ftp.py | FTP Server based on Python | FTP library
kandi X-RAY | ftp.py Summary
kandi X-RAY | ftp.py Summary
The FTP Server based on Python.
Support
Quality
Security
License
Reuse
Top functions reviewed by kandi - BETA
- Start the FTP connection .
- Parse command line options .
- fork stdin and stderr
- show usage
- Start child process .
- Serve forever .
- Get uid from username .
- Get a logger .
- set the uid for the given username
- Initialize FTP connection .
ftp.py Key Features
ftp.py Examples and Code Snippets
Community Discussions
Trending Discussions on ftp.py
QUESTION
I've been looking for SFTP python packages, ftpretty works fine for me: https://pypi.org/project/ftpretty/ but I want to use a more secure protocol.
PySftp is obviously a bit outdated (Edit: it seems that pysftp is still frequently used, about the error please see below): https://bitbucket.org/dundeemt/pysftp/src/master/
And throws me several errors on Win10, PyCharm, Python3.6:
...ANSWER
Answered 2021-Mar-07 at 17:24For the first error, it seems like a bug in pysftp
.
You can have a look at the Connection
class here on line 76, and the attribute _sftp_live
is defined on line 134, so this is definitely an error occurring at runtime without being validated correctly. I have also been able to find this related error, which likely explains the cause of this issue; the solution is also mentioned in the error if you want to explicitly fix it.
I would still consider using ftpretty
. It does use TLS
for security and a pretty safe wrapper, you can simply enable it by setting the secure parameter to True (secure=True
) - which by default is set as False
.
QUESTION
With this simplified dataframe :
...ANSWER
Answered 2020-Nov-24 at 14:53Do you mean:
QUESTION
I'm trying to upload mp4 file via ftp using ftplib
in Python. But I got UnicodeDecodeError.
Below is what I tried:
...ANSWER
Answered 2020-Oct-27 at 16:47open(file_path)
opens the file in text mode, so it'll treat the binary .mp4
file as if it was UTF8-encoded text.
You should open it in binary mode instead: open(file_path, 'rb')
.
QUESTION
Docker best practices state:
If a service can run without privileges, use USER to change to a non-root user.
In the case of cron
, that doesn't seem practical as cron
needs root privileges to function properly. The executable that cron
runs, however, does NOT need root privileges. Therefore, I run cron
itself as the root
user, but call my crontab
script to run the executable (in this case, a simple Python FTP download script I wrote) as a non-root user via the crontab -u
command.
The cron/Docker interactability and community experience still seems to be in its infancy, but there are some pretty good solutions out there. Utilizing lessons gleaned from this and this great posts, I arrived at a Dockerfile that looks something like this:
...ANSWER
Answered 2019-Sep-30 at 22:08The Alpine base images are based on a compact tool set called BusyBox and when you run crond
here you're getting the BusyBox cron and not any other implementation. Its documentation is a little sparse, but if you look at the crond source (in C) what you'll find is that there is not any redirection at all when it goes to run a job (see the non-sendmail version of start_one_job
); the job's stdout and stderr are crond's stdout and stderr. In Docker, since crond is the container primary process, that in turn becomes the container's output stream.
Anything that shows up in docker logs
definitionally went to stdout or stderr or the container's main process. If this cron implementation wrote your job's output directly there, there's nothing wrong or insecure with taking advantage of that.
In heavier-weight container orchestration systems, there is some way to run a container on a schedule (Kubernetes CronJobs, Nomad periodic jobs). You might find it easier and more consistent with these systems to set up a container that runs your job once and then exits, and then to set up the host's cron to run your container (necessarily, as root).
QUESTION
Edit: Using python 3.6 made matters worse. Completely reinstalling python 2.7.15 did not solve it. The same error appears.
Edit Again: I created a new user, test_user
and chown'd my ansible-web.txt file to that user (sudo chown test_user:test_user ansible-web.txt
). I ran the playbook again and those errors did not show. So, it's something wrong in my local files. I just don't even know where to begin to look.
I am trying to figure out what is causing this error. It's repeated over and over when I try to run any of my Ansible playbooks, but the playbook still runs successfully. There are two errors. Here is the entire output of running the play for a single remote host:
...ANSWER
Answered 2019-Dec-10 at 01:08Well, after my crash on boot this morning (got an initramfs
prompt, which led me to run fsck
while in grub), I decided to just go ahead an reinstall my OS. I reinstalled Ansible using apt, ran the playbook with ansible-playbook .yml
and it completed without the errors. This is not a solution, per se, since I obviously had other problems going on, but the python error is gone.
QUESTION
I'm trying to use progressbar to display the FTP file download progress (ftplib) but the progress does not update correctly. The speed starts out high then gradually decreases (down to bytes). The download ends up completing after a few seconds while the progress bar is still at 0%. It appears I am not updating the progress correctly and am not sure how to correct this.
I tried the solution I found here Show FTP download progress in Python (ProgressBar) using pbar += len(data)
but this gave me the following error:
ANSWER
Answered 2019-Sep-16 at 11:47The code is actually for progressbar2
library.
Its ProgressBar
class implements __iadd__
.
QUESTION
I am following this tutorial https://www.codementor.io/jamesezechukwu/how-to-deploy-django-app-on-heroku-dtsee04d4 for deploying my Django app to heroku. This is my first Django app:
Push rejected, failed to compile Python app.
I'm currently running python 3.7.0. this is set in my runtime.txt file:
python-3.7.0
stack trace:
...ANSWER
Answered 2019-Aug-28 at 17:06The version of paramiko (1.15.2) isn't compatible with python 3.7. Use a newer version. It uses the async
keyword which is reserved as argument in a function call.
If you look at the latest version, you'll see that this particular line of code is now using async_
.
You should always use the same version of python on your local machine as on production to spot these errors earlier.
I also suggest you look at some of the other dependencies, your version of gunicorn
is also more than 5 years old, it contains security vulnerabilities, so may some other (old) packages you're using.
Finally, use pip freeze
to compare what's installed in your local virtualenv compared to requirements.txt
. And make sure they are the same!
QUESTION
I'm running a Python3 script on the terminal, and on my terminal it writes out a lot of print()
statements that I added, as well as output from libraries I'm using.
I wish to save the output in a file, so I run the script like this.
python3 myscript.py > logfile.txt
Everything I print()
from within Python now goes into the file as I want, instead of being printed to my terminal.
However, within this script I'm also using the ftputil library to download files, and some output from ftputil still shows on the terminal where I ran the script. So now the output of my program is split in 2 parts, which makes it less useful.
...ANSWER
Answered 2019-Aug-22 at 20:47You're right, this deprecation warning is from ftputil.
ftputil uses the Python warnings
module to show this warning. The warnings
module not only outputs warnings (by default to stderr), but also offers ways to control which warnings are actually shown (see The Warnings Filter and Temporarily Suppressing Warnings).
You can suppress the ftputil warning with
QUESTION
I am trying to connect to connect to a FTP site to download some files:
...ANSWER
Answered 2019-Apr-30 at 02:09For anyone looking for an answer, the issue was that implicit FTPS connections need the socket to be ssl wrapped automatically.I used the below piece of code written by George Leslie-Waksman
QUESTION
I'm having a hard time trying to create an SFTP client using Paramiko (Python).
Code:
...ANSWER
Answered 2019-Apr-04 at 13:26Your server is using a keyboard-interactive authentication, not a simple password authentication.
Normally Paramiko is smart enough to fallback to the keyboard-interactive authentication, when the password authentication fails and the keyboard-interactive prompt has one field only (likely a password).
The problem is that your server behaves, as if the password authentication succeeded.
You can explicitly make Paramiko try the keyboard-interactive authentication using this code:
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install ftp.py
You can use ftp.py like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.
Support
Reuse Trending Solutions
Find, review, and download reusable Libraries, Code Snippets, Cloud APIs from over 650 million Knowledge Items
Find more librariesStay Updated
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page