I'm currently playing around with home networking and getting into servers and I found a python command that will make the directories of you python folder avaliable over your LAN. To execute it in windows, I've just been entering the following into Command Prompt:
cd..
cd..
cd Python33
python -m http.server 8000
This creates the server that can be accessed via x.x.x.x:8000. Instead of manually entering these commands every time, however, I tried to make a batch script that would do this automatically. Below is the script for the batch file. The problem I'm having, is When I run the batch file, command prompt is saying 'python' is not recognized as an internal or external command. I was wondering if someone could tell my why this is happening and offer a solution or reason as to why this couldn't work. Thanks.
#ECHO OFF
cd..
cd..
cd Python33
python -m http.server 8000
pause
EDIT:
Below is the code that works, this will also set the directory to the C drive.
#ECHO OFF
cd C:\
C:\Python33\python -m http.server 8000
Maybe it's because the batch file is located in a directory that is not 2- level deep (C:\path\to\batch.bat)
Specifying the path of the python executable will solve your problem.
#ECHO OFF
C:\python33\python -m http.server 8000
pause
Try this:
#ECHO OFF
cd..
cd..
cd Python33
python -m SimpleHTTPServer 8000
pause
Related
I'm trying to start SIMPLEHTTPSERVER from inside python via an os.system batch command to serve to a specific directory.
In a regular batch file, I have:
#echo off
pushd c:\MC\log
start C:\Python27\python.exe -m SimpleHTTPServer 8080
Which works great, but in Python if I use:
os.system('cmd /c pushd c:\MC\log start C:\Python27\python.exe -m SimpleHTTPServer 8080')
it starts the server, but seems to ignore the server directory.
Is it clear where I'm going wrong?
I have a simple python script which I want to start a daemon-service in background in docker container
/sbin/start-stop-daemon --start --user root --make-pidfile --pidfile /var/lock/subsys/my-application.pid --exec 'python /opt/app/uc/monitor/bin/my-application.py'
when I execute this command in a shell I get
/sbin/start-stop-daemon: unable to stat //python /opt/app/uc/monitor/bin/my-application.py (No such file or directory)
However when execute just the below command in shell it works
python /opt/app/uc/monitor/bin/my-application.py
I'm sure the python is installed and all the links have been setup.
Thanks for the help
That error message implies that start-stop-daemon is looking for a file to open (the stat operation is a check before it opens the file) and treating your 'python ... ' argument as if it was a file.
See this example which confirms this. You may need to read the man page for start-stop-daemon, for your Ubuntu version, to check what a valid command would be for your setup.
Simplest solution is probably to create a shell script (say /opt/app/uc/monitor/bin/run-my-application.sh), and put this into it:
#!/bin/bash
python /opt/app/uc/monitor/bin/my-application.py
Be sure to do chmod +x on this file. If python is not found, use which python to find the path to python and use that in the script.
Now try:
/sbin/start-stop-daemon --start --user root --make-pidfile --pidfile /var/lock/subsys/my-application.pid --exec '/opt/app/uc/monitor/bin/run-my-application.sh'
I have following bash script:
export MYCONFIG_CONFIG=TEST
/home/bmwant/projects/test/venv/bin/python3 /home/bmwant/projects/test/script.py
My python script needs environment variable which I have set, but running this script as
bash run_python.sh
shows an error
/home/bmwant/projects/test/venv/bin/python3: can't open file 'home/bmwant/projects/tes': [Errno 2] No such file or directory
What is wrong? I have set script as executable with chmod u+x run_python.sh
Have made script.py executable?
chmod u+x /home/bmwant/projects/test/script.py
Try this script:
#!/bin/bash
export MYCONFIG_CONFIG=TEST
pushd /home/bmwant/projects/test/venv/bin/
./python3 /home/bmwant/projects/test/script.py
popd
You can run it via:
./run_python.sh
So, the problem was in file format. I have created it in Windows and then copy via FileZilla to remote server where I was trying to run it. So just run
apt-get install dos2unix
dos2unix run_python.sh
bash run_python.sh
and all will work well. Thanks #Horst for the hint.
I have a short Python script that needs to run at startup - Ubuntu 13.10. I have tried everything I can think of but can't get it to run. The script:
#!/usr/bin/python
import time
with open("/home/username/Desktop/startup.txt", 'a') as f:
f.write(str(time.time()) + " It worked!")
(The actual script is a bit different, as I'm just using this for testing purposes, but you get the idea.)
I've tried all of the following, with no luck:
Put the command python startuptest.py in crontab, as #reboot
python /home/username/Documents/startuptest.py, both as the regular user and as sudo
Put the command python /home/username/Documents/startuptest.py in /etc/rc.local
Opened Ubuntu's Startup Applications and put the command there
Done all of the preceding, putting the command into a shell script
and calling that shell script instead
Nothing works. I get the feeling I'm missing something simple. Any ideas? (The script runs fine if I just run the command from a terminal.)
Instructions
Copy the python file to /bin:
sudo cp -i /path/to/your_script.py /bin
Add A New Cron Job:
sudo crontab -e
Scroll to the bottom and add the following line (after all the #'s):
#reboot python /bin/your_script.py &
The “&” at the end of the line means the command is run in the background and it won’t stop the system booting up.
Test it:
sudo reboot
Practical example:
Add this file to your Desktop: test_code.py (run it to check that it works for you)
from os.path import expanduser
import datetime
file = open(expanduser("~") + '/Desktop/HERE.txt', 'w')
file.write("It worked!\n" + str(datetime.datetime.now()))
file.close()
Run the following commands:
sudo cp -i ~/Desktop/test_code.py /bin
sudo crontab -e
Add the following line and save it:
#reboot python /bin/test_code.py &
Now reboot your computer and you should find a new file on your Desktop: HERE.txt
Put this in /etc/init (Use /etc/systemd in Ubuntu 15.x)
mystartupscript.conf
start on runlevel [2345]
stop on runlevel [!2345]
exec /path/to/script.py
By placing this conf file there you hook into ubuntu's upstart service that runs services on startup.
manual starting/stopping is done with
sudo service mystartupscript start
and
sudo service mystartupscript stop
If you are on Ubuntu you don't need to write any other code except your Python file's code , Here are the Steps :-
Open Dash (The First Icon In Sidebar).
Then type Startup Applications and open that app.
Here Click the Add Button on the right.
There fill in the details and in the command area browse for your Python File and click Ok.
Test it by Restarting System . Done . Enjoy !!
Create file ~/.config/autostart/MyScript.desktop
with
[Desktop Entry]
Encoding=UTF-8
Name=MyScript
Comment=MyScript
Icon=gnome-info
Exec=python /home/your_path/script.py
Terminal=false
Type=Application
Categories=
X-GNOME-Autostart-enabled=true
X-GNOME-Autostart-Delay=0
It helps me!
In similar situations, I've done well by putting something like the following into /etc/rc.local:
cd /path/to/my/script
./my_script.py &
cd -
echo `date +%Y-%b-%d_%H:%M:%S` > /tmp/ran_rc_local # check that rc.local ran
This has worked on multiple versions of Fedora and on Ubuntu 14.04 LTS, for both python and perl scripts.
nano /etc/rc.local
and edit in
python ~/path-to-script.py
worked for me
When I run python -m SimpleHTTPServer 8000 or python -m CGIHTTPServer 8000 in my shell I am hosting the content of my current directory to the internet.
I would like to make the following cgi_script.py work correctly using the above command in the command line when I browse to 192.xxx.x.xx:8000/cgi_script.py
#!/usr/bin/env python
print "Content-Type: text/html"
print
print """\
<html>
<body>
<h2>Hello World!</h2>
</body>
</html>
"""
But this script is displayed literally and not only the "Hello World!" part.
Btw I changed the file permissions to 755 for cgi_script.py as well as the folder I am hosting it from.
Try with python -m CGIHTTPServer 8000.
Note that you have to move the script to a cgi-bin or htbin directory in order to be runnable.
SO doesn't allow me to comment so I'm adding this as a separate answer, addition to rodrigo's.
You can use another parameter cgi_directories which defaults to ['/cgi-bin', '/htbin']. More info here
In Python3 the command line is simply
python3 -m http.server --cgi 8000
When I ran into this issue I found that depending on which directory you are in when you run the python -m CGIHTTPServer 8000 command yields different results. When attempting to run the command while in the cgi-bin directory the browser continued to return the raw script code. once I cd'ed one level higher and ran the python -m CGIHTTPServer 8000 command again my script began executing.
#Bentley4 -ifyou are still not able to do,
try importing cgi.
#!C:\Python34\python.exe -u
import cgi
print ("Content-type:text/html")
HTH
This work for me, run the python -m CGIHTTPServer 8000 command same menu level with cgi-bin,and move cgi_script.py into cgi-bin folder.In browser type http://localhost:8000/cgi-bin/cgi_script.py