[Introduction]
Hi! I'm working on a python script to automate installation of UWP-Apps, it's been a long time i'm not touching Python; until this day. The script uses Depedencies inside the script directory, i've looking up on my older scripts and found this specific code.
os.chdir(os.path.dirname(sys.argv[0]))
[Problematic]
However, using the above code doesn't work on my current script but it's working fine on older scripts. When using above, it shows:
OSError: [WinError 123] The filename, directory name, or volume label syntax is incorrect: ''
Already looking up on Internet about this topic; but most of them was talking about running the script from outer/different directory that leads me to dead end.
Any helps is appreciated :)
The easiest answer is probably to change your working directory, then call the .py file from where it is:
cd path/to/python/file && python ../.py
Of course you might find it even easier to write a script that does it all for you, like so:
Save this as runPython.sh in the directory where you're running the python script from, is:
#!/bin/sh
cd path/to/python/file
python ../script.py
Make it executable (for yourself):
chmod +x ./runPython.sh
Then you can simply enter your directory and run it:
./runPython.sh
If you want to only make changes to the python script:
mydir = os.getcwd() # would be the folder where you're running the file
mydir_tmp = "path/to/python/file" # get the path
mydir_new = os.chdir(mydir_tmp) # change the current working directory
mydir = os.getcwd()
The reason you got an error was because sys.argv[0] is the name of the file itself, instead you can pass the directory as a string and use sys.argv[1] instead.
import os
from os.path import abspath, dirname
os.chdir(dirname(abspath(__file__)))
You can use dirname(abspath(__file__))) to get the parent directory of the python script and os.chdir into it to make the script run in the directory where it is located.
Related
I'm having a strange issue. I'm just trying to do a basic show directory contents for relative path.
Created a test directory on the desktop
test directory contents
test.py
test1 -- folder
sample.txt
contents of test.py
import os
dataDir = "test1"
for file in os.listdir("/Users/username/Desktop/test"):
print(file)
Summary - absolute path
works - in visual studio code
works - macOS terminal python3 /Users/username/Desktop/test/test.py
however when use the variable I get an error:
contents of test.py
import os
dataDir = "test1"
for file in os.listdir(dataDir):
print(file)
Summary - relative path
works - in visual studio code
ERROR - macOS terminal python3 /Users/username/Desktop/test/test.py
Traceback (most recent call last):
File "/Users/username/Desktop/test/test.py", line 4, in
for file in os.listdir(dataDir):
FileNotFoundError: [Errno 2] No such file or directory: 'test1'
I all depends on what folder you are in when you launch the code.
Let's say your .py file is in folder1/folder2/file.py and your test1 folder is folder1/folder2/test1/.... You open a terminal in the folder1 and launch python3 /folder2/file.py. The program is going to check for files in folder1/test1/.
Try navigating to the actual file folder, that should work. If you want to use it wherever you launch the code you will have to do some checks on the current directory and manually point the code to the wanted path.
The following solution is inspired by that answer
A simple way to solve your problem is to create the absolute path. (I recommend using that always when you're working with different directories and files)
First of all, you need to care about your actual working directory. While using VS Code your working directory is in your desired directory (/Users/username/Desktop/test/).
But if you use the command line your actual working directory may change, depending where you're calling the script from.
To get the path where script is actually located, you can use the python variable __file__. __file__ is the full path to the directory where your script is located.
To use your script correctly and being able to call it using both ways, the following implementation can help you:
import os
dataDir = "test1"
# absolute path to the directory where your script is in
scriptDir = os.path.dirname(__file__)
# combining the path of your script and the 'searching' directory
absolutePath = os.path.join(scriptDir, dataDir)
for file in os.listdir(absolutePath):
print(file)
Basically, I want to call a python program from anywhere using shell so I added a shebang and copied it to /usr/local/bin with executable permission. The python program takes a command line argument which is the relative path of an input file.
I am stuck here, I have no idea what to do so that I can obtain the absolute path of the shell. I am assuming once I get the absolute path somehow, I can use sys.argv[1] to get the entered relative path of the file(which I will append to the absolute path of shell working directory) but please do correct me it won't work.
You can Print Working Directory via os.environ['PWD']. Content of your_script.py:
#!/usr/bin/python3.5
import os
print(os.environ['PWD'])
Usage:
sanyash#sanyash-ub16:/etc/nginx$ your_script.py
/etc/nginx
sanyash#sanyash-ub16:/etc/nginx$
I put a python script to my /usr/local/bin to use as a command in bash.
I want to use my bash working directory, but I only get the current directory where the script is executed.
Is there a way to use the current directory as a command does?
E.g. with convert2ogg (as python script in /usr/local/bin)
$ cd ~/Music
$ convert2ogg *.mp3
import os,sys
print("CWD: "+os.getcwd())
print("Script: "+sys.argv[0])
print(".EXE: "+os.path.dirname(sys.executable))
print("Script dir: "+ os.path.realpath(os.path.dirname(sys.argv[0])))
pathname, scriptname = os.path.split(sys.argv[0])
print("Relative script dir: "+pathname)
print("Script dir: "+ os.path.abspath(pathname))
This code was very clear for me. os.path was not the solution for all my problems ;) I look for an answer, and seems anybody ask it before. The first line with os.getcwd() solved my problem.
I'm making a compress script for my text editor, and it's all working up to the part where it needs to make the file Run. Inside of Run is just this code: python ./App.pyc. When I run the program by double-clicking on it in Finder, it says that it can't open file './App.pyc' [Errno 2] No such file or directory within Terminal.
And if I run it through Terminal after I've cd'd to the directory Run and App.pyc are in, it works. I'm assuming this is because we aren't in the right directory.
My question is, how can I make sure Run is being ran in the right directory? If I put cd in it, it'll work, but then if somebody moves the folder elsewhere it won't work anymore.
#!/usr/bin/python
### Compresser script.
# Compress files.
import App
import Colors
# Import modules
import os
# Clear the folder to put the compressed
# files in (if it exists).
try:
os.system('rm -rf BasicEdit\ Compressed')
except:
pass
# Remake the folder to put compressed files in.
os.system('mkdir BasicEdit\ Compressed')
# Move the compiled files into the BasicEdit
# Compressed folder.
os.system('mv App.pyc BasicEdit\ Compressed/')
os.system('mv Colors.pyc BasicEdit\ Compressed/')
# Create contents of run file.
run_file_contents = "python ./App.pyc\n"
# Write run file.
run_file = open("./BasicEdit Compressed/Run", 'w')
run_file.write(run_file_contents)
# Give permissions of run file to anybody.
os.system('chmod a+x ./BasicEdit\ Compressed/Run')
# Finally compress BasicEdit, and remove the old
# folder for BasicEdit Compressed.
os.system('zip -9r BasicEdit.zip BasicEdit\ Compressed')
os.system('rm -rf BasicEdit\ Compressed')
(PS, what's [Errno 1]? I've never seen it before.)
The Python script's current working directory can be modified with the os.chdir() call, after which references to . will be correct.
If you want to find the location of the source file currently being run rather than hardcoding a directory, you can use:
os.chdir(os.path.dirname(__file__))
The bash equivalent to this logic is:
cd "${BASH_SOURCE%/*}" || {
echo "Unable to change directory to ${BASH_SOURCE%/*}" >&2
exit 1
}
See BashFAQ #28 for more details and caveats.
As developed above together with #William Purcell, you have to retrieve the absolute path by os.pwd() and then use the absolute path for the python call.
I withdraw my proposal and go with #Charles Duffy's answer. However, I don't delete my attempt as the comments seem to be useful to others!
My Crontab -l
# m h dom mon dow command
SHELL=/bin/bash
PATH=/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games
00 8,20 * * * python /home/tomi/amaer/controller.py >>/tmp/out.txt 2>&1
My controller.py has config file settings.cfg also it uses other script in the folder it's located (I chmoded only controller.py)
The error
1;31mIOError^[[0m: [Errno 2] No such file or directory: 'settings.cfg'
I have no idea how to fix this? Please help me?
Edit: The part that read the config file
def main():
config=ConfigParser.ConfigParser()
config.readfp(open("settings.cfg"),"r")
As I initially wrote in my comment, this is because you are using relative path to the current working directory. However, that is not going to be the same when running all this via the cron executable rather than the python interpreter directly via the shebang.
Your current code would look for the "settings.cfg" in the current working directory which is where the cron executable resides, and not your script. Hence, you would need to change your code logic to using absolute paths by the help of the "os" built-in standard module.
Try to following line:
import os
...
def main():
config = ConfigParser.ConfigParser()
scriptDirectory = os.path.dirname(os.path.realpath(__file__))
settingsFilePath = os.path.join(scriptDirectory, "settings.cfg")
config.readfp(open(settingsFilePath,"r"))
This will get your the path of your script and then appends the "settings.cfg" with the appropriate dir separator for your operating system which is Linux in this particular case.
If the location of the config file changes any time in the future, you could use the argparse module for processing a command line argument to handle the config location properly, or even without it simply just using the first argument after the script name like sys.argv[1].
Your code is looking for settings.cfg in its current working directory.
This working directory will not be the same when cron executes the job, hence the error
You have two "easy" solutions:
Use an absolute path to the config file in your script (/home/tomi/amaer/config.cfg)
CD to the appropriate directory first in your crontab (cd /home/tomi/amaer/ && python /home/tomi/amaer/controller.py)
The "right" solution, though, would be to pass your script a parameter (or environment variable) that tells it where to look for the config file.
It's not exactly good practice to assume your config file will always be lying just next to your script.
You might want to have alook at this question: https://unix.stackexchange.com/questions/38951/what-is-the-working-directory-when-cron-executes-a-job