C:\\ftp os.chdir(ddir) f=ftplib.FTP(10.2.2.1, user, user123) f.cwd(/home/protocol/cccdb

remote file

How to open remote file ? example: \\server\share\file.txt thanks a lot -- http://mail.python.org/mailman/listinfo/python-list

Re: remote file

yqyq22 wrote: How to open remote file ? example: \\server\share\file.txt like you'd open any other file: f = open(filename) e.g. f = open(r\\server\share\file.txt) /F -- http://mail.python.org/mailman/listinfo/python-list